engineering mechanics: statics in si units, 12esource/home/bzhsieh/download/100... · mohr’s...

Moments of Inertia 10

Engineering Mechanics:

Statics in SI Units, 12e

Copyright © 2010 Pearson Education South Asia Pte Ltd


Chapter Outline

1. Definitions of Moments of Inertia for Areas

2. Parallel-Axis Theorem for an Area

3. Radius of Gyration of an Area

4. Moments of Inertia for Composite Areas

5. Product of Inertia for an Area

6. Moments of Inertia for an Area about Inclined Axes

7. Mohr’s Circle for Moments of Inertia

8. Mass Moment of Inertia


10.1 Definition of Moments of Inertia for Areas

面積慣性矩的定義

• 當分佈負載垂直作用在面積上，且其強度為線性變化，則計算分佈負載對一軸的力矩會包含一個「面積慣性矩」的量。

dAydMM

dAyydFdM

dAypdAdF

yp

2

2

)(

)(

壓力隨深度成線性變化



Moment of Inertia

• moments of inertia of the differential plane area dA

about the x and y axes

• For entire area, moments of

inertia are

Ay

Ax

dAxI

dAyI

2

2

dAxdI

dAydI

y

x

2

2



Polar moment of inertia 極慣性矩

• Polar moment of inertia for entire

area (about the pole O or z axis)

where r is perpendicular from the

pole (z axis) to the element dA

yxA

O IIdArJ 2


Example

Determine the moment of inertia for the rectangular area

with respect to the centroidal x’ axis.

求長方形面積對 x’ 軸 (形心軸 x’ 軸 ) 之慣性矩


Solution

the moment of inertia for the rectangular area with

respect to the centroidal x’ axis:

3

2/

2/

2

2/

2/

2

2

12

1

''

)'('

'

bh

dyyb

bdyy

dAyI

h

h

h

h

Ax

'

)( 22

bdydA

dAyIdAyI xx


Test


with respect to (a) the centroidal y’ axis and (b) the pole

or z’ axis perpendicular to the x’-y’ plane and passing

through the centroid C.

yxA

O IIdArJ 2

dAxI y

2

3

12

1bhI x

dAxId y

2


Test -- Answer

(a) For moment of inertia about y’ axis

3

12

1hbI y

3

2/

2/

2

2/

2/

2

2

12

1

''

)'('

'

hb

dxxh

hdxx

dAxI

b

b

b

b

Ay

dx’

x’


Test -- Answer

(b) For polar moment of inertia about

point C,

)(12

1

12

1

12

1

22

33

bhbh

hbbh

IIJ yxC


Chapter Outline










10.2 Parallel Axis Theorem for an Area

• For moment of inertia of an area known about an axis

passing through its centroid, determine the moment of

inertia of area about a corresponding parallel axis

using the parallel axis theorem 平行軸定理



• The fixed distance between the parallel x and x’ axes

is defined as dy

• For moment of inertia of dA about x axis

• For entire area

Ay

Ay

A

Ayx

yx

dAddAyddAy

dAdyI

dAdydI

22

2

2

'2'

'

'



2

yxx AdII

A

yA

yA

x dAddAyddAyI 22 '2'

0;0' ydAydAy

First integral represent the moment of inertia of the area about the centroidal axis

Third integral represents the total area A

Second integral = 0 since x’ passes through the area’s centroid C



• Similarly

• For polar moment of inertia about

an axis perpendicular to the x-y

plane and passing through pole

O (z axis)

2AdJJ CO

2

xyy AdII

2

yxx AdII


Example


with respect to (a) the centroidal x’ axis, (b) the axis xb

passing through the base of the rectangular.


Solution

Part (a) – to the centroidal x’ axis

Part (b) – to the axis xb

By applying parallel axis theorem,

32

12

1' bhdAyI

Ax

3

2

3

2

3

1

2)(

12

1

bh

hbhbh

AdII xxb

Test

• Determine the moment of inertia for the rectangular area

with respect to (a) the centroidal y’ axis and (b) the axis

yb passing through the base of the rectangular.


2

xyy AdIIb

3

12

1hbI y

c


Solution

Part (a) – to the centroidal y’ axis

Part (b) – to the axis yb

By applying parallel axis theorem,

3

2

3

2

3

1

212

1

hb

bhbhb

AdII xyyb

3

12

1hbI y

dx’

dx=b/2

yb


Chapter Outline










10.3 Radius of Gyration of an Area

面積的迴轉半徑

• A quantity used in the design of columns in structural

mechanics 常用於結構力學中的圓柱設計

• For radii of gyration 迴轉半徑

• Similar to finding moment of inertia of a differential

area about an axis

dAydIAkI

A

Jk

A

Ik

A

Ik

xxx

Oz

yy

xx

22


Chapter Outline










10.4 Moments of Inertia for Composite Areas

複合面積的慣性矩

• Composite area consist of a series of connected simpler parts or shapes

• 由簡單的元件或形狀構成

• Moment of inertia of the composite area = algebraic sum of the moments of inertia of all its parts

• 所有元件的慣性矩的代數合


10.4 Moments of Inertia for Composite Areas

複合面積的慣性矩

Procedure for Analysis 解題程序

1. Composite Parts (分解)

2. Parallel Axis Theorem (運用平行軸定理)

• Moment of inertia of each part is determined about its centroidal axis (先求對形心的慣性矩)

• When centroidal axis does not coincide with the

reference axis, the parallel axis theorem is used (運用平行軸定理)

3. Summation (加總)


Example

Compute the moment of inertia of the composite area

about the x axis.


Solution

1. Composite Parts


Solution

2. Parallel Axis Theorem

Rectangle

Circle

46

224

2

'

104.11

7525254

1

mm

AdII yxx

46

23

2

'

105.112

7515010015010012

1

mm

AdII yxx


Solution

Summation

For moment of inertia for the composite area

46

66

10.1101

104.11105.112

mm

I x

Test

• Determine the moment of inertia for the cross-sectional

area of the member about the y axis.


)( 2

' xyy AdII

Answer

• The cross section can be subdivided into three

composite rectangular areas A, B, and D.

• Rectangle A and D


49

23

23

2

'

)10(90.1

25010030010030012

1

))(()(12

1

mm

dbhbh

AdII

x

xyy

Answer

• Rectangle B


49

23

23

2

'

)10(80.1

060010060010012

1

))(()(12

1

mm

dbhbh

AdII

x

xyy

Answer


49

999

2

'

)10(60.5

)10(90.1)10(80.1)10(90.1

)(

mm

AdII xyy


Chapter Outline










10.5 Product of Inertia for an Area

面積的慣性積

• Product of inertia for an element of area dA located at

a point (x, y) is defined as

• Thus for product of inertia,

Axy xydAI

xydAdI xy


面積的慣性積

• The product of inertia may either be

positive, negative, or zero.

• The product of inertia Ixy for an area

will be zero if either the x or y axis

is an axis of symmetry for the area.


Axy xydAI



Parallel Axis Theorem

• For the product of inertia of dA with respect to the x

and y axes

• For the entire area,

• Forth integral represent the total area A,

dAdydxdIA

yxxy ''

Ayx

Ay

A Ax

Ayxxy

dAdddAxddAyddAyx

dAdydxdI

''''

''

yxyxxy dAdII ''


Example

Determine the product Ixy of the triangle.


Solution

Differential element has thickness

dx and area dA = y dx

Using parallel axis theorem,

locates centroid of the element

or origin of x’, y’ axes

yxdAIddI xyxy~~

yx ~,~


Solution

Due to symmetry,

Integrating we have

2/~,~0 yyxxId xy

82

22

0

3

2

2 hbdxx

b

hI

b

xy

dxxb

hx

b

hxxdx

b

hyxydxdI xy

3

2

2

222)(0

yxdAIddI xyxy~~

Test

• Determine the product of inertia for the cross-sectional

area of the member, about the x and y centroidal axes.


)( '' yxyxxy dAdII

Test -- Answer

• The cross section can be subdivided into three

composite rectangular areas A, B, and D.

• Rectangle A


49

''

)10(50.1

)200)(250)(100*300(0

mm

dAdII yxyxxy

Test -- Answer

• Rectangle B


4

''

0

00

mm

dAdII yxyxxy

Test -- Answer

• Rectangle D


49

''

)10(50.1

)200)(250)(100*300(0

mm

dAdII yxyxxy

Test -- Answer

• The product of inertia for the

entire cross section is:


49

99

)10(00.3

)]10(50.1[ ] 0 [ )]10(50.1[

mm

I xy


Chapter Outline










10.8 Mass Moment of Inertia 質量慣性矩

• Mass moment of inertia is defined as the integral of the

second moment about an axis of all the elements of

mass dm which compose the body

• For body’s moment of inertia

about the z axis,

• The axis that is generally chosen

for analysis, passes through the

body’s mass center G

mdmrI 2


10.8 Mass Moment of Inertia

• If the body consists of material having a variable

density ρ = ρ(x, y, z), the element mass dm of the body

may be expressed as dm = ρ dV

• Using volume element for integration,

• When ρ being a constant,

VdVrI 2

V

dVrI 2



Procedure for Analysis

Shell Element 殼元素分析法

• For a shell element having height z,

radius y and thickness dy, volume

dV = (2πy)(z)dy

Disk Element 盤元素分析法

• For disk element having radius y,

thickness dz, volume dV = (πy2) dz


Example

Determine the mass moment of inertia of the cylinder

about the z axis. The density of the material is constant.


Solution

Shell Element 殼元素

For volume of the element,

For mass,

Since the entire element lies at the same distance r from

the z axis, for the moment of inertia of the element,

drhrdmrdI

drrhdVdm

drhrdV

z

32 2

2

2


Solution

Integrating over entire region of the cylinder,

For the mass of the cylinder

So that

hRdrrhdmrIR

mz

4

0

32

22

2

02 hRrdrhdmm

R

m

2

2

1mRI z




• If the moment of inertia of the body about an axis

passing through the body’s mass center is known, the

moment of inertia about any other parallel axis may be

determined by using parallel axis theorem

• Using Pythagorean theorem,

r 2 = (d + x’)2 + y’2

• For moment of inertia of body

about the z axis,

mmm

mm

dmddmxddmyx

dmyxddmrI

222

222

'2''

''




• For moment of inertia about the z axis,

Radius of Gyration

• For moment of inertia expressed using k, radius of

gyration,

m

IkormkI 2

2mdII G


Example

If the plate has a density of 8000kg/m3 and a thickness of

10mm, determine its mass moment of inertia about an

axis perpendicular to the page and passing through point

O.


Solution

The plate consists of 2 composite

parts, the 250mm radius disk minus

the 125mm radius disk.

Disk

For moment of inertia of a disk,

Mass center of the disk is located

0.25m from point O

2

2

1mrIG

22222

2

.473.125.071.1525.071.152

1

2

1

71.1501.025.08000

mkgdmrmI

kgVm

ddddO

ddd


Solution

Hole

For moment of inertia of plate about point O,

222

22

2

.276.025.093.3125.093.32

1

2

1

93.301.0125.08000

mkg

dmrmI

kgVm

hhhhO

hhh

2.20.1276.0473.1 mkg

IIIhOdOO


Chapter Outline






6. Moments of Inertia for an Area about Inclined

Axes




10.6 Moments of Inertia for an Area about Inclined Axes

對一傾斜軸的面積慣性矩

• In structural and mechanical design, necessary to

calculate Iu, Iv and Iuv for an area with respect to a set

of inclined u and v axes when the values of θ, Ix, Iy

and Ixy are known

• Use transformation equations which relate the x, y and

u, v coordinates

sincos

sincos

xyv

yxu



dAxyyxuvdAdI

dAyxdAudI

dAxydAvdI

xyv

yxu

uv

v

u

)sincos)(sincos(

)sincos(

)sincos(

sincos

sincos

22

22



• Integrating,

• Simplifying using trigonometric identities (三角恆等式),

)sin(coscossincossin

cossin2cossin

cossin2sincos

22

22

22

xyyxuv

xyyxv

xyyxu

IIII

IIII

IIII

22 sincos2cos

cossin22sin



• Polar moment of inertia about the z axis passing

through point O is,

2cos2sin2

2sin2cos22

2sin2cos22

xy

yx

uv

xy

yxyx

v

xy

yxyx

u

III

I

IIIII

I

IIIII

I

yxvuO IIIIJ



Principal Moments of Inertia 主慣性矩

• Iu, Iv and Iuv depend on the angle of inclination θ of the

u, v axes (Iu, Iv and Iuv 與 u, v 軸的傾斜角度有關)

• 有一個特定的軸方位角，可以使得面積對其的慣性矩為最大及最小

• 這一組座標軸＝該面積的「主軸 (principle axes)」

• 面積對此「主軸」的慣性矩＝「主慣性矩」



• The angle θ = θp defines the orientation of the principal

axes for the area

2

)(2tan

02cos22sin2

2

yx

xy

p

p

xy

yxu

II

I

III

d

dI



2cos2sin2

2sin2cos22

2sin2cos22

xy

yx

uv

xy

yxyx

v

xy

yxyx

u

III

I

IIIII

I

IIIII

I

2

)(2tan

yx

xy

p II

I



• Substituting each of the sine and cosine ratios

• Result gives the max or min moment of inertia for the

area

• It can be shown that Iuv = 0, that is, the product of

inertia with respect to the principal axes is zero

• Any symmetric axis represent a principal axis of inertia

for the area

2

2

max

min22

xy

yxyxI

IIIII


Example

Determine the principal moments of inertia for the beam’s

cross-sectional area with respect to an axis passing

through the centroid.

求通過形心的主慣性軸及主軸方位

2

2

max

min22

xy

yxyxI

IIIII

2

)(2tan

yx

xy

p II

I


Solution

Moment and product of inertia of the cross-sectional area,

49

49

49

1000.3

1060.5

1090.2

mmI

mmI

mmI

xy

y

x


Solution

Using the angles of inclination of principal axes u and v,

(先求主軸角度)

22.2

2/1060.51090.2

]1000.3[

2/2tan

99

9

yx

xy

pII

I

1.57 9.32

2.1142 8.652

21

21

pp

pp

and

and


Solution

For principal of inertia with respect to the u and v axes

(主慣性矩)

49

min

49

max

10960.0

1054.7

mmI

mmI

29

299

99

2

2

max

min

1000.32

1060.51090.2

2

1060.51090.2

22

xy

yxyxI

IIIII

99max

min 1029.31025.4 I


Chapter Outline









10.7 Mohr’s Circle for Moments of Inertia

慣性矩的莫耳圓


2cos2sin2

2sin2cos22

2sin2cos22

xy

yx

uv

xy

yxyx

v

xy

yxyx

u

III

I

IIIII

I

IIIII

I

----- (式1)

----- (式3)

(式1) 及 (式3) 平方後相加



222

2

2

2

2

22

RIaI

III

III

I

uvu

xy

yx

uv

yx

u

以慣性矩為橫軸、慣性積為縱軸，繪圖，可畫出一個半徑為R的圓 (莫耳圓，Mohr’s circle)



莫耳圓的半徑為R，圓心位於 (a, 0) 處。

2

yx IIa

2

2

2xy

yxI

IIR

222

2

2

2

2

22

RIaI

III

III

I

uvu

xy

yx

uv

yx

u



2

2

max

min22

xy

yxyxI

IIIII

2

(a,0) intyx II

aOpo

2

2

2xy

yxI

IIR

2

2tanyx

xy

p II

I

xyx IIAporeference , int


Example

Using Mohr’s circle, determine the principle moments of

the beam’s cross-sectional area with respect to an axis

passing through the centroid.


Solution

1. Determine Ix, Iy and Ixy (先求 Ix, Iy and Ixy )

From previous examples,

49

49

49

1000.3

1060.5

1090.2

mmI

mmI

mmI

xy

y

x


Solution

2. Construct the Circle (畫莫耳圓)

Center of circle, O

Reference point

A (2.90, -3.00)

Distance OA ( radius, R )

25.4

2

)60.590.2(

2

yx IIa

2

2

2xy

yxI

IIR

xyx IIA ,

29.300.335.122OA

49

49

49

1000.3

1060.5

1090.2

mmI

mmI

mmI

xy

y

x


Solution

3. Principal Moments of Inertia (求主慣性矩)

Circle intersects I axis at points

(7.54, 0) and (0.96, 0)

49

min

49

max

10 96.0)29.325.4(

10 54.7)29.325.4(

mmI

mmI


Solution

4. Principal Axes (求主軸方位)

由半徑OA到正向x軸，這個角度代表「從x軸到最大慣性矩Imax軸之間夾角的兩倍」

Angle 2θp1 is determined from the

circle by measuring CCW (逆時針量測) from OA to the direction of

the positive I axis

2.114

29.3

00.3sin180 sin1802 11

1

OA

BAp


Solution

4. Principal Axes (求主軸方位)

從 x 軸 (the positive x axis) 逆時針轉 57.1° ，

為 Imax 的主軸 (the positive u axis)

v 軸則垂直於 u 軸

1.57

2.1142

1

1

p

p

The END OF CHAPTER 10


engineering mechanics: statics in si units, 12esource/home/bzhsieh/download/100... · mohr’s...

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