MOMENT OF INERTIA
BY
GP CAPT NC CHATTOPADHYAY
WHAT IS MOMENT OF
INERTIA?




IT IS THE MOMENT REQUIRED BY A SOLID
BODY TO OVERCOME IT’S RESISTANCE TO
ROTATION
IT IS RESISTANCE OF BENDING MOMENT
OF A BEAM
IT IS THE SECOND MOMENT OF MASS
(mr2) OR SECOND MOMENT OF AREA (Ar2)
IT’S UNIT IS m4 OR kgm2
PERPENDICULAR AXIS
THEOREM

The moment of inertia of a plane
area about an axis normal to the
plane is equal to the sum of the
moments of inertia about any two
mutually perpendicular axes lying in
the plane and passing through the
given axis.
Moment of Inertia: Iz = Ix+Iy
PARALLEL (TRANSFER)AXIS
THEOREM



THE MOMENT OF AREA OF AN
OBJECT ABOUT ANY AXIS
PARALLEL TO THE
CENTROIDAL AXIS IS THE
SUM OF MI ABOUT IT’S
CENTRODAL AXIS AND THE
PRODUCT OF AREA WITH THE
SQUARE OF DISTANCE OF CG
FROM THE REF AXIS
IXX= IG+Ad2
A is the cross-sectional area.
: is the perpendicuar distance between
the centroidal axis and the parallel
axis.
Moment of Inertia - Parallel
Axis Theorem
Parallel axis theorem:
Consider the moment of
inertia Ix of an area A
with respect to an axis
AA’. Denote by y the
distance from an element
of area dA to AA’.
I x   y dA
2
Moment of Inertia - Parallel
Axis Theorem
Consider an axis BB’ parallel
to AA’ through the centroid C
of the area, known as the
centroidal axis. The equation
of the moment inertia
becomes
2
I x   y dA    y  d  dA
2
  y dA  2 ydA  d
2
2
 dA
Moment of Inertia - Parallel
Axis Theorem
The first integral is the moment
of inertia about the centroid.
2

I x   y dA
The second component is the first moment area about the
centroid
yA   ydA  y  0
  ydA  0
Moment of Inertia - Parallel
Axis Theorem
Modify the equation obtained
with the parallel axis theorem.
I x   y dA  2  ydA  d
2
 Ix  d A
2
2
 dA
Example – Moment of
Inertia
Compute the moment of inertia in
the x about the AA` plane.
AA`
Example – Moment of
Inertia
Compute the moment of inertia in
the x about the AA` plane.
Ix 
h b

y 2 dA    y 2 dxdy
Area
0 0
h
 y 
bh
 b  
3
 3 0
3
3
AA`
Example – Moment of
Inertia
From earlier lecture, the moment of
inertia about the centroid
Ix 

h/2
y dA 
2

2
y bdy
 h/2
Area
h/2
3
 y 
bh
 b  
 3   h/2 12
3
Example – Moment of
Inertia
Using the parallel axis theorem
Ix  Ix  d A
2
2
bh  h 
 4 3

    bh     bh
12  2 
 12 
3
bh
AA`

3
3
Parallel Axis - Why?
Recall that the method of finding centroids of
composite bodies? Follow a Table technique
How would you be able to find the moment of
inertia of the body. Use a similar technique, table
method, to find the moment of inertia of the body.
Parallel Axis - Why?
Use a similar technique, table method, to find the
moment of inertia of the body.
Bodies
Ai
yi
yA

y
A
i
i
i
y i*Ai
Ii
di2Ai
di=y i-ybar

Ix   Ix  d A
2


i
  I xi   yi  y

2
Ai
Moment of Inertia
Use a set of standard tables:
Example - Moment of
Inertia
Find the moment of inertia of the body, Ix and the
radius of gyration, kx (rx)
I THINK…. I CAN THINK….
Example - Moment of
Inertia
Set up the reference axis at AB and find the centroid
Bodies
Ai
yi
y i*Ai
1
2
18
18
1
5
18
90
36
 yi Ai
Ii
di=y i-ybar
108
3
108 in
y

 3.0 in.
2
 Ai 36 in
di2Ai
Example - Moment
of Inertia
From the table to find the moment of
inertia
A
y
y *A
I
Bodies
1
2
i
i
18
18
1
5
36
ybar
I
i
di=y i-ybar
di2Ai
18
90
6
54
-2
2
72
72
108
60
i
i
3 in.
204 in4

I x   I xi   yi  y

2
Ai
 60 in 4  144 in 4  204 in 4
144
Example - Moment of
Inertia
Compute the radius of gyration, rx.
4
Ix
204 in
rx 

A
36 in 2
 2.38 in.
Example - Moment of
Inertia
Find the moment of inertia of the body, Ix and the
radius of gyration, kx (rx)
Example - Moment of
Inertia
The components of the two bodies
and subtract the center area from the
total area.
Bodies
Ai
yi
1
2
21600
-9600
90
90
12000
ybar
I
90 mm
46800000 mm4
y i*Ai
Ii
1944000 58320000
-864000 -11520000
1080000 46800000
di=y i-ybar
di2Ai
0
0
0
0
0
Example - Moment of
Inertia
Compute the radius of gyration, rx.
Ix
46800000 mm
rx 

A
12000 mm 2
 62.45 mm
4
PROBLEM….

FIND THE MI OF A CIRCULAR
LAMINA OF RADIUS “r“?