1

CHAPTER 9: Moments of Inertia! Moment of Inertia of Areas

! Second Moment, or Moment of Inertia, of anArea

! Parallel-Axis Theorem! Radius of Gyration of an Area! Determination of the Moment of Inertia of an

Area by Integration! Moments of Inertia of Composite Areas! Polar Moment of Inertia

2

9.1 Moment of Inertia: Definition

x

y

y

xdA=(dx)(dy)

∫=A

x dAyI 2)(

O

∫=A

y dAxI 2)(

3

x

y

x´= Centroidal axis

y´ = Centroidal axis

CG

dy

y´

dx x´dA

∫ +=A

yx dAdyI 2)'(

∫ ++=A

yy dAddyy ])())('(2)'[( 22

∫∫∫ ++=A

yA

yA

dAddAdydAy 22 )())('(2)'(

∫∫ ++=A

yA

yx dAddAydI 2'2

0, y´ = 0

AdII yxx20++=

AdII xyy20++=

9.2 Parallel-Axis Theorem of an Area

2AdJJ CO +=

O

4

y

x

kx

O

A

AJk

AI

kAIk O

Oy

yx

x ===

9.3 Radius of Gyration of an Area

2

The radius of gyration of an areaA with respect to the x axis isdefined as the distance kx, whereIx = kx A. With similar definitions forthe radii of gyration of A withrespect to the y axis and withrespect to O, we have

5

The rectangular moments of inertia Ixand Iy of an area are defined as

These computations are reduced to single integrations by choosing dA to be a thinstrip parallel to one of the coordinate axes. The result is

x

y

y

dx

x∫∫ == dAxIdAyI yx

22

dxyxdIdxydI yx23

31

==

9.4 Determination of the Moment of Inertia of anArea by Integration

6

x´

y´

b/2

h/2

� Moment of Inertia of a Rectangular Area.

x

y

b

hy

dy

∫=A

x dAyI 2

∫=h

bdyy0

2 )(

hby

0

3

3)(

=

ydy

∫==A

xx dAyII 2'

∫=h

dyby0

2 )2

(4

2/

0

3

3)

2(4

hyb

=

3

3bh=

12

3bh=

dA = bdydA = (b/2)dy

7

x´

y´

b/2

h/2

x

y b

h

∫=A

y dAxI 2

∫=b

hdxx0

2 )(

bhx

0

3

3)(

=

∫==A

yy dAxII 2'

∫=h

dxhx0

2 )2

(4

2/

0

3

3)

2(4

bxh

=

3

3hb=

12

3hb=

x dx

xdx

dA = hdxdA = (h/2)dx

8

x

y

b

h/2

h/212

3bhIx =

3

3bhIx =

2AdII xx +=

23

)2

)((12

hbhbh+=

412

33 bhbh+=

3

3bhIx =

9

dIx = y2 dA dA = l dy

Using similar triangles, we have

dyh

yhbdAh

yhblh

yhbl −

=−

=−

=

Integrating dIx from y = 0 to y = h, we obtain

∫∫

∫−=

−=

=hh

x

dyyhyhbdy

hyhby

dAyI

0

32

0

2

2

)(

12]

43[

3

0

43 bhyyhhb h =−=

� Moment of Inertia of a Triangular Area.

2AdII xx +=

36)

3)(

2(

12

32

3

2

bhhbhbhAdII xx

=−=

−=

y

x

h

b/2

h-y

b/2

dy

yl

10

Example 9.1

Determine the moment of inertia of the shaded area shown with respect toeach of the coordinate axes.

x

y

a

y = kx2

b

11

x

y

a

y = kx2

b

� Moment of Inertia Ix.

dydA = (a-x)dy

2kxy =2kab =

2abk =

2/12/1

22 y

baxorx

aby ==

Substituting x = a and y=b

∫=A

x dAyI 2

∫ −=b

dyxay0

2 )(

∫ −=b

dyybaay

0

2/12/1

2 )(

dyybadyya

bb

∫∫ −=0

2/52/1

0

2

bb

ybaay

0

2/72/1

0

3

)72(

3−=

)72(

32/7

2/1

3

bbaab

−=

72

3

33 abab−=

21

3ab=

12

x

y

a

y = kx2

b

� Moment of Inertia Iy.

22 x

aby = ∫=

Ay dAxI 2

∫=a

ydxx0

2

dx

dA = ydx

∫=a

dxxabx

0

22

2 )(

∫=a

dxxab

0

42

ax

ab

0

5

2 )5

)((=

)5

)((5

2

aab

=

5

3ba=

13

Example 9.2

Determine the moment of inertia of the shaded area shown with respect toeach of the coordinate axes.

x

y

y2 = x2

y1 = x

(a,b)

14

x

y

y2 = x2

y1 = x

(a,b)

dy

dA = (x2 - x1)dy

∫=A

x dAyI 2

∫ −=b

dyxxy0

2 )( 12

∫ −=b

dyyyy0

2/12 )(

∫∫ −=bb

dyydyy0

3

0

2/5 )()(

bb yy0

4

0

2/7

472

−=

� Moment of Inertia Ix.

472 4

2/7 bb −=

15

x

y

y2 = x2

y1 = x

(a,b)

� Moment of Inertia Iy.

∫=A

y dAxI 2

∫ −=a

dxyyx0

12 )( 2

dx

dA = (y1 - y2)dx ∫ −=a

dxxxx0

22 )(

∫∫ −=aa

dxxdxx0

4

0

3 )()(

aaxx

0

5

0

4

54−=

54

54 aa−=

16

The parallel-axis theorem is used very effectively to compute the moment ofinertia of a composite area with respect to a given axis.

d

c

ocentroid are related to the distance d between points C and O by the relationship

2AdJJ CO +=

9.5 Moment of Inertia of Composite Areas

A similar theorem can be used withthe polar moment of inertia. Thepolar moment of inertiaJO of an area about O and the polarmoment of inertia JC of the areaabout its

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Example 9.3

Compute the moment of inertia of the composite area shown.

75 mm

75 mm

100 mm

25 mm

x

18

SOLUTION

75 mm

75 mm

100 mm

x

25 mm

= (dy)Cir

Ciryxctx AdIbhI )()3

( 2Re

3

+−=

Circt ])75)(25()25(41[])150)(100(

31[ 224

Re3 ×+−= ππ

= 101x106 mm4

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Example 9.4

Determine the moments of inertia of the beam�s cross-sectional areashown about the x and y centroidal axes.

400

Dimension in mm

100

400

100 600

100

x

y

C

20

400

Dimension in mm

100

400

100 600

100

x

y

C

SOLUTION

AdyA

BdyD

D

CyxByxAyxx AdIAdIAdII )()()( 222 +++++=

])200)(300100()300)(100(121[

]0)100)(600(121[])200)(300100()300)(100(

121[

23

323

×++

++×+=

= 2.9x109 mm4

0

21

400

Dimension in mm

100

400

100 600

100

x

y

C

CxyBxyAxyy AdIAdIAdII )()()( 222 +++++=

C

BA

])250)(300100()100)(300(121[

]0)600)(100(121[])250)(300100()100)(300(

121[

23

323

×++

++×+=

= 5.6x109 mm4

AdxA

dxDD

0

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Example 9.5 (Problem 9.31,33)

Determine the moments of inertia and the radius of gyration of theshaded area with respect to the x and y axes.

6 mm

O x

y

24 mm

24 mm

6 mm

12 mm12 mm

24 mm 24 mm

8 mm

23

6 mm

O x

y

24 mm

24 mm

6 mm

12 mm12 mm

24 mm 24 mm

8 mm

SOLUTION

A

dyA

B

Ix = 390x103 mm4

mmAIk x

x 9.21)]648()488()624[(

10390 3

=×+×+×

×==

CyxByxAyxx AdIAdIAdII )()()( 222 +++++=

Iy = 64.3x103 mm4

CBA ])48)(6(121[])8)(48(

121[])24)(6(

121[ 333 ++=

0 00

CxyBxyAxyy AdIAdIAdII )()()( 222 +++++=

mmAI

k yy 87.8

)]648()488()624[(103.64 3

=×+×+×

×==

0

C

B

A

])27)(648()6)(48(121[

]0)48)(8(121[

])27)(624()6)(24(121[

23

3

23

×++

++

×+=

C

dyC

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Example 9.6 (Problem 9.32,34)

Determine the moments of inertia and the radius of gyration of theshaded area with respect to the x and y axes.

2 m

O1 m1 m1 m1 m

0.5 m0.5 m 2 m 2 m

0.5 m0.5 m

x

y

25

2 m

O1 m1 m1 m1 m

0.5 m0.5 m 2 m 2 m

0.5 m0.5 m

x

y

A

BdyB

CdyC

Ix = 46 m4

mAIk x

x 599.1)]14()24()65[(

46=

×−×−×==

142

242

652 )()()( ××× +−+−+= CyxByxAyxx AdIAdIAdII

0

C

BA

])5.1)(14()1)(4(121[

])2)(42()2)(4(121[]0)6)(5(

121[

23

233

×+−

×+−+=

Iy = 46.5 m4

CBA ])4)(1(121[])4)(2(

121[])5)(6(

121[ 333 −−=

0 00

CxyBxyAxyy AdIAdIAdII )()()( 222 +−+−+=

mAI

k yy 607.1

)]14()24()65[(5.46

=×−×−×

==

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Example 9.7

Determine the moments of inertia and the radius of gyration of theshaded area with respect to the x and y axes and at the centroidal axes.

1 cm 1 cm

5 cm

1 cm

5 cm

x

y

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1 cm 1 cm

5 cm

1 cm

5 cm

x

y

CG

Y

∑∑ = AyAY

cm

Y

5.2)15(3

)51)(5.0()]15)(5.3[(2

=×

×+×=

4

2

2

25.51)5.2)(15(145

cm

AdII yxx

=

−=

−=

� Moments of inertia about centroid

4

23

23

25.51

])2)(15()1)(5(121[(

])1)(15()5)(1(121[(2

cm

I

OR

x

=

×++

×+=

cmAIkk x

yx 848.115

25.51====

4

323

145

)1)(5(31])5.3)(15()5)(1(

121[(2

cm

Ix

=

+×+=

� Moments of inertia about x axis4

323

25.51

)5)(1(121])2)(15()1)(5(

121[(2

cm

II yy

=

+×+==

0.5

3.5

2

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Example 9.8

The strength of a W360 x 57 rolled-steel beam is increased by attaching a229 mm x 19 mm plate to its upper flange as shown. Determine themoment of inertia and the radius of gyration of the composite section withrespect to an axis which is parallel to the plate and passes through thecentroid C of the section.

C

229 mm

19 mm

358 mm

172 mm

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229 mm

172 mm

SOLUTION19 mm

358 mm xO

C x´Y

188.5 mm

The wide-flange shape of W360 x 57found by referring to Fig. 9.13A = 7230 mm2 42.160 mmI x =

AyAY Σ=Σ

Aplate = (229)(19) = 4351 mm2

)7230)(0()4351)(5.188()72304351( +=+Y

mmY 8.70=

� Centroid

� Moment of Inertia

flangewidexplatexx III −+= )()( '''

flangewidexplatex YAIAdI −+++= )()( 2'

2'

[ ]46

26

23

108.256)8.70)(7230(102.160

)8.705.188)(4351()19)(229(121

mm×=

+×+

−+=

� Radius of Gyration

)72304351(108.256 6

'2' +

×==

AIk x

x

mmkx 149' =

46' 10257 mmI x ×=

d

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The polar moment of inertia ofan area A with respect to the poleO is defined as

The distance from O to the element of area dA is r. Observing that r 2 =x 2 + y 2 , weestablished the relation

y

xx

yr

A

dA

O ∫= dArJO2

yxO IIJ +=

9.6 Polar Moment of Inertia

31

Example 9.9

(a) Determine the centroidal polar moment of inertia of a circular area bydirect integration. (b) Using the result of part a, determine the moment ofinertia of a circular area with respect to a diameter.

y

x

r

O

32

udu

O

y

x

r

SOLUTIONa. Polar Moment of Inertia.

duudAdAudJO π22 ==

∫∫ ∫ ===rr

OO duuduuudJJ0

3

0

2 2)2( ππ

4

2rJO

π=

b. Moment of Inertia with Respect to a Diameter.

xyxO IIIJ 2=+=

xIr 22

4 =π

4

4rII xdiameter

π==