engineering mechanics: statics chapter 10: moments of inertia chapter 10: moments of inertia

Engineering Mechanics: StaticsEngineering Mechanics: Statics

Chapter 10: Moments of

Inertia

Chapter 10: Moments of

Inertia

Chapter ObjectivesChapter Objectives

To develop a method for determining the moment of inertia for an area.

To introduce the product of inertia and show how to determine the maximum and minimum moments of inertia for an area.

To discuss the mass moment of inertia.

Chapter OutlineChapter Outline

Definitions of Moments of Inertia for Areas

Parallel-Axis Theorem for an AreaRadius of Gyration of an AreaMoments of Inertia for an Area by

IntegrationMoments of Inertia for Composite

AreasProduct of Inertia for an Area

Chapter OutlineChapter Outline

Moments of Inertia for an Area about Inclined Axes

Mohr’s Circle for Moments of InertiaMass Moment of Inertia

10.1 Moments of Inertia10.1 Moments of Inertia

Definition of Moments of Inertia for Areas Centroid for an area is determined by

the first moment of an area about an axis

Second moment of an area is referred as the moment of inertia

Moment of inertia of an area originates whenever one relates the normal stress σ or force per unit area, acting on the transverse cross-section of an elastic beam, to applied external moment M, that causes bending of the beam


Definition of Moments of Inertia for Areas Stress within the beam varies linearly

with the distance from an axis passing through the centroid C of the beam’s cross-sectional area

σ = kz For magnitude of the force acting

on the area element dA dF = σ dA = kz dA


Definition of Moments of Inertia for Areas Since this force is located a distance z from

the y axis, the moment of dF about the y axis

dM = dF = kz2 dA Resulting moment of the entire stress

distribution = applied moment M

Integral represent the moment of inertia of area about the y axis

dAzM 2


Moment of Inertia Consider area A lying in the x-y plane Be definition, moments of inertia of the

differential plane area dA about the x and y axes

For entire area, moments of inertia are given by

Ay

Ax

yx

dAxI

dAyI

dAxdIdAydI

2

2

22


Moment of Inertia Formulate the second moment of dA

about the pole O or z axis This is known as the polar axis

where r is perpendicular from the pole (z axis) to the element dA

Polar moment of inertia for entire area,

yxAO

O

IIdArJ

dArdJ

2

2


Moment of Inertia Relationship between JO, Ix and Iy is

possible since r2 = x2 + y2

JO, Ix and Iy will always be positive since they involve the product of the distance squared and area

Units of inertia involve length raised to the fourth power eg m4, mm4

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

Consider moment of inertia of the shaded area

A differential element dA is located at an arbitrary distance y’ from the centroidal x’ axis



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

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

AyAyA

A yx

yx

dAddAyddAy

dAdyI

dAdydI

22

2

2

'2'

'

'



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

Third integral represents the total area A

Similarly

For polar moment of inertia about an axis perpendicular to the x-y plane and passing through pole O (z axis)

2

2

2

0;0'

AdJJ

AdII

AdII

ydAydAy

CO

xyy

yxx



Moment of inertia of an area about an axis = moment of inertia about a parallel axis passing through the area’s centroid plus the product of the area and the square of the perpendicular distance between the axes

10.3 Radius of Gyration of an Area

10.3 Radius of Gyration of an Area

Radius of gyration of a planar area has units of length and is a quantity used in the design of columns in structural mechanics

Provided moments of inertia are known For radii of gyration

Similar to finding moment of inertia of a differential area about an axisdAydIAkI

AJ

kA

Ik

AI

k

xxx

Oz

yy

xx

22

10.4 Moments of Inertia for an Area by

Integration


Integration When the boundaries for a planar area

are expressed by mathematical functions, moments of inertia for the area can be determined by the previous method

If the element chosen for integration has a differential size in two directions, a double integration must be performed to evaluate the moment of inertia

Try to choose an element having a differential size or thickness in only one direction for easy integration


Integration


IntegrationProcedure for Analysis If a single integration is performed to

determine the moment of inertia of an area bout an axis, it is necessary to specify differential element dA

This element will be rectangular with a finite length and differential width

Element is located so that it intersects the boundary of the area at arbitrary point (x, y)

2 ways to orientate the element with respect to the axis about which the axis of moment of inertia is determined


Integration


IntegrationProcedure for AnalysisCase 1 Length of element orientated parallel to the

axis Occurs when the rectangular element is

used to determine Iy for the area Direct application made since the element

has infinitesimal thickness dx and therefore all parts of element lie at the same moment arm distance x from the y axis


Integration


IntegrationProcedure for AnalysisCase 2 Length of element orientated

perpendicular to the axis All parts of the element will not lie at the

same moment arm distance from the axis For Ix of area, first calculate moment of

inertia of element about a horizontal axis passing through the element’s centroid and x axis using the parallel axis theorem


Integration


IntegrationExample 10.1Determine the moment of inertia for the rectangular area with respect to (a) the centroidal x’ axis, (b) the axis xb passing through the base of the rectangular, and (c) the pole or z’ axis perpendicular to the x’-y’ plane and passing through the centroid C.

SolutionPart (a) Differential element chosen, distance y’

from x’ axis Since dA = b dy’

3

2/

2/

22

12

1

''

bh

dyydAyIh

hAx


Integration


Integration

SolutionPart (b) Moment of inertia about an axis

passing through the base of the rectangle obtained by applying parallel axis theorem

32

3

2

3

1

212

1bh

hbhbh

AdII xxb


Integration


Integration

SolutionPart (c) For polar moment of inertia about point

C

)(12

1

12

1

22

'

3'

bhbh

IIJ

hbI

yxC

y


Integration


Integration


Integration


IntegrationExample 10.2Determine the moment

of inertia of the shaded

area about the x axis


Integration


IntegrationSolution A differential element of area that is

parallel to the x axis is chosen for integration

Since element has thickness dy and intersects the curve at arbitrary point (x, y), the area

dA = (100 – x)dy All parts of the element lie at the same

distance y from the x axis

Solution

46

200

0

200

0

42

2200

0

2

2

2

)10(107

400

1100

400100

)100(

mm

dyydyy

dyy

y

dyxy

dAyI

A

Ax


Integration


Integration


Integration


IntegrationSolution A differential element

parallel to the y axis is chosen for integration

Intersects the curve at arbitrary point (x, y)

All parts of the element do not lie at the same distance from the x axis


Integration


IntegrationSolution Parallel axis theorem used to determine

moment of inertia of the element For moment of inertia about its centroidal

axis,

For the differential element shown

Thus, 3

3

121

121

dxyId

yhbxb

bhI

x

x

Solution For centroid of the element from the x axis

Moment of inertia of the element

Integrating

46

100

0

2/33

32

32

10107

4003

1

3

1

3

1

212

1~

2/~

mm

dxxdxydII

dxyy

ydxdxyydAIddI

yy

Axx

xx


Integration


Integration


Integration


IntegrationExample 10.3Determine the moment of inertia with

respect to the x axis of the circular area.

SolutionCase 1 Since dA = 2x dy

4

2

)2(

4

222

2

2

a

dyyay

dyxy

dAyI

a

a

A

Ax


Integration


Integration

SolutionCase 2 Centroid for the element

lies on the x axis Noting

dy = 0 For a rectangle,

3' 12

1bhI x


Integration


Integration

Solution

Integrating with respect to x

4

3

2

3

2

212

1

4

2/322

3

3

a

dxxaI

dxy

ydxdI

a

ax

x


Integration


Integration


Integration


IntegrationExample 10.4Determine the moment of inertia of

the shaded area about the x axis.

SolutionCase 1 Differential element parallel to x axis

chosen Intersects the curve at (x2,y) and (x1, y)

Area, dA = (x1 – x2)dy All elements lie at the same distance y

from the x axis 41

0

42/7

1

0

21

0 2122

0357.04

1

7

2myyI

dyyyydyxxydAyI

x

Ax


Integration


IntegrationView Free Body Diagram


Integration


IntegrationSolutionCase 2 Differential element

parallel to y axis chosen Intersects the curve at (x,

y2) and (x, y1) All elements do not lie at

the same distance from the x axis

Use parallel axis theorem to find moment of inertia about the x axis


Integration


IntegrationSolution

Integrating

41

0

74

1

0

63

6331

32

2

12112

312

2

3'

0357.021

1

12

13

13

1

3

1

212

1~

12

1

mxx

dxxxI

dxxxdxyy

yyydxyyyydxydAIddI

bhI

x

xx

x

10.5 Moments of Inertia for Composite Areas


A composite area consist of a series of connected simpler parts or shapes such as semicircles, rectangles and triangles

Provided the moment of inertia of each of these parts is known or can be determined about a common axis, moment of inertia of the composite area = algebraic sum of the moments of inertia of all its parts



Procedure for AnalysisComposite Parts Using a sketch, divide the area into its

composite parts and indicate the perpendicular distance from the centroid of each part to the reference axis

Parallel Axis Theorem Moment of inertia of each part is

determined about its centroidal axis, which is parallel to the reference axis



Procedure for AnalysisParallel Axis Theorem If the centroidal axis does not coincide with

the reference axis, the parallel axis theorem is used to determine the moment of inertia of the part about the reference axis

Summation Moment of inertia of the entire area about

the reference axis is determined by summing the results of its composite parts



Procedure for AnalysisSummation If the composite part has a hole, its

moment of inertia is found by subtracting the moment of inertia of the hole from the moment of inertia of the entire part including the hole



Example 10.5Compute the moment of inertia of the composite area about the x axis.



SolutionComposite Parts Composite area

obtained by subtracting the circle form the rectangle

Centroid of each area is located in the figure



SolutionParallel Axis

Theorem Circle

Rectangle

4623

2'

46224

2'

105.1127515010015010012

1

104.117525254

1

mm

AdII

mm

AdII

yxx

yxx



SolutionSummation For moment of inertia for the composite

area,

46

66

10101

105.112104.11

mm

I x



Example 10.6Determine the moments of inertia of the beam’s cross-sectional area about the x and y centroidal axes.



SolutionComposite PartsConsidered as 3

composite areas A, B, and D

Centroid of each area is located in the figure

View Free Body Diagram




Theorem Rectangle A

4923

2'

4923

2'

1090.125030010010030012

1

10425.120030010030010012

1

mm

AdII

mm

AdII

xyy

yxx




Theorem Rectangle B

493

2'

493

2'

1080.160010012

1

1005.010060012

1

mm

AdII

mm

AdII

xyy

yxx




Theorem Rectangle D

4923

2'

4923

2'

1090.125030010010030012

1

10425.120030010030010012

1

mm

AdII

mm

AdII

xyy

yxx



SolutionSummation For moment of inertia for the entire

cross-sectional area,

49

999

49

999

1060.5

1090.11080.11090.1

1090.2

10425.11005.010425.1

mm

I

mm

I

y

x

engineering mechanics: statics chapter 10: moments of inertia chapter 10: moments of inertia

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