6161103 10.4 moments of inertia for an area by integration

10.4 Moments of Inertia for an Area by Integration10.4 Moments of Inertia

for an Area by 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 � 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


for an Area by 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 � 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


for an Area by IntegrationProcedure for AnalysisCase 1� Length of element orientated parallel to the axis� Occurs when the rectangular element is used to

determine Iy for the areadetermine 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


for an Area by IntegrationProcedure for AnalysisCase 2� Length of element orientated perpendicular to

the axis� All parts of the element will not lie at the same � 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



Example 10.1Determine the moment of inertia for the rectangular area with respect to (a) the

centroidal 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.

Solution

Part (a)

� Differential element chosen, distance y’ from x’ axis



x’ axis

� Since dA = b dy’

3

2/

2/

22

12

1

''

bh

dyydAyIh

hAx

=

== ∫∫ −

Solution

Part (b)

� Moment of inertia about an axis passing through the base of the rectangle obtained



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

32

3

2

3

1

212

1bh

hbhbh

AdII xxb

=

+=

+=

Solution

Part (c)

� For polar moment of inertia about point C

1



)(12

1

12

1

22

'

3'

bhbh

IIJ

hbI

yxC

y

+=

+=

=



Example 10.2

Determine the moment of

inertia of the shaded area inertia of the shaded area

about the x axis



Solution

� A differential element of area that is parallel to the x axis is chosen for integration

� Since element has thickness dy and � 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

2

2

)100( dyxy

dAyI

A

Ax

−=

=

∫∫



46

200

0

200

0

42

2200

0

2

)10(107

400

1100

400100

mm

dyydyy

dyy

y

A

=

−=

−=

∫ ∫

∫

∫



Solution

� A differential element parallel to the y axis is chosen for integrationintegration

� Intersects the curve at arbitrary point (x, y)

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

10.4 Moments of Inertia for an Area by

Integration

10.4 Moments of Inertia for an Area by

IntegrationSolution

� Parallel axis theorem used to determine moment of inertia of the element

� For moment of inertia about its centroidal axis,� 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

2/~ yy =



� Moment of inertia of the element

� Integrating

( )

( ) 46

100

0

2/33

32

32

10107

4003

1

3

1

3

1

212

1~

mm

dxxdxydII

dxyy

ydxdxyydAIddI

Axx

xx

=

===

=

+=+=

∫∫∫



Example 10.3

Determine the moment of inertia with respect

to the x axis of the circular area.to the x axis of the circular area.

Solution

Case 1

� Since dA = 2x dy2= ∫



( )

4

2

)2(

4

222

2

2

a

dyyay

dyxy

dAyI

a

a

A

Ax

π=

−=

=

=

∫

∫∫

−

Solution

Case 2

� Centroid for the element lies on the x axis



on the x axis

� Noting

dy = 0

� For a rectangle,

3' 12

1bhI x =

Solution

Integrating with respect to x

( )

3

2

212

1

3

3

dxy

ydxdI x

=

=



Integrating with respect to x

( )

4

3

2

4

2/322

a

dxxaIa

ax

π=

−= ∫−



Example 10.4

Determine the moment of inertia of the

shaded area about the x

axis.

Solution

Case 1

� Differential element parallel to x axis chosen

Intersects the curve at (x ,y) and (x , y)



� 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

=−=

−=−== ∫∫∫


for an Area by IntegrationSolutionCase 2� Differential element parallel

to y axis chosen� Intersects the curve at (x, y ) � 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

10.4 Moments of Inertia for an Area by Integration

10.4 Moments of Inertia for an Area by Integration

Solution

� Integrating

( ) ( )2

3

3'

1~

12

1

yy

bhI x

−

=

( ) ( )

( ) ( )

( )41

0

74

1

0

63

6331

32

12112

312

2

0357.021

1

12

13

13

1

3

1

212

1~

mxx

dxxxI

dxxxdxyy

yyydxyyyydxydAIddI

x

xx

=−=

−=

−=−=

−+−+−=+=

∫

6161103 10.4 moments of inertia for an area by integration

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