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VECTOR MECHANICS FOR ENGINEERS:

STATICS

Eighth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

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9 Distributed Forces:

Moments of Inertia

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9 - 2

Moment of Inertia of a Mass

• Angular acceleration about the axis AA’ of the

small mass m due to the application of a

couple is proportional to r2 m.

r2 m = moment of inertia of the

mass m with respect to the

axis AA’

• For a body of mass m the resistance to rotation

about the axis AA’ is

inertiaofmomentmassdmr

mrmrmrI

2

23

22

21

• The radius of gyration for a concentrated mass

with equivalent mass moment of inertia is

m

IkmkI 2

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9 - 3

Moment of Inertia of a Mass

• Moment of inertia with respect to the y coordinate

axis is

dmxzdmrI y222

• Similarly, for the moment of inertia with respect to

the x and z axes,

dmyxI

dmzyI

z

x

22

22

• In SI units, 22 mkgdmrI

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I is the Inertia matrix which has a size of 3 x3 and is

symmetric

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

Eig

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

Eig

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

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•Since r is independent of inclination of

the coordinate axes and depends only on

the position of the origin. Therefore

sum of moment of inertia at a point in

space for a given body is an invariant

with respect to rotation of axes.

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Mirror Symmetry & mixed moments of Inertia

• Ixz of each half gives the contribution of same magnitude

but opposite sign.

• This conclusion is also true for Ixy .

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• Similarly Ixy = 0

• If two axes form a plane of symmetry for the mass

distribution of a body, the products of inertia having as an

index the coordinate that is normal to the plane of

symmetry will be zero.

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Body of Revolution

9 - 16

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Body of Revolution

• Let z axis coincide with the axis of symmetry.

This is true all possible xy axis formed by rotating

the z axis at O.

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Radius of gyration

• kx, ky and kz are radius of gyration.

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9 - 19

Parallel Axis Theorem

•For the

rectangular axes

with origin at O

and parallel

centroidal axes

x’y’z’,

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9 - 20

The coordinates of any point are (x,y,z)

xbar, ybar, zbar (the quantities with bar

on them give the coordinates of CM

from O

x’, y’, z’ are distances of any point

fromCM

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9 - 21

2 2

x

2 2

I y z dm

y y z z dm

x

2 2 2 2

I

y z dm 2y y dm 2z z dm y z dm

22 zymII xx

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9 - 22

x

2 2 2 2

I

y z dm 2y y dm 2z z dm y z dm

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9 - 23

y dm 0 z dm 0

Definition of CM

Ycm and Zcm are the coordinates of CM from origin. Here CM is

at the origin itself.

cmy dm y

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9 - 24

Parallel Axis Theorem

22 zymII xx

22

22

yxmII

xzmII

zz

yy

• Generalizing for any axis AA’ and a

parallel centroidal axis,

2mdII

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Parallel axis for product moments

9 - 25

'xy x yI I mx y

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9 - 26

Moments of Inertia of Thin Plates

• For a thin plate of uniform thickness t and homogeneous

material of density , the mass moment of inertia with

respect to axis AA’ contained in the plate is

areaAA

AA

It

dArtdmrI

,

22

• Similarly, for perpendicular axis BB’ which is also

contained in the plate,

areaBBBB ItI ,

• For the axis CC’ which is perpendicular to the plate,

BBAA

areaBBareaAAareaCCC

II

IItJtI ,,,

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9 - 27

t is mass per unit area

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9 - 28

Moments of Inertia of Thin Plates

• For the principal centroidal axes on a rectangular plate,

21213

121

, mabatItI areaAAAA

21213

121

, mbabtItI areaBBBB

22121

,, bamIII massBBmassAACC

• For centroidal axes on a circular plate,

2414

41

, mrrtItII areaAABBAA

221 mrIII BBAACC

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9 - 29

Moments of Inertia of a 3D Body by Integration

• Moment of inertia of a homogeneous body

is obtained from double or triple

integrations of the form

dVrI 2

• For bodies with two planes of symmetry,

the moment of inertia may be obtained

from a single integration by choosing thin

slabs perpendicular to the planes of

symmetry for dm.

• The moment of inertia with respect to a

particular axis for a composite body may

be obtained by adding the moments of

inertia with respect to the same axis of the

components.

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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Vector Mechanics for Engineers: Statics

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Part b

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Vector Mechanics for Engineers: Statics

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Part c

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9 - 35

Moments of Inertia of Common Geometric Shapes

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Moments of Inertia of Common Geometric Shapes

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Sample Problem 9.12

9 - 37

Determine the moments of inertia of

the steel forging with respect to the

xyz coordinate axes, knowing that

the density of steel is 7850 kg/m3.

SOLUTION:

• With the forging divided into a prism and

two cylinders, compute the mass and

moments of inertia of each component

with respect to the xyz axes using the

parallel axis theorem.

• Add the moments of inertia from the

components to determine the total moments

of inertia for the forging.

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Sample Problem 9.12

9 - 38

Determine the moments of inertia of the steel

forging with respect to the xyz coordinate

axes, knowing that the density of steel is 7850

kg/m3.

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Sample Problem 9.12

9 - 39

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Sample Problem 9.12

9 - 40

•With the forging divided into a prism

and two cylinders, compute the mass

and moments of inertia of each

component with respect to the xyz axes

using the parallel axis theorem.

•Add the moments of inertia from the

components to determine the total

moments of inertia for the forging.

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Sample Problem 9.12

9 - 41

kg161

)kg/m7850)(m1047301(

m104731

m)0750(m)0250(

:cylindereach

334-

34-

2

.

.m

.

..V

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Sample Problem 9.12

9 - 42

23

2

1000502

100025

21

22

21

mkg102363

161m161

.

..

ymmaIx

cylinders :mm50,mm5.62,mm75,mm25 yxLa

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Sample Problem 9.12

9 - 43

23

2

10005622

1000752

100025

121

222

121

mkg102565

1613161

3

.

..

xmLamI

.

y

cylinders

:mm50,mm5.62,mm75,mm25 yxLa

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Sample Problem 9.12

9 - 44

23

2

1000502

10005622

1000752

100025

121

2222

121

kg.m101568

1613161

3

.

..

yxmLamI

.

z

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Sample Problem 9.12

9 - 45

prism (a = 50 mm., b =

150 mm, c = 50 mm):

23

2

1000502

1000150

12122

121

mkg 10125.6

kg94.2cbmII zx

23

2

1000502

100050

12122

121

m kg102251

kg942

.

.acmI y

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Sample Problem 9.12

9 - 46

• Add the moments of inertia from the components

to determine the total moments of inertia.

33 1026332101256 ..Ix

23 mkg106512.Ix

33 1025652102251 ..I y

23 mkg107411.I y

33 1015682101256 ..Iz

23 mkg104422.Iz

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Moment of Inertia With Respect to an Arbitrary Axis

kk is the arbitrary axis

r is the vector from

origin to point (x,y,z)

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Transformation of product MI

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

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Find Iz’z’ and Ix’z’

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© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: Statics

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