bölüm 4 eğik eğilme2014
TRANSCRIPT
1Unsymmetric Bending
Simetrik olmayan (Eğik) EğilmeUnsymmetric Bending
Unsymmetric Bending 2
Chapter Outline
• Pure Bending• Unsymmetric Bending for symmetric cross-section• Unsymmetric Bending for unsymmetric cross-section• Inertia moments – principal inertia moments
Unsymmetric Bending 3
• Analysis of pure bending has been limited to members subjected to bending couples acting in a plane of symmetry.
• Members remain symmetric and bend in the plane of symmetry.• The neutral axis of the cross section coincides with the axis of the
couple.
Pure Bending Basit Eğilme
Unsymmetric Bending 4
• We will now consider situations in which the bending couples do not act in a plane of symmetry.
• Cannot assume that the member will bend in the plane of the couples.
Unsymmetric Bending for Symmetric Cross-section
Simetrik Kesitli Kirişlerde Ansimetrik Eğilme
yM
Unsymmetric Bending for symmetric cross-section
• We wish to determine the conditions under which the neutral axis of
a cross section of arbitrary shape does not coincides with the axis of
the couple as shown.
• The resultant force and moments from the distribution of elementary
forces in the section must satisfy
0xF
• Neutral axis passes through centroid.
00 0 dAzanddAydAx
6
Determination of stress distribution due to Mz
zIdAy 2
zMm
y
Mxz MdA
cyydAyM zz
yccy y
MmM
xy
Mm
Mx
zz
zz
yI
Mσy
Iσc
IσMz
zMx
zMx
y
zMm
zz
zz
inertiaofmoment
yc
yM
7
Determination of stress distribution due to My
yIdAz 2
yMm
z
Mxy MdA
czzdAzM yy
zccz z
MmM
xz
Mm
Mx
yy
yy
zI
Mσ
zIσ
cIσ
My
yMx
yMx
z
yMm
yy
yy
inertiaofmoment
yM
zc
8
Resultant stress: Superposition method
yI
MzI
M
σσ
z
z
y
yx
Mx
Mxx
zy
yM
Unsymmetric Bending 4 - 9
Unsymmetric Bending for symmetric «T» cross-section
• Superposition is applied to determine stresses in the most general
case of unsymmetrical bending.
• Resolve the couple vector into components along the principle centroidal axes.
sincos MMMM yz kMjMM zy
10
• Superpose the component stress distributions
zI
M
y
yMx
y yI
M
z
zMx
z
Combined stress, zy Mx
Mxx
zI
MyI
MzI
My
IM
yzy
y
z
zx
sincos
Unsymmetric Bending 11
• Along the neutral axis, normal stress must be zero. Thus
axisNeutralzyII
zIIy
II
II
zy
zI
MyI
M
zI
My
IM
y
z
y
z
y
z
y
z
yzx
y
y
z
zx
tantantan
tantancossin
0sincos
0
Unsymmetric Bending 12
Example 4.08
A 180 Nm couple is applied to a rectangular
wooden beam in a plane forming an angle of
30 deg. with the vertical. Determine
(a) the maximum stress in the beam,
(b) the angle that the neutral axis forms with
the horizontal plane.
z
yy
z
Unsymmetric Bending 13
• Resolve the couple vector into components along the principle
centroidal axes and calculate the corresponding maximum stresses.
sincos MMMM yz
• Combine the stresses from the component stress distributions.
zI
My
IM
y
y
z
zx
Solution
z
y
MM
tan
Unsymmetric Bending 14
• Determine the angle of the neutral axis.
zIIy
MM
zMM
IIy
y
z
z
y
z
y
y
z
tantan
zyzy tanor tan
0 zI
My
IM
y
y
z
zx
orzy
II
y
z tantan the angle of the neutral axis
Unsymmetric Bending 4 - 15
• Resolve the couple vector into components and calculate the
corresponding maximum stresses.
NmNmM
NmNmMo
y
oz
9030sin180
9.15530cos180
463
121
463121
1048.04090
1043.29040
mmmmmmI
mmmmmmI
y
z
Unsymmetric Bending 16
• Determine the angle of the neutral axis.
o
y
z
z
y
II
MM
35.72
92.25772.01048.01043.2tantan
5773.09.155
90tan
6
6
Unsymmetric Bending 4 - 17
.20;45at occurs todue
stress nsilelargest te The
AM
NmNmM
NmNmMo
y
oz
9030sin180
9.15530cos180
mmmm
NmmmmmmNmm
zI
My
IM
A
Ay
yA
z
zA
201048.0
1090451043.2109.155
46
3
46
3
max64.6 MPaA
Unsymmetric Bending 4 - 18
.20;45at occurs todue
stress normalsmallest The
EM
mmmm
NmmmmmmNmm
zI
My
IM
E
Ey
yE
z
zE
201048.0
1090451043.2109.155
46
3
46
3
min64.6 MPaE
Unsymmetric Bending 4 - 19
• Distribution of the streeses in the cross-section.
MPa64.6
MPa64.6
z
x
y
b=60
mm
h=80
mm
L=1 m
F=10 kN
A
B
Örnek: Şekildeki kirişin A, B, C ve D noktalarında meydana gelen
gerilmeleri hesaplayınız.
C
D
Çözüm: Kesit özellikleri (atalet momentleri):
46
31213
121
1056.2
8060
mmI
hbI
x
x
A b=60 mm
x
y
h=80
mm
F=10 kN
M
B
46
31213
121
1044.1
6080
mmI
bhI
y
y
Unsymmetric Bending 22
Eğilme momenti bileşenleri:
NmmM
kNmmkNLFM61010
10110
siyo
o
87.36
87.368060tan
NmmkNmM
kNmMM
x
x
61088
87.36cos10cos
b=60 mm
x
y
h=80
mm
F=10 kN
M
A
B
NmmkNmM
kNmMM
y
y
61066
87.36sin10sin
Unsymmetric Bending 23
o
y
x
II
13.53333,1tan
87.36tan44.156.2tantan
b=60 mm
x
y
h=80
mm
F=10 kN
M
A
B
TETarafsız Eksen (TE)
Unsymmetric Bending 24
b=60 mm
x
y
h=80
mm
F=10 kN
M
A
B
TEAy
yA
x
xA x
IM
yI
M
)40;30();( AyxA AA
MPaA
A
250
301044.110640
1056.2108
6
6
6
6
A noktasındaki gerilme:
Unsymmetric Bending 25
b=60 mm
x
y
h=80
mm
F=10 kN
M
A
B
TE
B noktasındaki gerilme:
MPa
xI
My
IM
B
B
By
yB
x
xB
250
301044.110640
1056.2108
6
6
6
6
)40;30();( ByxB BB
Unsymmetric Bending 26
b=60 mm
x
y
h=80
mm
F=10 kN
M
A
B
TE
C noktasındaki gerilme:
401044.110630
1056.2108
6
6
6
6
C
Cy
yC
x
xC x
IM
yI
M
)40;30();( CyxC CC C
MPaC 92.72
Unsymmetric Bending 27
x
y
F=10 kN
M
A
B
TE
D noktasındaki gerilme:
01044.110630
1056.2108
6
6
6
6
D
Dy
yD
x
xD x
IM
yI
M
)0;30();( DyxD DD C
MPaD 75.96
D
Unsymmetric Bending 28
Unsymmetric Bending 29
Unsymmetric Bending 30
Unsymmetric Bending 31
Unsymmetric Bending 32
• In general, the neutral axis of the section will not coincide with the axis of the couple.
Bending for unsymmetric cross-section
33
Simetrik Olmayan Kirişlerde Eğilme
)1( ve A yzA xz MdAxMdAy
Unsymmetric Bending 34
(2) 21 xCyCzzz
(3) 21 xCyCEE zz
(4) 21 xCyCz
Unsymmetric Bending 35
(5) - ve 2121 AyAx dAxCyCxMdAxCyCyM
(6) 21
221
2122
1
yxyAAy
xyxAAx
ICICdAxCxydACM
ICICdAxyCdAyCM
(6) 21
21
yxyy
xyxx
ICICM
ICICM
2
1
CC
IIII
MM
yxy
xyx
y
xveya
Unsymmetric Bending 36
22xyyx
xyxxy
IIIIMIM
C
21
xyyx
xyyyx
IIIIMIM
C
ve
7 22 xIII
IMIMy
IIIIMIM
xyyx
xyxxy
xyyx
xyyyxz
7 2xyyx
xyxxyxyyyxz III
xIMIMyIMIM
veya
21 xCyCz
Unsymmetric Bending 37
8 tan veya xyxIMIMIMIM
yxyyyx
xyxxy
xyyyx
xyxxy
IMIMIMIM
tan
Tarafsız Eksenin x ekseni ile yaptığı açı:
xyyyx
xyxxy
IMIMIMIM
1tan
σz gerilme fonksiyonu sıfıra eşitlenirse Tarafsız Eksen (TE) denklemi ve Tarafsız Eksenin açısı bulunur.
TE
Unsymmetric Bending 38
Unsymmetric Bending 39
Unsymmetric Bending 40
Örnek:
Şekilde kesiti ve yükleme durumu verilen kiriş için:a) Ağırlık merkezinin yerini belirleyiniz.b) Ağırlık merkezinden geçen eksenlere göre
atalet momentlerini bulunuz.c) Kesitteki eğilme momentinin değeri M=20
kNm olduğuna göre bileşenlerini bulunuz.d) Tarafsız Ekseni belirleyiniz.e) A, B ve C noktalarındaki gerilmeleri
hesaplayınız.f) Kiriş kesitindeki ekstremum gerilmeleri
belirleyerek emniyetli olup olmadığını irdeleyiniz.
M
A
B C
G
MPaem 300o30
Unsymmetric Bending 41
x
Kesitin ağırlık merkezinin yeri:
Unsymmetric Bending 42
Atalet Momentleri: Paralel Eksen Teoremi
X
Y
C
y
x
O
Ci
x
yyi
xi
2iyiixx dAII
2ixiiyy dAII
ii yxiixyxy ddAII
xxd
yyd
ix
iy
i
i
Unsymmetric Bending
Ağırlık merkezinden geçen x ve y eksenlerine kesitin atalet momentleri:
Paralel Eksen Teoremi ile
Y
X
mmymmy
1590
2
1
mmxmmx
7515
2
1
mmxxd
mmxxd
x
x
403575
203515
2
1
2
1
mmxxd
mmyyd
y
y
506515
256590
2
1
2
1
xI
46101.85040309002520180300 mmI xy
Unsymmetric Bending 44
M
A
B C
GxM
yM
kNmMM ox 32.1730cos20cos
kNmMM oy 1030sin20sin
Eğilme momentinin bileşenleri:
Unsymmetric Bending 45
yIII
IMIMx
IIIIMIM
xyyx
xyyyx
xyyx
xyxyxz 22
Gerilme Denklemi:
66
66
22 10101010
1.8708.891.241.810708.832.17
1.8708.891.2491.24101.832.17
yxz
yxz 3.15182.69
3.1518.108
yxz 462.0719.0
Unsymmetric Bending 46
Tarafsız Eksen Denklemi:
Tarafsız Eksen Denklemi, Gerilme denklemi sıfıra eşitlenerek bulunur:
xxy tan556.1
Tarafsız Eksenin eğim açısı :
o28.57556.1tan
T.E.
0462.0719.0 yxz
47
T.E.
Kesitteki gerilmeler:
A
B C
A(-35;115) noktası MPaA 3.78115462.035719.0
MPaB 865.465462.035719.0
MPaC 145.9165462.085719.0
MPaMPa emC 300145.91max
Olduğundan kiriş emniyetlidir.
B(-35;-65) noktası
C(85;-65) noktası
Kesit kontrolü
Unsymmetric Bending 48
Örnek: şekildeki ‘L’ kesitli kirişin C noktasına P=4 kN luk eğik bir
kuvvet uygulanmaktadır. Ҩ=60 olduğuna göre:
a) Kesitin ağırlık merkezinin yerini belirleyiniz.
b) Atalet momentlerini hesaplayınız.
c) T.E. Tarafsız ekseni bulunuz.
d) Maximum çeki ve bası gerilmelerini hesaplayınız.
xyyx III ,,
Unsymmetric Bending 49
Çözüm:
Ağırlık merkezinin yeri
Atalet momentleri
Ix =2.783x106 mm4
Iy =1.003x106 mm4
Ixy =-0.973x106 mm4
mmy
mmx
74.398001100
5800651100
74.198001100
4080051100
Unsymmetric Bending 50
Momentin bileşenleri
kNmRMkNR
PRM
A
A
AB
8.44.2224.2
0350
max
kNmM
MMkNmM
MM
y
y
x
x
4.2
60cos8.4cos157.4
60sin8.4sin
max
max
Unsymmetric Bending 51
Gerilme fonksiyonu
yxyxz 9945.0428.1,
66
662 10101010973.0003.1783.2
973.0157.4783.24.2973.04.2003.1157.4
xyz
2xyyx
xyxxyxyyyxz III
xIMIMyIMIM
Unsymmetric Bending 52
Tarafsız Eksen Denklemi ve Tarafsız Eksenin eğim Açısı
xyxyz
4362.10428.19945.0
oxy 15.55tan
Unsymmetric Bending 53
Gerilmeler
yxz 9945.0428.1
MPaA 6.125)74.39(9945.0)26.60(428.1
MPaB 108)26.80(9945.0)74.19(428.1
T.E. den en uzak noktalar:
26.80;74.19ve
74.39;26.60
B
A
Unsymmetric Bending 54
Örnek: Yükleme durumu ve kesiti görülen kirişin A, B ve C noktalarında meydana gelen gerilmeleri hesaplayınız ve σem=200 MPa için kirişin kontrolünü yapınız.
Fya b
z
y10
40
10
60 mm10
y
xC
F
A
B
C
o mbakNF 60 ,1 ,2
55
Unsymmetric Bending 56
Unsymmetric Bending 57
Unsymmetric Bending 58
Unsymmetric Bending 59
B
C
Unsymmetric Bending 60
0105.15.1 6
y
oz
MNmmkNmMM
2yzyz
yzzzyyzyyzx III
zIMIMyIMIM
2yzyz
yzzyzx III
zIMyIM
o
zyzy
44.41883.0tan
883.0912.0805.0
0x
N.A.
B
C
zy
zy
x
x
805.0912.0
1010101087.218.425.3
87.25.125.35.1 66662
46
46
46
1087.2
1018.4
1025.3
mmI
mmI
mmI
yz
z
y
for neutral axis
Solution:
Unsymmetric Bending 61
zyzyx 805.0912.0,
74 ,50A
MPaA
A
97.1374805.050912.0
N.A.
B
C
74 ,38B
MPaB
B
91.2474805.038912.0
6 ,50C
max43.50
6805.050912.0
MPaC
C
Stresses at points A, B and C:
Unsymmetric Bending 62
Unsymmetric Bending 63
Unsymmetric Bending 64
Unsymmetric Bending 65
Unsymmetric Bending 66
Unsymmetric Bending 67
Unsymmetric Bending 68
Unsymmetric Bending 69
Unsymmetric Bending 70