轉動慣量 moments of inertia - cct慣性矩之莫爾圓(mohr’s circle for moment of inertia) 7...
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轉動慣量或慣性矩Moments of Inertia
Chen-Ching Ting(丁振卿)
Mechanical Engineering National Taipei University of Technology(國立台北科技大學機械系)
Homepage httpcctmentutedutwE-mail chchtingntutedutw
CCT Group
助教 陳育煒
May 17 2012
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 1 69
課程大綱
1 轉動慣量或慣性矩(Moment of Inertia)
2 面積慣性矩(Moment of Inertia for Area)面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)面積慣性矩之迴轉半徑(Radius of Gyration for Area)
3 複合面積之慣性矩(Moment of Inertia for Composite Area)
4 面積之慣性積(Product of Inertia for Area)平行軸定理(Parallel-Axis Theorem)
5 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)主軸慣性矩(Principal Moment of Inertia)
6 慣性矩之莫爾圓(Mohrrsquos Circle for Moment of Inertia)
7 質量慣性矩(Mass Moment of Inertia)平行軸定理(Parallel-Axis Theorem)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 2 69
參考文獻
RC Hibbeler Statics Pearson Education Inc 歐亞書局有限公司ISBN=978-986-154-861-6 Chapter 10 2009
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 3 69
轉轉轉動動動慣慣慣量量量或或或慣慣慣性性性矩矩矩Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 4 69
慣性(Inertia)
慣性(Inertia)簡單來說乃物體持續維持不變的行為
伽利略(1632)提出一個不受任何外力(或者合外力為0)的物體將保持靜止或勻速直線運動
牛頓(1687)提出所有物體都將一直處於靜止或者勻速直線運動狀態直到出現施加其上的力改變它的運動狀態止
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 5 69
轉動慣量或慣性矩 I
轉動慣量也稱慣性矩(Moment of Inertia)為物體對旋轉運動的慣性比較直線運動與旋轉運動得
F = ma = mdv
dt(1)
τ = Iα = Idω
dt(2)
其中F為力(N)m為質量(kg)a為加速度(ms2)v為速度(ms)τ為扭矩(N middotm)I為轉動慣量或慣性矩(kg middotm2)α為角加速度(rads2)ω為角速度(rads)
v = rω (3)
I = mr2 (4)
τ = rF = rmdv
dt= mr2 dω
dt= Iα
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 6 69
轉動慣量或慣性矩 II
一般物件的動能K用轉動力學的定義取代
K =1
2mv2 (5)
=1
2m(rω)2 (6)
=1
2Iω2 (7)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 7 69
面面面積積積慣慣慣性性性矩矩矩Moment of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 8 69
面積慣性矩
由左圖液面下y深度的壓力p為
p = γy (8)
其中γ = ρg為液體比重(SpecificWeight)液面下y深度之微小面積dA的受力dF為
dF = pdA = γydA (9)其扭矩dM為
dM = ydF = γy2dA (10)ˆdM = γ
ˆy2dA = γIx (11)
其中Ix =acutey2dA稱為x軸上之面積慣性矩
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 9 69
極慣性矩(Polar Moment of Inertia)
極點
由左圖x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy如下
dIx = y2dA (12)
Ix =
ˆAy2dA (13)
dIy = x2dA (14)
Iy =
ˆAx2dA (15)
當微小面積dA對極點(O)或z軸旋轉時其面積極慣性矩JO如下
dJO = r2dA (16)
r2 = x2 + y2 (17)
JO =
ˆAr2dA =
ˆA
(x2 + y2)dA = Ix + Iy (18)
其中r為微小面積dA到極點(O)的距離極點(O)簡單說乃z軸上的任意點由於x2 y2 r2及面
積(A)為正因此JO Ix Iy 均為正面積慣性矩的單位為m4mm4
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 10 69
面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 11 69
面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 12 69
面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 13 69
面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 14 69
面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 15 69
面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 16 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 18 69
解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 19 69
解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 21 69
課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課程大綱
1 轉動慣量或慣性矩(Moment of Inertia)
2 面積慣性矩(Moment of Inertia for Area)面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)面積慣性矩之迴轉半徑(Radius of Gyration for Area)
3 複合面積之慣性矩(Moment of Inertia for Composite Area)
4 面積之慣性積(Product of Inertia for Area)平行軸定理(Parallel-Axis Theorem)
5 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)主軸慣性矩(Principal Moment of Inertia)
6 慣性矩之莫爾圓(Mohrrsquos Circle for Moment of Inertia)
7 質量慣性矩(Mass Moment of Inertia)平行軸定理(Parallel-Axis Theorem)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 2 69
參考文獻
RC Hibbeler Statics Pearson Education Inc 歐亞書局有限公司ISBN=978-986-154-861-6 Chapter 10 2009
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 3 69
轉轉轉動動動慣慣慣量量量或或或慣慣慣性性性矩矩矩Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 4 69
慣性(Inertia)
慣性(Inertia)簡單來說乃物體持續維持不變的行為
伽利略(1632)提出一個不受任何外力(或者合外力為0)的物體將保持靜止或勻速直線運動
牛頓(1687)提出所有物體都將一直處於靜止或者勻速直線運動狀態直到出現施加其上的力改變它的運動狀態止
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 5 69
轉動慣量或慣性矩 I
轉動慣量也稱慣性矩(Moment of Inertia)為物體對旋轉運動的慣性比較直線運動與旋轉運動得
F = ma = mdv
dt(1)
τ = Iα = Idω
dt(2)
其中F為力(N)m為質量(kg)a為加速度(ms2)v為速度(ms)τ為扭矩(N middotm)I為轉動慣量或慣性矩(kg middotm2)α為角加速度(rads2)ω為角速度(rads)
v = rω (3)
I = mr2 (4)
τ = rF = rmdv
dt= mr2 dω
dt= Iα
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 6 69
轉動慣量或慣性矩 II
一般物件的動能K用轉動力學的定義取代
K =1
2mv2 (5)
=1
2m(rω)2 (6)
=1
2Iω2 (7)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 7 69
面面面積積積慣慣慣性性性矩矩矩Moment of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 8 69
面積慣性矩
由左圖液面下y深度的壓力p為
p = γy (8)
其中γ = ρg為液體比重(SpecificWeight)液面下y深度之微小面積dA的受力dF為
dF = pdA = γydA (9)其扭矩dM為
dM = ydF = γy2dA (10)ˆdM = γ
ˆy2dA = γIx (11)
其中Ix =acutey2dA稱為x軸上之面積慣性矩
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極慣性矩(Polar Moment of Inertia)
極點
由左圖x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy如下
dIx = y2dA (12)
Ix =
ˆAy2dA (13)
dIy = x2dA (14)
Iy =
ˆAx2dA (15)
當微小面積dA對極點(O)或z軸旋轉時其面積極慣性矩JO如下
dJO = r2dA (16)
r2 = x2 + y2 (17)
JO =
ˆAr2dA =
ˆA
(x2 + y2)dA = Ix + Iy (18)
其中r為微小面積dA到極點(O)的距離極點(O)簡單說乃z軸上的任意點由於x2 y2 r2及面
積(A)為正因此JO Ix Iy 均為正面積慣性矩的單位為m4mm4
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面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
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面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 12 69
面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 13 69
面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
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面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
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面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
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課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
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課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
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課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
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課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
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Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
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課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
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面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
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面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
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平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
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平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
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平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
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課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
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課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
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課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
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結結結論論論
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結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
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- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
參考文獻
RC Hibbeler Statics Pearson Education Inc 歐亞書局有限公司ISBN=978-986-154-861-6 Chapter 10 2009
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轉轉轉動動動慣慣慣量量量或或或慣慣慣性性性矩矩矩Moment of Inertia
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慣性(Inertia)
慣性(Inertia)簡單來說乃物體持續維持不變的行為
伽利略(1632)提出一個不受任何外力(或者合外力為0)的物體將保持靜止或勻速直線運動
牛頓(1687)提出所有物體都將一直處於靜止或者勻速直線運動狀態直到出現施加其上的力改變它的運動狀態止
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轉動慣量或慣性矩 I
轉動慣量也稱慣性矩(Moment of Inertia)為物體對旋轉運動的慣性比較直線運動與旋轉運動得
F = ma = mdv
dt(1)
τ = Iα = Idω
dt(2)
其中F為力(N)m為質量(kg)a為加速度(ms2)v為速度(ms)τ為扭矩(N middotm)I為轉動慣量或慣性矩(kg middotm2)α為角加速度(rads2)ω為角速度(rads)
v = rω (3)
I = mr2 (4)
τ = rF = rmdv
dt= mr2 dω
dt= Iα
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轉動慣量或慣性矩 II
一般物件的動能K用轉動力學的定義取代
K =1
2mv2 (5)
=1
2m(rω)2 (6)
=1
2Iω2 (7)
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面面面積積積慣慣慣性性性矩矩矩Moment of Inertia for Area
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面積慣性矩
由左圖液面下y深度的壓力p為
p = γy (8)
其中γ = ρg為液體比重(SpecificWeight)液面下y深度之微小面積dA的受力dF為
dF = pdA = γydA (9)其扭矩dM為
dM = ydF = γy2dA (10)ˆdM = γ
ˆy2dA = γIx (11)
其中Ix =acutey2dA稱為x軸上之面積慣性矩
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極慣性矩(Polar Moment of Inertia)
極點
由左圖x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy如下
dIx = y2dA (12)
Ix =
ˆAy2dA (13)
dIy = x2dA (14)
Iy =
ˆAx2dA (15)
當微小面積dA對極點(O)或z軸旋轉時其面積極慣性矩JO如下
dJO = r2dA (16)
r2 = x2 + y2 (17)
JO =
ˆAr2dA =
ˆA
(x2 + y2)dA = Ix + Iy (18)
其中r為微小面積dA到極點(O)的距離極點(O)簡單說乃z軸上的任意點由於x2 y2 r2及面
積(A)為正因此JO Ix Iy 均為正面積慣性矩的單位為m4mm4
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面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
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面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
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面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
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面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
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面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
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面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
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課課課堂堂堂練練練習習習
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解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
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課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
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課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
轉轉轉動動動慣慣慣量量量或或或慣慣慣性性性矩矩矩Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 4 69
慣性(Inertia)
慣性(Inertia)簡單來說乃物體持續維持不變的行為
伽利略(1632)提出一個不受任何外力(或者合外力為0)的物體將保持靜止或勻速直線運動
牛頓(1687)提出所有物體都將一直處於靜止或者勻速直線運動狀態直到出現施加其上的力改變它的運動狀態止
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 5 69
轉動慣量或慣性矩 I
轉動慣量也稱慣性矩(Moment of Inertia)為物體對旋轉運動的慣性比較直線運動與旋轉運動得
F = ma = mdv
dt(1)
τ = Iα = Idω
dt(2)
其中F為力(N)m為質量(kg)a為加速度(ms2)v為速度(ms)τ為扭矩(N middotm)I為轉動慣量或慣性矩(kg middotm2)α為角加速度(rads2)ω為角速度(rads)
v = rω (3)
I = mr2 (4)
τ = rF = rmdv
dt= mr2 dω
dt= Iα
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 6 69
轉動慣量或慣性矩 II
一般物件的動能K用轉動力學的定義取代
K =1
2mv2 (5)
=1
2m(rω)2 (6)
=1
2Iω2 (7)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 7 69
面面面積積積慣慣慣性性性矩矩矩Moment of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 8 69
面積慣性矩
由左圖液面下y深度的壓力p為
p = γy (8)
其中γ = ρg為液體比重(SpecificWeight)液面下y深度之微小面積dA的受力dF為
dF = pdA = γydA (9)其扭矩dM為
dM = ydF = γy2dA (10)ˆdM = γ
ˆy2dA = γIx (11)
其中Ix =acutey2dA稱為x軸上之面積慣性矩
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 9 69
極慣性矩(Polar Moment of Inertia)
極點
由左圖x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy如下
dIx = y2dA (12)
Ix =
ˆAy2dA (13)
dIy = x2dA (14)
Iy =
ˆAx2dA (15)
當微小面積dA對極點(O)或z軸旋轉時其面積極慣性矩JO如下
dJO = r2dA (16)
r2 = x2 + y2 (17)
JO =
ˆAr2dA =
ˆA
(x2 + y2)dA = Ix + Iy (18)
其中r為微小面積dA到極點(O)的距離極點(O)簡單說乃z軸上的任意點由於x2 y2 r2及面
積(A)為正因此JO Ix Iy 均為正面積慣性矩的單位為m4mm4
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面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 11 69
面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
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面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
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面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
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面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
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面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
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課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
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課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
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傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
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傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
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課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
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結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
慣性(Inertia)
慣性(Inertia)簡單來說乃物體持續維持不變的行為
伽利略(1632)提出一個不受任何外力(或者合外力為0)的物體將保持靜止或勻速直線運動
牛頓(1687)提出所有物體都將一直處於靜止或者勻速直線運動狀態直到出現施加其上的力改變它的運動狀態止
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轉動慣量或慣性矩 I
轉動慣量也稱慣性矩(Moment of Inertia)為物體對旋轉運動的慣性比較直線運動與旋轉運動得
F = ma = mdv
dt(1)
τ = Iα = Idω
dt(2)
其中F為力(N)m為質量(kg)a為加速度(ms2)v為速度(ms)τ為扭矩(N middotm)I為轉動慣量或慣性矩(kg middotm2)α為角加速度(rads2)ω為角速度(rads)
v = rω (3)
I = mr2 (4)
τ = rF = rmdv
dt= mr2 dω
dt= Iα
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轉動慣量或慣性矩 II
一般物件的動能K用轉動力學的定義取代
K =1
2mv2 (5)
=1
2m(rω)2 (6)
=1
2Iω2 (7)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 7 69
面面面積積積慣慣慣性性性矩矩矩Moment of Inertia for Area
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面積慣性矩
由左圖液面下y深度的壓力p為
p = γy (8)
其中γ = ρg為液體比重(SpecificWeight)液面下y深度之微小面積dA的受力dF為
dF = pdA = γydA (9)其扭矩dM為
dM = ydF = γy2dA (10)ˆdM = γ
ˆy2dA = γIx (11)
其中Ix =acutey2dA稱為x軸上之面積慣性矩
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極慣性矩(Polar Moment of Inertia)
極點
由左圖x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy如下
dIx = y2dA (12)
Ix =
ˆAy2dA (13)
dIy = x2dA (14)
Iy =
ˆAx2dA (15)
當微小面積dA對極點(O)或z軸旋轉時其面積極慣性矩JO如下
dJO = r2dA (16)
r2 = x2 + y2 (17)
JO =
ˆAr2dA =
ˆA
(x2 + y2)dA = Ix + Iy (18)
其中r為微小面積dA到極點(O)的距離極點(O)簡單說乃z軸上的任意點由於x2 y2 r2及面
積(A)為正因此JO Ix Iy 均為正面積慣性矩的單位為m4mm4
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面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
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面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
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面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
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面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
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面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
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面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
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課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 21 69
課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
轉動慣量或慣性矩 I
轉動慣量也稱慣性矩(Moment of Inertia)為物體對旋轉運動的慣性比較直線運動與旋轉運動得
F = ma = mdv
dt(1)
τ = Iα = Idω
dt(2)
其中F為力(N)m為質量(kg)a為加速度(ms2)v為速度(ms)τ為扭矩(N middotm)I為轉動慣量或慣性矩(kg middotm2)α為角加速度(rads2)ω為角速度(rads)
v = rω (3)
I = mr2 (4)
τ = rF = rmdv
dt= mr2 dω
dt= Iα
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 6 69
轉動慣量或慣性矩 II
一般物件的動能K用轉動力學的定義取代
K =1
2mv2 (5)
=1
2m(rω)2 (6)
=1
2Iω2 (7)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 7 69
面面面積積積慣慣慣性性性矩矩矩Moment of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 8 69
面積慣性矩
由左圖液面下y深度的壓力p為
p = γy (8)
其中γ = ρg為液體比重(SpecificWeight)液面下y深度之微小面積dA的受力dF為
dF = pdA = γydA (9)其扭矩dM為
dM = ydF = γy2dA (10)ˆdM = γ
ˆy2dA = γIx (11)
其中Ix =acutey2dA稱為x軸上之面積慣性矩
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極慣性矩(Polar Moment of Inertia)
極點
由左圖x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy如下
dIx = y2dA (12)
Ix =
ˆAy2dA (13)
dIy = x2dA (14)
Iy =
ˆAx2dA (15)
當微小面積dA對極點(O)或z軸旋轉時其面積極慣性矩JO如下
dJO = r2dA (16)
r2 = x2 + y2 (17)
JO =
ˆAr2dA =
ˆA
(x2 + y2)dA = Ix + Iy (18)
其中r為微小面積dA到極點(O)的距離極點(O)簡單說乃z軸上的任意點由於x2 y2 r2及面
積(A)為正因此JO Ix Iy 均為正面積慣性矩的單位為m4mm4
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面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
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面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
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面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
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面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
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面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
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面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
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課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
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課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
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課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
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課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
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課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
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結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
轉動慣量或慣性矩 II
一般物件的動能K用轉動力學的定義取代
K =1
2mv2 (5)
=1
2m(rω)2 (6)
=1
2Iω2 (7)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 7 69
面面面積積積慣慣慣性性性矩矩矩Moment of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 8 69
面積慣性矩
由左圖液面下y深度的壓力p為
p = γy (8)
其中γ = ρg為液體比重(SpecificWeight)液面下y深度之微小面積dA的受力dF為
dF = pdA = γydA (9)其扭矩dM為
dM = ydF = γy2dA (10)ˆdM = γ
ˆy2dA = γIx (11)
其中Ix =acutey2dA稱為x軸上之面積慣性矩
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 9 69
極慣性矩(Polar Moment of Inertia)
極點
由左圖x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy如下
dIx = y2dA (12)
Ix =
ˆAy2dA (13)
dIy = x2dA (14)
Iy =
ˆAx2dA (15)
當微小面積dA對極點(O)或z軸旋轉時其面積極慣性矩JO如下
dJO = r2dA (16)
r2 = x2 + y2 (17)
JO =
ˆAr2dA =
ˆA
(x2 + y2)dA = Ix + Iy (18)
其中r為微小面積dA到極點(O)的距離極點(O)簡單說乃z軸上的任意點由於x2 y2 r2及面
積(A)為正因此JO Ix Iy 均為正面積慣性矩的單位為m4mm4
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面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
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面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
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面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
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面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
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面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
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面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 16 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
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課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
面面面積積積慣慣慣性性性矩矩矩Moment of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 8 69
面積慣性矩
由左圖液面下y深度的壓力p為
p = γy (8)
其中γ = ρg為液體比重(SpecificWeight)液面下y深度之微小面積dA的受力dF為
dF = pdA = γydA (9)其扭矩dM為
dM = ydF = γy2dA (10)ˆdM = γ
ˆy2dA = γIx (11)
其中Ix =acutey2dA稱為x軸上之面積慣性矩
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 9 69
極慣性矩(Polar Moment of Inertia)
極點
由左圖x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy如下
dIx = y2dA (12)
Ix =
ˆAy2dA (13)
dIy = x2dA (14)
Iy =
ˆAx2dA (15)
當微小面積dA對極點(O)或z軸旋轉時其面積極慣性矩JO如下
dJO = r2dA (16)
r2 = x2 + y2 (17)
JO =
ˆAr2dA =
ˆA
(x2 + y2)dA = Ix + Iy (18)
其中r為微小面積dA到極點(O)的距離極點(O)簡單說乃z軸上的任意點由於x2 y2 r2及面
積(A)為正因此JO Ix Iy 均為正面積慣性矩的單位為m4mm4
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 10 69
面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 11 69
面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 12 69
面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 13 69
面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
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面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 15 69
面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 16 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 18 69
解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
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平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
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課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
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結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
面積慣性矩
由左圖液面下y深度的壓力p為
p = γy (8)
其中γ = ρg為液體比重(SpecificWeight)液面下y深度之微小面積dA的受力dF為
dF = pdA = γydA (9)其扭矩dM為
dM = ydF = γy2dA (10)ˆdM = γ
ˆy2dA = γIx (11)
其中Ix =acutey2dA稱為x軸上之面積慣性矩
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 9 69
極慣性矩(Polar Moment of Inertia)
極點
由左圖x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy如下
dIx = y2dA (12)
Ix =
ˆAy2dA (13)
dIy = x2dA (14)
Iy =
ˆAx2dA (15)
當微小面積dA對極點(O)或z軸旋轉時其面積極慣性矩JO如下
dJO = r2dA (16)
r2 = x2 + y2 (17)
JO =
ˆAr2dA =
ˆA
(x2 + y2)dA = Ix + Iy (18)
其中r為微小面積dA到極點(O)的距離極點(O)簡單說乃z軸上的任意點由於x2 y2 r2及面
積(A)為正因此JO Ix Iy 均為正面積慣性矩的單位為m4mm4
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 10 69
面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
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面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
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面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
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面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 14 69
面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
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面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
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課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
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課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
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課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
極慣性矩(Polar Moment of Inertia)
極點
由左圖x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy如下
dIx = y2dA (12)
Ix =
ˆAy2dA (13)
dIy = x2dA (14)
Iy =
ˆAx2dA (15)
當微小面積dA對極點(O)或z軸旋轉時其面積極慣性矩JO如下
dJO = r2dA (16)
r2 = x2 + y2 (17)
JO =
ˆAr2dA =
ˆA
(x2 + y2)dA = Ix + Iy (18)
其中r為微小面積dA到極點(O)的距離極點(O)簡單說乃z軸上的任意點由於x2 y2 r2及面
積(A)為正因此JO Ix Iy 均為正面積慣性矩的單位為m4mm4
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 10 69
面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 11 69
面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 12 69
面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 13 69
面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 14 69
面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 15 69
面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
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課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
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平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
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平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
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結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
面面面積積積慣慣慣性性性矩矩矩之之之平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 11 69
面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
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面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
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面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
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面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 15 69
面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 16 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
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課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
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課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
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課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
面積慣性矩之平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)則x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy可如下換算
Ix =
ˆA
(y prime + dy )2dA (19)
=
ˆA
y prime2dA + 2dy
ˆA
y primedA + d2y
ˆA
dA
Iy =
ˆA
(x prime + dx)2dA (20)
=
ˆA
x prime2dA + 2dx
ˆA
x primedA + d2x
ˆA
dA
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 12 69
面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 13 69
面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 14 69
面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 15 69
面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 16 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 18 69
解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 19 69
解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 21 69
課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
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課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
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傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
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傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
面積慣性矩之平行軸定理- Continued
由於面心在xrsquo軸上之等價力臂為y primeyrsquo軸上之等價力臂為x prime當以面心為轉軸點時即x prime = y prime = 0則x軸上之面積慣性矩Ix為ˆ
Ay prime2dA = macrIx prime (21)
ˆy primedA = y prime
ˆAdA = 0 (22)
d2y
ˆAdA = Ad2
y (23)
Ix =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA (24)
= macrIx prime + Ad2y (25)
同理可得y軸上之面積慣性矩Iy
Iy =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA (26)
= macrIy prime + Ad2x
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 13 69
面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
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面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
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面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 16 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 19 69
解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
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課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
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平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
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平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
面積之極慣性矩
極點
新極點
由於通過面心之極慣性矩JC與極點(O)到面心(C)之距離d為
d2 = d2x + d2
y (27)
JC = macrIx prime + macrIy prime (28)
JO = Ix + Iy (29)
= macrIx prime + Ad2y + macrIy prime + Ad2
x (30)
= macrIx prime + macrIy prime + A(d2x + d2
y ) (31)
= JC + Ad2 (32)
也就是面積A之極慣性矩JO等於通過面積A之面心極慣性矩JC 與面積乘上極點(O)到面心(C)之距離d的平方和
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面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
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面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
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課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
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課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
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課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
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課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
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Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
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課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
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面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
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面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
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平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
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平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
面面面積積積慣慣慣性性性矩矩矩之之之迴迴迴轉轉轉半半半徑徑徑Radius of Gyration for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 15 69
面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 16 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
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課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
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複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
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Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
面積慣性矩之迴轉半徑
當x軸上之面積慣性矩Ix與y軸上之面積慣性矩Iy為已知時面積慣性矩之迴轉半徑可得知
dIx = y2dA (33)
Ix = kxA kx =
radicIxA
(34)
dIy = x2dA (35)
Iy = kyA ky =
radicIyA
(36)
dJO = r2dA (37)
JO = kOA kO =
radicJOA
(38)
其中(kx ky kO)分別為x軸y軸極點(O)之面積慣性矩迴轉半徑
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 16 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 19 69
解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 21 69
課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
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平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 17 69
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
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解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
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解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
解題技巧 I
極點
新極點
繪圖並標出面積(A)之面心座標位置(C)接著以面心座標位置(C)為原點畫出新座標軸(xrsquoyrsquo)其中xrsquo軸平行於原(舊)座標x軸yrsquo軸平行於原(舊)座標y軸並寫出新舊座標軸的平移量dx與dy其中dx為yrsquo軸與y軸的平移量dy為xrsquo軸與x軸的平移量取微小面積dA其在原(舊)座標之點座標為(xy)在新座標之點座標為(xrsquoyrsquo)帶入x軸上之面積慣性矩Ix及y軸上之面積慣性矩Iy還有xrsquo軸上之面積慣性矩 macrIx prime及y軸上之面積慣性矩 macrIy prime可如下換算
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 18 69
解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 19 69
解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
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課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
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課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
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質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
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平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
解題技巧 II
Ix =
ˆAy2dA =
ˆAy prime2dA + 2dy
ˆAy primedA + d2
y
ˆAdA = macrIx prime + Ad2
y
macrIx prime =
ˆAy prime2dA (39)
Iy =
ˆAx2dA =
ˆAx prime2dA + 2dx
ˆAx primedA + d2
x
ˆAdA = macrIy prime + Ad2
x
macrIy prime =
ˆAx prime2dA (40)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 19 69
解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 21 69
課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
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課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
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質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
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課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
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課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
解題技巧 III
另外微小面積dA的面積慣性矩dIx dIy也可寫成
dIx = d macrIx prime + dAd2y (41)
Ix =
ˆdIx =
ˆd macrIx prime +
ˆdAd2
y (42)
dIy = d macrIy prime + dAd2x (43)
Iy =
ˆdIy =
ˆd macrIy prime +
ˆdAd2
x (44)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 20 69
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 21 69
課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 I
如左圖
1 求矩形面積(A)作用在面心C(xrsquoyrsquo)之xrsquo軸上的慣性矩
2 求矩形面積(A)作用在xb軸上的慣性矩
3 求矩形面積(A)作用在面心C(xrsquoyrsquo)之極點或z軸方向上的慣性矩(垂直x-y平面)
解解解答答答1 首先在矩形面積(A)上取一微小面積dA以(xrsquoyrsquo)為座標軸yrsquo由minus h
2sim h
2則
dA = bdy prime (45)
macrIxprime =
ˆAy prime2dA (46)
=
ˆ h2
minus h2
y prime2bdy prime (47)
= b
ˆ h2
minus h2
y prime2dy prime (48)
=1
12bh3 (49)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 21 69
課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
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複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 II
2 由於面積A=bh又dy = h2得
Ixb = macrIxprime + Ad2y (50)
=1
12bh3 + bh(
h
2)2 (51)
=1
3bh3 (52)
另解以矩形面積(A)之左下角(O)為原點座標軸(xy)則所取的微小面積
dA = bdy (53)
Ixb =
ˆAy2dA (54)
=
ˆ h
0y2bdy (55)
=1
3by3|h0 (56)
=1
3bh3 (57)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 22 69
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
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迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 III
3 長方形之邊長b與h則面心慣性矩 macrIxprime macrIyprime與長方形之邊軸慣性矩Ix Iy為
macrIxprime =1
12bh3 (58)
macrIyprime =1
12hb3 (59)
Ix =1
3bh3 (60)
Iy =1
3hb3 (61)
JC = macrIxprime + macrIyprime =1
12bh3 +
1
12hb3 (62)
=1
12bh(h2 + b2) (63)
JO = Ix + Iy =1
3bh3 +
1
3hb3 (64)
=1
3bh(h2 + b2) (65)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 23 69
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 I
如左圖求陰影面積(A)作用在x軸上的慣性矩(Ix )解解解答答答如左上圖首先在陰影面積(A)上取一微小面積dAy由0 sim 200mm則
y2 = 400x (66)
x =y2
400(67)
dA = (100minus x)dy (68)
Ix =
ˆAy2dA (69)
=
ˆ 200mm
0y2(100minus x)dy (70)
=
ˆ 200mm
0y2(100minus
y2
400)dy (71)
=
ˆ 200mm
0100y2 minus
y4
400)dy (72)
= 107times 106mm4 (73)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 24 69
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 II
另解如前頁左下圖取微小面積dA=ydx之長方形其邊長b=dx與h=ydy = y2所以
y2 = 400x (74)
y =radic
400x (75)
d macrIxprime = d(1
12bh3) =
1
12y3dx (76)
dIx = d macrIxprime + dAd2y =
1
12y3dx + ydx
( y2
)2=
1
3y3dx (77)
Ix =
ˆ 100
0
1
3y3dx =
ˆ 100
0
1
3
(radic400x
)3dx (78)
= 107times 106mm4 (79)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 25 69
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
Example 10-3 解答
取微小面積dA=2ydx之長方形其邊長b=dx與h=2y所以
x2 + y2 = a2 (80)
y =radic
a2 minus x2 (81)
d macrIx prime = d(1
12bh3) =
1
12dx(2y)3 =
2
3y3dx (82)
macrIx prime =
ˆ a
minusa
2
3y3dx =
ˆ a
minusa
2
3
(radica2 minus x2
)3dx =
πa4
4(83)
圓之半徑a則面心慣性矩 macrIx prime macrIy prime為
macrIx prime =πa4
4(84)
macrIy prime =πa4
4(85)
JC = macrIx prime + macrIy prime =πa4
2(86)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 26 69
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
複複複合合合面面面積積積之之之慣慣慣性性性矩矩矩Moment of Inertia for Composite Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 27 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (87)
k =
radicI
m(88)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (89)
= I1 + I2 + + In (90)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 28 69
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
複合面積之慣性矩
複合面積之慣性矩(Ixc Iyc)為各局部面積之慣性矩(Ixi Iyi )合
Ixc =i=nsumi=1
Ixi (91)
= Ix1 + Ix2 + + Ixn (92)
Iyc =i=nsumi=1
Iyi (93)
= Iy1 + Iy2 + + Iyn (94)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 29 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
長方形與圓形複合
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 30 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性矩後再進行合併首先A與D矩形面積作用在x與y軸上的慣性矩為
macrIxprimeyprime =1
12bh3 (95)
IxAD = macrIxprime + Ad2y (96)
=1
12times 100times 3003 + 100times 300times 2002 (97)
= 1425times 109mm4 (98)
IyAD = macrIyprime + Ad2x (99)
=1
12times 300times 1003 + 100times 300times 2502 (100)
= 19times 109mm4 (101)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 31 69
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 II
因為(xy)座標原點即為B矩形面積之面心也就是B矩形面積作用在x與y軸上的慣性矩為
IxB = macrIxprime (102)
=1
12times 600times 1003 (103)
= 5times 107mm4 (104)
IyB = macrIyprime (105)
=1
12times 100times 6003 (106)
= 18times 109mm4 (107)
所以
Ix = IxA + IxB + IxD (108)
= 2(1425times 109) + 5times 107mm4 = 29times 109mm4 (109)
Iy = IyA + IyB + IyD (110)
= 2(19times 109) + 18times 109mm4 = 56times 109mm4 (111)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 32 69
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
面面面積積積之之之慣慣慣性性性積積積Product of Inertia for Area
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 33 69
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
面積之慣性積(Product of Inertia for Area)
為了下個章節求解慣性矩之極值(極大或極小值)方便數學運算在此先定義面積之慣性積(Ixy )為
dIxy = xydA (112)
Ixy =
ˆAxydA (113)
面積之慣性積的符號可為正負或零當慣性積之面積相對轉軸對稱時則面積之慣性積和為零
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 34 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 35 69
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
平行軸定理
極點
新極點
如左圖面積慣性矩之平行軸定理乃以面心(C)為新極點設定成新座標系統(xrsquoyrsquo)面心(C)在舊座標系統的座標是(dx dy )在新座標系統的座標是(00)其中xrsquo軸平行於原座標x軸yrsquo軸平行於原座標y軸取微小面積dA其在新座標之點座標為(xrsquoyrsquo)面積之慣性積(Ixy )為
dIxy = (x prime + dx)(y prime + dy )dA (114)
當以面心為轉軸點時即x prime = y prime = 0ˆA
x primedA = x primeˆA
dA = 0
ˆA
y primedA = y primeˆA
dA = 0 (115)
Ixy =
ˆA
(x prime + dx)(y prime + dy )dA (116)
=
ˆA
x primey primedA + dx
ˆA
y primedA + dy
ˆA
x primedA + dxdy
ˆA
dA (117)
= macrIxprimey prime + Adxdy
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 36 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
Engineering Mechanics Statics Twelfth EditionRussell C Hibbeler
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 37 69
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 I
如左圖求截面積作用在截面積的面心(C)之x與y軸上的慣性積(Ixy )解解解答答答如左下圖將截面積分成ABD三塊矩形分別求出矩形面積作用在x與y軸上的慣性積後再進行合併A矩形面積作用在x與y軸上的慣性積為
IxyA = macrIxprimeyprime + Adxdy (118)
= 0 + 300times 100times (minus250)times 200 = minus15times 109mm4(119)
B矩形面積作用在x與y軸上的慣性積為
IxyB = macrIxprimeyprime + Adxdy = 0 + 0 = 0mm4 (120)
D矩形面積作用在x與y軸上的慣性積為
IxyD = macrIxprimeyprime + Adxdy (121)
= 0 + 300times 100times 250times (minus200) = minus15times 109mm4(122)
所以 Ixy = IxyA + IxyB + IxyD (123)
= minus15times 109 + 0minus 15times 109mm4 = minus3times 109mm4 (124)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 38 69
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
傾傾傾斜斜斜軸軸軸面面面積積積之之之慣慣慣性性性矩矩矩
Moment of Inertia for Area about Inclined Axis
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 39 69
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
傾斜軸面積之慣性矩
當轉動軸傾斜θ角時也就是座標軸旋轉θ角而Ix Iy Ixy為已知則新的座標軸(uv)可寫為
u = xcosθ + ysinθ (125)
v = ycosθ minus xsinθ (126)
由u軸上之面積慣性矩Iuv軸上之面積慣性矩Iv及慣性積Iuv得知
dIu = v2dA = (ycosθ minus xsinθ)2dA (127)
dIv = u2dA = (xcosθ + ysinθ)2dA (128)
dIuv = uvdA = (xcosθ + ysinθ)(ycosθ minus xsinθ)dA (129)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 40 69
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
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傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
傾斜軸面積之慣性矩 I
由x軸上之面積慣性矩Ixy軸上之面積慣性矩Iy及慣性積Ixy得知
Ix =
ˆy2dA Iy =
ˆx2dA Ixy =
ˆxydA (130)
Iu =
ˆ(ycosθ minus xsinθ)2dA (131)
=
ˆ(y2cos2θ minus 2xycosθsinθ + x2sin2θ)dA (132)
= cos2θ
ˆy2dAminus 2cosθsinθ
ˆxydA + sin2θ
ˆx2dA (133)
= Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (134)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 41 69
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
傾斜軸面積之慣性矩 II
Iv =
ˆ(xcosθ + ysinθ)2dA (135)
=
ˆ(x2cos2θ + 2xycosθsinθ + y2sin2θ)dA (136)
= cos2θ
ˆx2dA + 2cosθsinθ
ˆxydA + sin2θ
ˆy2dA (137)
= Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (138)
Iuv =
ˆ(xcosθ + ysinθ)(ycosθ minus xsinθ)dA (139)
=
ˆ(xycos2θ + y2cosθsinθ minus x2cosθsinθ minus xysin2θ)dA (140)
= (cos2 minus sin2θ)
ˆxydA + cosθsinθ
ˆy2dAminus cosθsinθ
ˆx2dA
= Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 42 69
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
傾斜軸面積之慣性矩 III
再由
sin2θ + cos2θ = 1 (141)
sin(αplusmn β) = sinαcosβ plusmn cosαsinβ (142)
sin2θ = sin(θ + θ) = 2sinθcosθ (143)
cos(αplusmn β) = cosαcosβ ∓ sinαsinβ (144)
cos2θ = cos(θ + θ) = cos2θ minus sin2θ (145)
= 2cos2θ minus 1 = 1 minus 2sin2θ (146)
所以
Iu = Ixcos2θ minus 2Ixycosθsinθ + Iy sin
2θ (147)
= Ix(cos2θ + 1
2) + Iy (
1 minus cos2θ
2) minus Ixy sin2θ (148)
=Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 43 69
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
傾斜軸面積之慣性矩 IV
Iv = Iycos2θ + 2Ixycosθsinθ + Ixsin
2θ (149)
= Iy (cos2θ + 1
2) + Ix(
1 minus cos2θ
2) + Ixy sin2θ (150)
=Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (151)
Iuv = Ixy (cos2 minus sin2θ) + Ixcosθsinθ minus Iycosθsinθ (152)
= Ixycos2θ +1
2Ixsin2θ minus 1
2Iy sin2θ (153)
=Ix minus Iy
2sin2θ + Ixycos2θ (154)
所以
JO = Iu + Iv = Ix + Iy (155)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 44 69
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
主主主軸軸軸慣慣慣性性性矩矩矩Principal Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 45 69
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
主軸慣性矩 I
所謂主軸慣性矩(Principal Moment ofInertia)乃定義發生在慣性矩的極值(極大或極小值)上也就是慣性矩之一次微分式為零或斜率為零
dIu
dθ=
d(Ix+Iy
2+
IxminusIy2
cos2θ minus Ixy sin2θ)
dθ(156)
= minus2Ix minus Iy
2sin2θ minus 2Ixy cos2θ (157)
= 0 (158)
其中dsin2θ = cos2θ dcos2θ = minussin2θ
當慣性矩之一次微分式為零時傾斜軸與主軸之夾角θ = θp
minus2Ix minus Iy
2sin2θp minus 2Ixycos2θp = 0 (159)
tan2θp =sin2θpcos2θp
=minusIxy
(Ix minus Iy )2
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 46 69
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
主軸慣性矩 II
cos2θp =IxminusIy
2radic(IxminusIy
2 )2 + I 2xy
sin2θp =minusIxyradic
(IxminusIy
2 )2 + I 2xy
(160)
(Iu)max =Ix + Iy
2+
Ix minus Iy2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
minus IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2+
radic(Ix minus Iy
2)2 + I 2
xy (161)
(Iv )min =Ix + Iy
2minus Ix minus Iy
2
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
+ IxyminusIxyradic
(IxminusIy
2 )2 + I 2xy
=Ix + Iy
2minusradic
(Ix minus Iy
2)2 + I 2
xy (162)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 47 69
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
主軸慣性矩 III
由tan2θp = tan(π + 2θp)可得到θp1與θp2二個解即2θp2 = π + 2θp1也就是θp2 minus θp1 = π
2
tan2θp = tan(π + 2θp) (163)
2θp2 = π + 2θp1 (164)
θp2 minus θp1 =π
2(165)
所以
Imaxmin
=Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (166)
Iuv =Ix minus Iy
2
minusIxyradic(IxminusIy
2 )2 + I 2xy
+ Ixy
IxminusIy2radic
(IxminusIy
2 )2 + I 2xy
(167)
= 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 48 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 49 69
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 I如左圖求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (168)
Iy = 56times 109mm4 (169)
Ixy = minus3times 109mm4 (170)
則主軸與x軸的夾角θp
tan2θp =minusIxy
(Ix minus Iy )2= minus222 (171)
θp = minus329 571 (172)
由於Iy gt Ix因此主軸慣性矩(Imax )的主軸比較靠近y軸
Imaxmin =
Ix + Iy
2plusmnradic
(Ix minus Iy
2)2 + I 2
xy (173)
Imax = 754times 109mm4 (174)
Imin = 96times 108mm4 (175)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 50 69
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
慣慣慣性性性矩矩矩之之之莫莫莫爾爾爾圓圓圓Mohrrsquos Circle for Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 51 69
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
慣性矩之莫爾圓
由於 Iu =Ix + Iy
2+
Ix minus Iy2
cos2θ minus Ixy sin2θ (176)
Iv =Ix + Iy
2minus Ix minus Iy
2cos2θ + Ixy sin2θ (177)
Iuv =Ix minus Iy
2sin2θ + Ixycos2θ (178)
將式(176)與式(178)各自分別平方並相加可得
(Iu minusIx + Iy
2)2 + I 2
uv = (Ix minus Iy
2)2 + I 2
xy (179)
其中Ix Iy Ixy均為已知的常數式(179)可進一步寫成
(Iu minus a)2 + I 2uv = R2 (180)
其中a =Ix+Iy
2 R =radic
(IxminusIy
2 )2 + I 2xycos2θp1 =
aRsin2θp1 =
minusIxyR 2θp1為OA轉向I軸正方向順時針
轉動為負逆時針轉動為正即此處2θp1為負
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 52 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 53 69
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
解題技巧
首先確定轉動系統的Ix Iy Ixy接著算出莫爾圓之圓心座標a及半徑R並畫出莫爾圓再由莫爾圓上點出A(Ix Ixy )並由OA轉向I軸找出兩倍夾角2θp1及2θp2其中θp1及θp2為主軸(Principle Axis)與Ix軸的夾角由於主軸(Principle Axis)為極值之軸Imax = Iu及Imin = Iv決定θp1及θp2何者為Imax = Iu與Ix軸的夾角之方法乃比較Ix及Iy之大小Imax = Iu會靠近Ix及Iy之較大者也就是Imax = Iu與Ix及Iy之較大者的夾角比較小
a =Ix + Iy
2(181)
R =
radic(Ix minus Iy
2)2 + I 2
xy (182)
2θp2 = π + 2θp1 θp2 minus θp1 =π
2(183)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 54 69
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 I
如左圖以慣性矩之莫爾圓方法求截面積作用在截面積的面心(C)之主軸慣性矩(Imax )解解解答答答首先截面積作用在截面積的面心(C)之x與y軸上的慣性矩(Ix Iy )及慣性積(Ixy )分別為
Ix = 29times 109mm4 (184)
Iy = 56times 109mm4 (185)
Ixy = minus3times 109mm4 (186)
莫爾圓之圓心座標O(a0)圓上點座標A(Ix Ixy )圓半徑R = OA
a =Ix + Iy
2= 425 (187)
Ix + Iy
2+
Ix minus Iy
2= Ix = 29times 109 (188)
Ixy = minus3times 109 (189)
OA =radic
(135)2 + (minus3)2 times 109 = 329times 109(190)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 55 69
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 II
所以
Imax = a + R (191)
= (425 + 329)times 109 = 754times 109mm4 (192)
Imin = aminus R (193)
= (425minus 329)times 109 = 96times 108mm4 (194)
OA到I軸的夾角為2θp1
2θp1 = 180 minus sinminus1 BA
OA(195)
= 1142 (196)
θp1 = 571 (197)
由於Iy gt Ix所以Imax靠近y軸
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 56 69
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
質質質量量量慣慣慣性性性矩矩矩Mass Moment of Inertia
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 57 69
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
質量慣性矩
由
τ = Iα
其中τ為旋轉扭矩I為質量慣性矩α為旋轉角速度
I =
ˆmr2dm
=
ˆVr2ρdV
= ρ
ˆVr2dV
其中r為旋轉半徑或轉軸到旋轉體的距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 58 69
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
平平平行行行軸軸軸定定定理理理Parallel-Axis Theorem
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 59 69
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
平行軸定理
當通過物體質心(G)並在一轉軸上的質量慣性矩(IG )為已知時可透過座標軸轉換計算物體之質量慣性矩(I)
I =
ˆmr2dm (198)
=
ˆm
[(d + x prime)2 + y prime2]dm (199)
=
ˆm
(x prime2 + y prime2)dm + 2d
ˆmx primedm + d2
ˆmdm (200)
I = IG + md2 (201)
其中IG為在zrsquo軸上通過質心的質量慣性矩m為質量d為通過質心之軸zrsquo並平行原軸z之距離
acutem x primedm = x
acutem dm = 0 since x = 0
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 60 69
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
迴旋半徑及複合體
迴轉半徑(k)是一個可以用來計算轉動慣量的物理量當一力矩作用於一個物體上時物體會依照轉動慣量(I) 呈現應有的旋轉運動物體對於一直軸或質心的迴轉半徑(k)是此物體所有粒子對於此直軸或質心的均方根距離
I = mk2 (202)
k =
radicI
m(203)
複合體之質量慣性矩(Ic)為各局部物體之質量慣性矩(Ii )合
Ic =i=nsumi=1
Ii (204)
= I1 + I2 + + In (205)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 61 69
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課課課堂堂堂練練練習習習
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 62 69
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 I
如左上圖圓柱密度(ρ = const)為常數求z軸的質量慣性矩(Iz )解解解答答答如左下圖由圓柱殼體積(dV)對圓柱半徑進行積分可得圓柱體積所以
dIz = r2dm = r2ρdV (206)
dV = 2πr times h times dr (207)
dm = ρdV = ρ2πr times h times dr (208)
m = 2πρh
ˆ R
0rdr (209)
= πR2hρ (210)
dIz = r2ρtimes 2πr times h times dr (211)
= 2πρhr3dr (212)
Iz = 2πρh
ˆ R
0r3dr (213)
= 2πρh times1
4R4 (214)
=1
2mR2 (215)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 63 69
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 I如左上圖錐盤密度(ρ = 5slugft3)為常數求y軸的質量慣性矩(Iy )解解解答答答如左下圖取圓盤薄片體積(dV)對y軸進行積分可得錐盤體積所以
x = y2 (216)
dV = πx2dy (217)
dm = ρdV = ρπx2dy (218)
dIy =1
2dmx2 (219)
=1
2(ρπx2dy)x2 (220)
=1
2ρπx4dy (221)
=1
2ρπy8dy (222)
Iy =1
2ρπ
ˆ 1
0y8dy (223)
=1
2ρπ
1
9y9|10 (224)
= 0873slug middot ft2 (225)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 64 69
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 I
如左圖中空圓圈密度(ρ = 8000kgm3)與厚度(10mm)為常數求垂直中空圓圈並通過O點的質量慣性矩(IO)解解解答答答如下圖中空圓圈質量慣性矩(IO)等於大圓盤質量慣性矩((IO)d )減去小圓盤質量慣性矩((IO)h)所以
md = ρVd = 1571kg (226)
(IO)d =1
2md r
2d + mdd
2 = 1473kg middotm2 (227)
mh = ρVh = 393kg (228)
(IO)h =1
2mhr
2h + mhd
2 = 0276kg middotm2 (229)
IO = (IO)d minus (IO)h = 12kg middotm2 (230)
其中mdVdrd分別為大圓盤質量體積半徑mhVhrh分
別為小圓盤質量體積半徑d為大小圓盤質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 65 69
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 I
如左圖搖鎚由兩重均為10lb的細桿互為垂直組成求搖鎚在O點與質心(G)的質量慣性矩(IO IG )解解解答答答如左下圖由細桿OA在O點的質量慣性矩((IOA)O)與細桿BC在O點的質量慣性矩((IBC )O)得到
(IOA)O =1
3ml2 =
1
3
10
32222 = 0414slug middot ft2 (231)
另解
(IOA)O = (IOA)G + md2 =1
12ml2 + md2 = 0414slug middot ft2(232)
(IBC )O = (IBC )G + md2 =1
12ml2 + md2 = 1346slug middot ft2(233)
IO = (IOA)O + (IBC )O = 176slug middot ft2 (234)
其中l(IOA)G(IBC )Gmd分別為細桿OA及BC長度質心質量
慣性矩質量質心到轉軸點O距離
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 66 69
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
課堂練習 II
另外由複合結構之質心y
y =
sumymsumm
=1 times (10322) + 2 times (10322)
10322= 15ft (235)
IO = IG + md2 (236)
IG = IO minusmd2 = 176 minus (10322) times 152 = 0362slug middot ft2 (237)
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 67 69
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-
結結結論論論
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 68 69
結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
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- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
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- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
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- 平行軸定理(Parallel-Axis Theorem)
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- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
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- 主軸慣性矩(Principal Moment of Inertia)
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- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
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- 平行軸定理(Parallel-Axis Theorem)
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結論
轉動慣量或慣性矩為物體對旋轉運動的慣性而所謂慣性乃物體持續維持不變的行為也就是定義慣性矩可表達物體對本身旋轉運動的維持或持續性好壞更簡單的說慣性矩越大之物體其維持旋轉運動趨勢的能力越大例如惰輪儲能裝置惰輪之慣性矩越大當其轉動以後其維持旋轉運動趨勢的能力越大
Chen-Ching Ting (丁振卿) Mechanical Engineering National Taipei University of Technology (國立台北科技大學機械系) Homepage httpcctmentutedutw E-mail chchtingntutedutw CCT Group[2mm] 助教 陳育煒 ()轉動慣量或慣性矩Moments of Inertia May 17 2012 69 69
- 轉動慣量或慣性矩(Moment of Inertia)
- 面積慣性矩(Moment of Inertia for Area)
-
- 面積慣性矩之平行軸定理(Parallel-Axis Theorem for Area)
- 面積慣性矩之迴轉半徑(Radius of Gyration for Area)
-
- 複合面積之慣性矩(Moment of Inertia for Composite Area)
- 面積之慣性積(Product of Inertia for Area)
-
- 平行軸定理(Parallel-Axis Theorem)
-
- 傾斜軸面積之慣性矩(Moment of Inertia for Area about Inclined Axis)
-
- 主軸慣性矩(Principal Moment of Inertia)
-
- 慣性矩之莫爾圓(Mohrs Circle for Moment of Inertia)
- 質量慣性矩(Mass Moment of Inertia)
-
- 平行軸定理(Parallel-Axis Theorem)
-