EGR 231 Engineering Statics:
Spring 2014
Lecture 34: Moment of Inertia: Composite Body Method
Today:
Chap 9:
-- Introduction to Moment of Inertia
-- Method of Composite Bodies
-- Parallel Axis Theorem
Final Exam Topics
-- Trusses
-- Frames and Machines
-- Friction
-- Centroids
-- Moment of Inertia
y
2a
x
a
Homework Problem Assignment 37:
Problems 9.36
Determine the moment of inertia of the shaded
area with respect
to the x and y axis
4.8 in
4.8 in
O
2.3 in
Problem 9.47
Determine the polar moment
of inertia of the area with respect to
a) point O
b) the centroid of the area.
4 in
Problem 9:50:
Two channels and two plates are used to form
the column section shown. Determine the
moments of inertia and the radii of gyration
of the combined section with respect to
the centroidal axes shown.
Problem 9.56:
A channel and an angle are welded to
an a x 0.75 in steel plate. Knowing that
the centroidal y axis is located as shown,
determine
a) the width a
b) the moments of inertia with respect
to the centroidal x and y axes.
a
O
6 in
y
6 in
C 250 x 30
x
200 mm
10 mm
y
375 mm
L 6 x 6 x 3/4
C6x10.5
x
C
0.75
a
Centroid:
When we worked to calculate the centroid of an area, one of the steps was to
calculate the first moment of area.
Centroid:
A   dA
x
 x dA
y
A
 y dA
z
A
 z dA
A
where
 x dA
 y dA
 z dA
are the 1st moment of area. The first moment of area will show up again in
the study of strengths of material when shear stresses in beams are to be
calculated.
Another geometric calculation you will be expected to work with will be the
2nd moment of area. This is found in a way very similar to the calculation
of the first moment of area, except that the offset distance will be squared.
x
2
dA
y
2
dA
z
2
dA
The second moment of area shows up in the theory that determine the
bending stresses of beam and also in buckling theory, both of which are
studied in strength of materials. Since bending stresses will be the dominant
factor in determining beam and column strength, you will need to be able to
find values of the 2nd moment of area for cross sectional areas to analyze
beams and columns.
Over the years, engineers have traditionally called this 2nd of moment of area
by the term, Moment of Inertia such that.
Iy   x 2 dA
I x   y 2 dA
where Iy is the moment of inertia with respect to the y-axis and
Ix is the moment of inertia with respect to the x-axis.
Example 1: How inertia is found by integration.
Find the moment of inertia of the
y
area about the x centroidal axis.
Solution:
h/2
xc
Find:
h/2
xc
I x   y 2 dA
where
b
y
dA  b dy
Therefore:
Ix  
h/2
 h/2
h/2
 b
 h/2
dy
y
2
y (b dy ) v
h/2
y 2 dy
-h/2
y h/2
1 
 b y3 
3  y  h/2
b
3
1 h 1  h
 b   b  
3 2 3  2
3

1
1
1
bh3 
bh3 
bh3
3 8
3 8
12
Therefore the moment of inertia of a rectangular area about the centroidal
axis is given by
Ix 
1
bh3
12
If the inertia had been found through a y axis through the centroid it would
be
Iy 
1 3
bh
12
y
yc
Also notice that if the inertia is calculated about axis
at the end of the rectangle you’d get
Ix 
1 3
bh
3
and
Iy 
1 3
bh
3
In other words, the inertia depends upon where the axis
is located.
xc
x
Polar Moment of Inertia:
The polar moment of inertia is the
inertia about the axis perpendicular to the
x and y axis and through the origin.

JO   r 2 dA
y
dA
r

  x 2  y 2 dA
  x 2 dA
JO  Iy

 Ix
O
y
2
y
x
x
dA
Notice that the polar moment of inertia for this Area is simply the sum of the
Ix and Iy inertias of planar bodies.
Radii of Gyration:
If all the area of the body were concentrated at the same distance from the
axis and it still had the same inertia, this distance is defined as the radius of
gyration.
Ix  kx2 A
Iy  ky 2 A
y
JO  kO2 A
y
dA
r
O
A
y
x
x

k
O
x
Centroid by Composite Body Method:
The previous example showed how integration is used to define the moment
of inertia of an area. However, for most problems, (80% to 90% of the
time) you will not need to use integration to find moment of inertia. Instead
you will look up the moment inertia from a table of common shapes, such as
the table inside the back cover of your text.
A moment of
inertia table usually
gives the moment
of inertia through
the centroid of
a shape.
This information
combined with
the parallel
axis theorem
allows you to
find the moment
of inertia about
any axis at a known
distance away from
the centroidal axis.
A
Parallel Axis Theorem.
While the table gives the centroidal
moment of inertia of many shapes,
other times you will need to find the
moment of inertia about an axis which
does not pass through the centroid.
x
dy
The Parallel Axis Theorem may be used
to find this moment of inertia
x
To find the moment of inertia about any
other axis parallel to the centroidal axis you can use
I x  I x  A dy 2
where
I x= inertia about the centroidal axis
Ix = inertia about a non-centroidal axis
A = Area
dy = perpendicular distance of non-centroidal axis to centroidal axis.
Notice that inertia of a shape increases as you get farther away from the
centroidal axis. Therefore, expect the minimum inertia of a body to occur
about an axis passing through the area’s centroid.
-----------------------------------------------------------------------------------------Example 1: Find the moment of inertia about the x'-x' axis for the circular
area.
R=3 in
Use:
I x '  I x  A dy 2
where
I x x 
x
1
1
81
πR 4  π(3in) 4 
π=63.62in 2
4
4
4
πD 2 π(3in )2 9
A

 π  7.069in 2
4
4
4
dy = 5 in
Therefore:
I x ' x '  I x x  A d y 2
 63.62  7.069(5)2
 63.62  706.86  770.48in 4
x'
O
x
d=5 in
x'
Important Note: The parallel axis theorem must be applied from the
centroidal axis. Applying the parallel axis theorem with respect to any other
axis will result in an incorrect inertial value.
Example 2:
From the inertia table you find that the moment of inertia for a rectangular
area is given about both the centroidal axis and an axis along one end of the
rectangle. Can either one be used to find the inertia about another axis a
distance away from the rectangle?
b
xcentroid
b
h
h
xend
a + h/2
a
x'
Case 1
Case 2
Case 1: Correct method
Case 2: Incorrect method
I x '  I xend  A d y 2
1
 bh 3  (bh )(a )2
3
1
 bh 3  bha 2
3
I x '  I x  A dy 2
1
1
bh 3  (bh )(a  h )2
12
2
1

bh 3  (bh )(a 2  ah  h 2 )
12
1

bh 3  bha 2  bah 2  bh 3
12
13 3
bh  bha 2  bah 2
Note that
12

≠
1 3
bh  bha 2
3
Conclusion: Don't apply parallel axis theorem from any axis other than a
centroidal one.
Consider the formula for the half circle. Notice that the moment of inertia is
not given about the centroid.
Ix 
1
 R4
8
Since Ix is not the centroidal
axis you cannot use it for offsets of the
parallel axis theorem unless you first
determine the centroidal axis
inertia from it.
The centroidal inertia may be found from:
Ix  I  A d 2
so
1
1
4R
 R4  (  R2 )( )2
8
2
3
1
8
1
8

 4
4
I   R4 
R4    
 R  0.1976R
8
9
9 
8
I  Ix  A d 2 
4R
3
centroidal axis
x
Example 3: Find the Ix:
A1 
 R2
2
A2  bh
 8
I1   
 8 9
I2 
 4
R

1
bh3
12
d1
ii IIbar
bar
11 0.56
22 16
AA
3.53
12
dd
4.63
2
Ad2 2
Ad
IIi i
75.80 76.36
48
64
4 in
d2
x
Ixtotal
= = 140.36 in4
3 in
Example 2: Find the Iy:
A1 
 R2
A1 
2
A2 
I1 
A2  bh
I2 
i i Ibar
Ibar
11 1.99
22 9
AA
3.53
12
d1
I 
1 4
R
8
I2 
d2
1 3
bh
12
dd
1.5
1.5
2
Ad
Ad2
7.94
27
Ii Ii
9.93
36
Iy == 45.93 in4
Itotal
4 in
x
3 in
Example 4: Find Ix and Iy of the body shown:
Assume units are in mm.
y
40
30
180
30
x
120
i
Ibar_x
A
dy
Ady2
Ix
i
Ibar_y
A
dx
Adx2
Iy
y
Example 4: Find Ix and Iy of the body shown:
Assume units are in mm.
40
Solution:
Area 1: Large Positive Rectangle
Area 2: Small Negative rectangle
30
180
Moment of inertia about the x axis:
1
b1h13
12
1

b2h23
12
A1  b1h1
I 1x 
y 1  h1 / 2
A2  b2h2
I 2x
y 2  h2 / 2

 
I x  I 1x  A1d1y 2  I 2x  A2d 2y 2
30
x
120

where b1 = 120 h1 = 180 d1y = 90
b2 = 80 h2 = 120 d2y= 90
i
1
2
Ibar_x
A
6
58.32x10 21600
-11.52x106 -9600
Ady2
174.96x106
-77.76x106
dy
90
90
Ix
233.28x106
-89.28x106
144.00x106
total
Moment of inertia about the y axis:
1 3
b1 h1
x 1  b1 / 2
12
1 3

b2 h2
x 2  b2 / 2
12
 I 2y  A2d 2x 2
A1  b1h1
I 1y 
A2  b2h2
I 2y

I y  I 1y  A1d1x 2
i
1
2
 

where b1 = 120 h1 = 180 d1x = 60
b2 = 80 h2 = 120 d2x = 80
Ibar_y
A
dx
Adx2
Iy
6
6
25.92x10
21600
60
77.76x10 103.68x106
-5.12x106
-9600
80
-61.44x106 -66.56x106
37.12x106
total
so:
I x  144000000mm 4
I y  37120000mm 4
Moment Inertias of Common Structural Elements:
y
Start setting up HW Problem 9:50:
Two channels and two plates are used to form
the column section shown. Determine the
moments of inertia and the radii of gyration
of the combined section with respect to
the centroidal axes shown.
C 250 x 30
x
200 mm
Areas:
Area 1:
Area 2:
Area 3:
Area 4:
upper rectangle 375 x 10
lower rectangle 375 x 10
left C channel 250 x 30
right C-channel 250 x 30
Centroidal Inertia of Rectangle:
I 
i
1 3
bh
12
Ibar_x
[mm4]
10 mm
375 mm
Centroidal Inertia of C-channel:
IC-Chan read from table
A
dy
[mm2]
[mm]
A
dx
[mm2]
[mm]
Ady2
[mm3]
Ix
[mm4]
1
2
3
4
Total
i
1
2
3
4
Total
Ibar_y
[mm4]
Adx2
[mm3]
Iy
[mm4]
y
Start setting up HW Problem 9:50:
Two channels and two plates are used to form
the column section shown. Determine the
moments of inertia and the radii of gyration
of the combined section with respect to
the centroidal axes shown.
C 250 x 30
x
200 mm
Areas:
Area 1:
Area 2:
Area 3:
Area 4:
upper rectangle 375 x 10
lower rectangle 375 x 10
left C channel 250 x 30
right C-channel 250 x 30
Centroidal Inertia of Rectangle:
I 
i
1
2
3
4
Total
2
Centroidal Inertia of C-channel:
IC-Chan read from table
Ibar_x
[mm4]
i
1
375 mm
1 3
bh
12
1
(375)(10)3  31250
12
1
(375)(10)3  31250
12
32.6 x106
32.6 x106
A
dy
[mm2]
[mm]
Ady2
[mm3]
Ix
[mm4]
3750
132
6.534 x106
65.37125 x106
3570
-132
6.534 x106
65.37125 x106
3780
3780
0
0
Ibar_y
[mm4]
1
(375)3 (10)  43.945 106
12
1
(375)3 (10)  43.945 106
12
3
1.14 x106
4
1.14 x106
Total
10 mm
32.6 x106
32.6 x106
195.94 x106
0
0
[mm]
Adx2
[mm3]
[mm4]
0
0
43.945 x106
A
dx
[mm2]
3570
3570
0
0
3780
3780
-115.3
115.3
50.25x106
50.25x106
Iy
43.945 x106
51.39 x106
50.25x106
190.67x106