5.4 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 is a quantity
that relates the normal stress σ or force per
unit area, acting on the transverse crosssection of an elastic beam, to applied
external moment M, that causes bending of
the beam.
1
K.W.LIEW
5.4 Moments of Inertia
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
 The magnitude of the force acting
on the area element dA,
dF = σ dA = kz dA
K.W.LIEW
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1
5.4 Moments of Inertia
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 = dFz = kz2 dA
 The resulting moment of the entire stress
distribution = applied moment M,
M  k  z 2 dA

Integral represent the moment of inertia of area
about the y axis.
3
K.W.LIEW
5.4 Moments of Inertia
Moment of Inertia

Consider area A which lies in the x-y plane.

By definition, moments of inertia for the
differential planar area dA about the x and y
axes,
dI x  y 2dA dI y  x 2 dA

For entire area, moments of
inertia are given by
I x   y 2dA
A
I y   x 2dA
A
K.W.LIEW
4
2
5.4 Moments of Inertia
Moment of Inertia

Formulate the second moment of dA about
the pole O or z axis

This is known as the polar axis
dJ O  r 2 dA

where r is perpendicular from the pole (z
axis) to the element dA
Polar moment of inertia for entire area,
J O   r 2 dA  I x  I y
A
5
K.W.LIEW
5.4 Moments of Inertia
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, e.g. m4, mm4
K.W.LIEW
6
3
5.5 Parallel-Axis Theorem for an
Area



If the moment of inertia for an area known about an
axis passing through its centroid, it is convenient to
determine the moment of inertia of area about a
corresponding parallel axis using the parallel-axis
theorem
Consider finding moment of inertia
of the shaded area about x axis
A differential element dA is
located at an arbitrary distance
y’ from the centroidal x’ axis
7
K.W.LIEW
5.5 Parallel-Axis Theorem for an
Area


The fixed distance between the parallel x and x’ axes is
defined as dy
For moment of inertia of dA about x axis
dI x  y ' d y  dA
2

For entire area
I x    y ' d y  dA
2
A
  y '2 dA  2d y  y ' dA  d y2  dA
A

K.W.LIEW
A
A
First integral represent the moment of inertia of the area
about the centroidal axis
8
4
5.5 Parallel-Axis Theorem for an
Area

Second integral = 0 since x’ passes through the
area’s centroid C
 y' dA  y  dA  0;


y0
Third integral represents the total area A
I x  I x  Ad y2
Similarly
I y  I y  Ad x2

For polar moment of inertia about an axis
perpendicular to the x-y plane and passing through
pole O (z axis)
J O  J C  Ad 2
9
K.W.LIEW
5.5 Parallel-Axis Theorem for an
Area
Moment of
inertia of an
area about
an axis
K.W.LIEW
Moment of inertia about a
parallel axis passing through the area’s
centroid
=
+
The product of the area and the square
of the perpendicular distance between
the axes
10
5
5.6 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
kx 

Ix
A
ky 
Iy
A
kz 
JO
A
Similar to finding moment of inertia of a
differential area about an axis
I x  k x2 A dI x  y 2 dA
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K.W.LIEW
5.7 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
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.
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6
5.7 Moments of Inertia for an
Area by Integration
Procedure 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
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K.W.LIEW
5.7 Moments of Inertia for an
Area by Integration
Procedure for Analysis
Case 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
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7
5.7 Moments of Inertia for an
Area by Integration
Procedure for Analysis
Case 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
15
K.W.LIEW
5.7 Moments of Inertia for an
Area by Integration
Example 5.7
Determine 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.
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8
5.7 Moments of Inertia for an
Area by Integration
Solution
Part (a)
 Differential element chosen, distance y’ from
x’ axis
 Since dA = b dy’
I x   y '2 dA  
A

h/2
h / 2
y '2 dy
1 3
bh
12
17
K.W.LIEW
5.7 Moments of Inertia for an
Area by Integration
Solution
Part (b)
 Moment of inertia about an axis passing
through the base of the rectangle obtained
by applying parallel axis theorem
I xb  I x  Ad 2
2
1
h 1
 bh3  bh   bh3
12
3
 2
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18
9
5.7 Moments of Inertia for an
Area by Integration
Solution
Part (c)
 For polar moment of inertia about point C
1 3
hb
12
JC  I x  I y'
I y' 

1
bh(h 2  b 2 )
12
19
K.W.LIEW
5.7 Moments of Inertia for an
Area by Integration
Example 5.8
Determine the moment of
inertia of the shaded area
about the x axis
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10
5.7 Moments of Inertia for an
Area by Integration
Solution
 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
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K.W.LIEW
5.7 Moments of Inertia for an
Area by Integration
Solution
I x   y 2 dA
A
  y 2 (100  x )dy
A

y2 
 dy
y 2 100 
0
400 

200
1 200 4
 100 y 2 dy 
y dy
0
400 0
 107(106 )mm 4

K.W.LIEW
200
22
11
5.7 Moments of Inertia for an
Area by Integration
Solution
 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
23
K.W.LIEW
5.7 Moments of Inertia for an
Area by Integration
Solution
 Parallel axis theorem used to determine moment
of inertia of the element
 For moment of inertia about its centroidal axis,
Ix 

1 3
bh
12
For the differential element shown
b  bx h  y

Thus,
dI x 
K.W.LIEW
1
dxy 3
12
24
12
5.7 Moments of Inertia for an
Area by Integration
Solution
 For centroid of the element from the x axis
~
y  y/2

Moment of inertia of the element
2
1
 y 1
dI x  dI x  dA~
y 2  dxy 3  ydx   y 3 dx
12
2 3

Integrating
100 1
1 3
y dx  
400 x 3 / 2 dx
A3
0
3
 107 106 mm 4
I x   dI x  
 
25
K.W.LIEW
5.7 Moments of Inertia for an
Area by Integration
Example 5.9
Determine the moment of inertia with respect
to the x axis of the circular area.
K.W.LIEW
26
13
5.7 Moments of Inertia for an
Area by Integration
Solution
Case 1
 Since dA = 2x dy
I x   y 2 dA
A
  y 2 (2 x)dy
A

a

  y 2 2 a 2  y 2 dy
a

a 4
4
27
K.W.LIEW
5.7 Moments of Inertia for an
Area by Integration
Solution
Case 2
 Centroid for the element lies
on the x axis
 Noting
dy = 0
 For a rectangle,
1
I x '  bh3
12
K.W.LIEW
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14
5.7 Moments of Inertia for
an Area by Integration
Solution
1
3
dx2 y 
12
dI x 

2 3
y dx
3
Integrating with respect to x
Ix  
a
a

2 2
a  x2
3


3/ 2
dx
a 4
4
29
K.W.LIEW
5.8 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
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30
15
5.8 Moments of Inertia for
Composite Areas
• The moment of inertia of a composite area A about a given axis is obtained by
adding the moments of inertia of the component areas A1, A2, A3, ... , with respect
to the same axis.
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K.W.LIEW
5.8 Moments of Inertia for
Composite Areas
Procedure for Analysis
Composite 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
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16
5.8 Moments of Inertia for
Composite Areas
Procedure for Analysis
Parallel 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
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K.W.LIEW
5.8 Moments of Inertia for
Composite Areas
Procedure for Analysis
Summation
 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
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17
5.8 Moments of Inertia for
Composite Areas
Example 5.10
Compute the moment of
inertia of the composite
area about the x axis.
35
K.W.LIEW
5.8 Moments of Inertia for
Composite Areas
Solution
Composite Parts
 Composite area obtained
by subtracting the circle
form the rectangle
 Centroid of each area is
located in the figure
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36
18
5.8 Moments of Inertia for
Composite Areas
Solution
Parallel Axis Theorem
 Circle
I x  I x '  Ad y2
1
4
2
2
  25   25 75  11.4 10 6 mm 4
4
 

Rectangle
I x  I x '  Ad y2

1
1001503  100150752  112.5 106 mm 4
12
 
37
K.W.LIEW
5.8 Moments of Inertia for
Composite Areas
Solution
Summation
 For moment of inertia for the composite
area,
I x  11.4106   112.5106 
 
 101 106 mm 4
K.W.LIEW
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19
5.8 Moments of Inertia for
Composite Areas
Example 5.11
Determine the moments
of inertia of the beam’s
cross-sectional area
about the x and y
centroidal axes.
39
K.W.LIEW
5.8 Moments of Inertia for
Composite Areas
View Free Body Diagram
Solution
Composite Parts
 Considered as 3 composite
areas A, B, and D
 Centroid of each area is
located in the figure
K.W.LIEW
40
20
5.8 Moments of Inertia for
Composite Areas
Solution
Parallel Axis Theorem
 Rectangle A
I x  I x '  Ad y2
1
100 300 3  100 3002002  1.425 109 mm 4
12
I y  I y '  Ad x2
 


1
300100 3  100 3002502  1.90 109 mm 4
12
 
41
K.W.LIEW
5.8 Moments of Inertia for
Composite Areas
Solution
Parallel Axis Theorem
 Rectangle B
I x  I x '  Ad y2
1
6001003  0.05 109 mm 4
12
I y  I y '  Ad x2


K.W.LIEW
 
1
1006003  1.80 109 mm 4
12
 
42
21
5.8 Moments of Inertia for
Composite Areas
Solution
Parallel Axis Theorem
 Rectangle D
I x  I x '  Ad y2
1
100 300 3  100 3002002  1.425 109 mm 4
12
I y  I y '  Ad x2
 


1
300100 3  100 3002502  1.90 109 mm 4
12
 
43
K.W.LIEW
5.8 Moments of Inertia for
Composite Areas
Solution
Summation
 For moment of inertia for the entire crosssectional area,
 
 
 
 2.9010 mm
I  1.9010   1.8010   1.9010 
 5.6010 mm
I x  1.425 109  0.05 109  1.425 109
9
4
9
9
9
y
9
K.W.LIEW
4
44
22