Republic of the Philippines
DON HONORIO VENTURA STATE UNIVERSITY
Villa de Bacolor, Pampanga
STATICS OF RIGID BODIES
A. Course Code / Title
:
Statics 213 / Statics of Rigid Bodies
B. Module Number
:
Module 17 – Moment of Inertia
C. Time Frame
:
D. Description:
This module looks into the definition of Moment of Inertia and the formulas relating to moment of
inertia.
E. Objectives:
At the end of this module, the learner should be able to:
1. Have a better understanding about moment of inertia.
2. Memorize the formulas relating to moment of inertia of each geometric shapes.
3. Develop techniques and be able to solve problems relating to moment of inertia of an area.
F. Contents:
 Moment of inertia is a measure of an object’s resistance to changes to its rotation. It is the
capacity of a cross-section to resist bending.
Y
Y
dA = (a –x) dy
dA = dx dy
x
dx
dy
dy
x
y
y
X
X
a
dIx = y2 dA
2
dIy = x dA
dIx = y2 dA
Y
dA = y dx
y
x
X
dy
dIy = x2 dA
*choose dA to be a thin strip parallel to one of the coordinate axes, so that all points of the strip
are at the same distance from the axis.
I x   y 2 dA
I y   x 2 dA
 Polar moment of inertia gives an object’s ability to resist torsion (i.e. twisting) about a given
axis due to an applied torque.
Y
Jo   r dA
dA
2
r 2  x2  y 2
r
x
Jo   ( x 2  y 2 )dA
O
y
X
A
Jo   x dA   y dA
2
2
Jo  Iy  Ix
 Radius of Gyration is the distance from the axis of rotation to a point where the total mass of the
body is supposed to be concentrated, so that the moment of inertia about the axis may remain the
same. Gyration is the distribution of the components of an object.
Consider an area which has a moment of inertia Ix with respect to x – axis.
Y
A
O
X
If this area is concentrated into a thin strip parallel to the x – axis and have same moment
of inertia with respect to x- axis, the strip should be placed at a distance k x from the x – axis, where
kx is defined by the relation
Y
A
𝐼𝑥 = 𝑘𝑥 2 𝐴
kx
X
O
𝐼𝑥
𝐴
𝑘𝑥 = √
Thus;
Y
𝐼𝑦
𝐴
𝑘𝑦 = √
ky
X
O
A
Problems:
1. A rectangle has a base b and height d. Determine the moment of inertia of the rectangle with respect
to its centroidal x-axis.
Required:
Moment of inertia about Xc, I X C or I X
Figure:
YC
dY
d
/2
y
d
d
XC
/2
b
b
/2
/2
b
Solution:
I X C   y 2 dA
I X C   y 2 bdy
d
2
d

2
I X C  b  y 2 dy
d
I XC
 y3  2
 b 
 3 d
2
3
b  d   d  


 
  
3  2   2  
3
I XC 
I XC
I XC
I XC
b d3 d3 
   
3 8
8
b  2d 3 
 

3 8 
bd 3

12
Note:
I XC
bd 3

12
Formula for Moment of Inertia of a rectangle about its centroidal x - axis
2. Determine the moment of inertia of a triangle (base, b and altitude, h) with respect to its base.
Required:
Moment of Inertia about X - axis, Ix
Figure:
dY
h
X
y
X
b
Given:
Base of the triangle lies on X - axis.
Solution:
𝐼𝑥 = ∫ 𝑦 2 𝑑𝐴
𝐼𝑥 = ∫ 𝑦 2 [ 𝑥 𝑑𝑦]
𝐼𝑥 = ∫ 𝑥𝑦 2 𝑑𝑦
→ 𝑒𝑞. 1
Solve for x in terms of y:
𝑥
𝑏
=
𝑥=
ℎ−𝑦
ℎ
𝑏(ℎ−𝑦)
→ 𝑒𝑞. 2
ℎ
Substitute eq. 2 in eq. 1:
𝐼𝑥 = ∫
𝐼𝑥 =
𝐼𝑥 =
𝐼𝑥 =
𝐼𝑥 =
𝐼𝑥 =
𝑏
ℎ
𝑏(ℎ−𝑦)
ℎ
ℎ
∫0 (ℎ𝑦 2 − 𝑦 3 )𝑑𝑦
𝑏 ℎ𝑦 3
ℎ
𝑏
ℎ
𝑏
ℎ
𝑦 2 𝑑𝑦
[
3
{[
[
𝑏ℎ3
12
−
ℎ (ℎ 3 )
3
𝑦4
4 0
−
4ℎ4 −3ℎ4
12
ℎ
]
ℎ4
4
] − [0]}
]
(The moment of inertia of a triangle about its base)
 Parallel-Axis Theorem for an Area (Transfer Formula)
The parallel-axis theorem can be used to find the moment of inertia of an area about any axis that
is parallel to an axis passing through the centroid and about which the moment of inertia is known.
Ix  Ix  Ad 2
Iy  Iy  Ad 2
 Moments of Inertia for Composite Areas
The moment of inertia of a composite area can be determined using the following procedure.

Procedure for analysis

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.

If the centroidal axis for each part does not coincide with the reference axis, the parallel2
axis theorem I  I  Ad , should be used to determine the moment of inertia of the part
about the reference axis. For the calculation of I , use the table given below.

The moment of inertia of the entire area about the reference axis is determined by summing
the results of its composite parts about this axis.

If a composite part has an empty region (hole), its moment of inertia is found by subtracting
the moment of this region from the moment of inertia of the entire part including the region.
 Moment of Inertia of Geometric Figures:

Rectangle:
Y
Yc
bh3
12
hb3
IY 
12
bh
J C  b 2  h 2 
12
IX 
C
h
Xc
X
b

Triangle:
bh3
IX 
36
h
Xc
C
h
/3
X
b

Circle:
Yc
r 4
I X  IY 
r
Xc
C

JC 
r
4
4
2
Semi-circle:
Yc
I X  0.1098r 4
IX  IY 
Xc
C
X
JO 
r
r
r 4
8
4
4
Quarter circle
Y
Yc
I X  I Y  0.05488r 4
I X  IY 
C
Xc
O
JO 
X
r
r
r 4
16
4
8
Ellipse:
Y
IX 
b
C
X
a
ab3
4
ba3
IY 
4
ab(a 2  b 2 )
JO 
4
Problems:
1. Determine the moment of inertia of the shaded area with respect to the x-axis.
Y
12m
m
12m
m
6mm
24mm
8mm
X
24mm
6mm
24mm
24mm
Required:
Ix
Solution:
a. Divide the shaded into parts. (rectangle 1, rectangle 2 and rectangle 3)
Shape
Area, A
Distance, d
A1  bh  24mm6mm 
Rectangle 1
A1  144mm 2
Rectangle 3
1
6mm 
2
d1  27mm
A2  8mm48mm 
Rectangle 2
d1  24mm 
d2  0
A2  384mm 2
(Coincides with x - axis)
A3  48mm6mm 
d 3  24mm 
A3  288mm 2
1
6mm 
2
d 3  27mm
Note:
Since we are solving moment of inertia about the given x - axis, all distances must be measured from
x - axis.
Y
12mm 12mm
1
6mm
2
d1 = 27mm
24mm
8mm
X
24mm
d3 = 27mm
3
24mm
24mm
6mm
b. Solve for Ix:
Let
Ix1 = Moment of inertia of rectangle 1 about x-axis
Ix2 = Moment of inertia of rectangle 2 about x-axis
Ix3 = Moment of inertia of rectangle 3 about x-axis
Ix  Ix1  Ix2  Ix3
Ix1  I x1  A1d1
2
bh3
2
Ix1 
 A1d1
12
3
24mm6mm 
2
Ix1 
 144mm 2 27mm 
12
Ix1  432mm 4  104976mm 4
I X 2  I X  Ad 2
Ix1  105408mm 4
I X2
bh3
 Ad 2
12
2
8mm48mm 
2

 384mm20
12
 73728mm 4
I X2 
I X2
I X 3  I X  Ad 2
bh3
 Ad 2
12
3
48mm6mm 
2

 288mm27mm 
12
 210816mm 4
I X3 
I X3
I X3
I X  I X1  I X 2  I X 3
I X  105408 73728  210816
I X  389952mm 4
2. Determine the moment of inertia of cross-section shown about its centroidal axes.
90 mm
Y
20 mm
70 mm
10 mm
20 mm
X
50 mm
Required:
I X C , I YC
Solution:
a. Locate Centroidal Axes, YC and XC:
Based on the given figure, Y-axis is located at the center, thus, YC coincides with Y-axis. There’s
no need to locate YC, only XC.
To locate XC, solve for Ῡ:
Divide the shaded area into parts. (rectangle 1, rectangle 2, rectangle 3)
Shape
Area
Y
A1  bh
Rectangle 1
A1  90mm20mm 
y1  20  70 
A1  1800mm
y1  100mm
2
A2  bh
A2  700mm
2
A3  bh
A3  1000mm
y
y
d1  y1  y
d1  100mm  65.29mm
d1  34.71mm
1
70mm 
2
y2  55mm
d 2  y  y2
1
20mm 
2
y3  10mm
d 3  y  y3
y3 
A3  50mm20mm 
Rectangle 3
1
20
2
y2  20 
A2  10mm70mm 
Rectangle 2
d
2
d 2  65.29mm  55mm
d 2  10.29mm
d 3  65.29mm  10mm
d 3  55.29mm
A1 y1  A2 y2  A3 y3
AT
1800100  70055  100010
1800  700  1000
y  65.29mm
b. Solve for “d”. (Please refer to the table)
d is the distance from the centroid of the shape in consideration up to the centroidal axis.
90 mm
Y
20 mm
1
d1 =34.71mm
XC
2
d2 =10.29mm
70 mm
Y1 =100mm
10 mm
d3 =55.29mm
Y2 =55mm
20 mm
3
Y3 =10mm
50 mm
X
c. Solve for I X C :
I X C  I X C1  I X C 2  I X C 3
I X C1  I x  Ad 2
bh3
 Ad 2
12
3
90mm20mm 
2

 1800mm 2 34.71mm 
12
 2228611.38mm 4
I X C1 
I X C1
I X C1
I X C 2  I x  Ad 2
bh3
 Ad 2
12
3
10mm70mm 
2

 700mm 2 10.29mm 
12
 359952.20mm 4
I XC2 
I XC2
I XC2
I X C 3  I x  Ad 2
I XC3
I XC3
I XC3
bh3

 Ad 2
12
3
50mm20mm 
2

 1000mm 2 55.29mm 
12
 3090317.43mm 4
I X C  I X C1  I X C 2  I X C 3
I X C  2228611.38  359952.2  3090317.43
I X C  5678881.013mm 4
3. Determine the moment of inertia of the shaded area shown about x- and y-axis.
Y
100mm
100mm
150mm
150mm
75mm
150mm
X
Required:
I X , IY
Solution:
a. The shaded area is composed of a rectangle(1), triangle(2) and a circle(3).
There’s no need to locate the centroidal axes, since the required are the moment of inertia about
the given x ans y axes.
Area
dX
dY
A1  200mm 300mm 
d X1  100mm
dY1  150mm
Shape
A1  bh
Rectangle
A1  60000mm 2
Triangle
1
A2  bh
2
1
A2  150mm 300mm 
2
A2  22500mm 2
1
150
3
 250mm
d X 2  200 
d X2
1
300mm 
3
dY2  100mm
dY2 
A3  r 2
Circle
A3   75mm 
d X 3  100mm
2
dY3  150mm
A3  5625  17671.46mm 2
Note:
Distances dX and dY were measured from the centroid of each area up to the axis where you will
take the moment of inertia. In this case, the given X and Y axes.
b. Solve for Ix:
I X  I X1  I X 2  I X 3
I X1  I X  Ad 2
bh3
 Ad 2
12
3
200300
2

 60000150
12
 1.80 x109 mm 4
I X 2  I X  Ad 2
bh3
 Ad 2
36
3
150300
2

 22500100
36
 337.5 x106 mm 4
I X1 
I X2 
I X1
I X2
I X1
I X2
I X 3  I X  Ad 2
I X3 
I X3
I X3
r 4
 Ad 2
4
 754
2

 17671.46150
4
 422.46x106 mm 4
I X  I X1  I X 2  I X 3
I X  1.80x109  337.5 x106  422.46x106
I X  1715.04x106 mm 4
c. Solve for IY:
IY  IY1  IY2  IY3
I Y1  I Y  Ad 2
I Y2  I Y  Ad 2
hb3
 Ad 2
12
3
300200
2
I Y1 
 60000100
12
I Y1  800x106 mm 4
I Y2 
hb3
 Ad 2
36
3
300150
2
I Y2 
 22500250
36
I Y2  1.434x109 mm 4
I Y1 
I Y3  I Y  Ad 2
I Y3 
r 4
 Ad 2
4
 754
2
I Y3 
 17671.46100
4
I Y3  201.57 x106 mm 4
I Y  I Y1  I Y2  IY3
I Y  800x106  1.434x109  201.57 x106
I Y  2032.43x106 mm 4
4. Locate the centroid ӯ of the cross section shown and determine the moment of inertia about
the x’-axis.
300
mm
200
mm
300
mm
200
mm
300
mm
Ῡ
400 mm
50 mm
x'
5. Determine the moment of inertia of the shaded area shown about the centroidal axes
50 mm 100 mm
300 mm
G. References:
1. Vector Mechanics for Engineers 10th Edition, Beer, F.P., Johnston, E.R.Jr, Mazurek, D.F., Cornwell,
P.J.2013
2. Engineering Mechanics Statics 13th Edition, Hibbeler, R.C.,2013
3. Engineering Mechanics Statics 14th Edition, Hibbeler, R.C.,2016
4. Engineering Mechanics Statics 3rd Edition, Pytel, A., Kiusalaas, J., 2010