MODULE - 2
Centre of Gravity
Each and every particle forming a body is attracted by the earth towards its centre. This
forces of attraction form a system of parallel forces. The resultant of these forces is the weight of
the body. The point of application of this resultant force is called centre or gravity of the body. It
is the point at which the entire weight of the body can be assumed to be concentrated. The centre
of mass of a body is the point at which the entire mass of the body can be assumed to be
concentrated. The centre of mass and centre of gravity of a body will be different points only
when the gravitational force field is not uniform and parallel. In engineering applications it is
assumed that the gravitational force is uniform and is parallel and hence the centre of mass and
centre of gravity are the same point. Geometric figures and curves do not possess weight. The
term centre of gravity as applied to geometric figures and curves is something of a misnomer.
For this reason the term centroid is used instead of centre of gravity for plane areas and curves.
Centroid is the geometrical centre of an area, curve or an object. When the material of the body
is homogeneous, the geometrical centre that is the centroid coincides with the centre of gravity of
the body.
Consider a body of weight W as shown in fig. 2.1. OX, OY and OZ are the three
reference axes and x, y and z are the co-ordinates of centre of gravity (G) of the body. Let the
co-ordinates of elemental bodies of weights dw1, dw2 etc be x1, y1, z1 and x 2, y 2, z 2 etc.
Total weight of the body, W = dw1  dw 2  dw 3  .......... Sum of moments of forces
dw1,dw 2 ,dw 3 etc. about any axis is equal to the moment of force W about the same axis.
dw1x1  dw 2 x 2  dw 3x3  ............ = dw1  dw 2  dw3  ........ x
x
x1dw1  x 2dw 2  x 3dw 3  .........
dw1  dw 2  dw 3  .........
x
=
 xdw
 dw
 (xdw)
 dw
Similarly,
y
y1dw1  y 2dw 2  y3dw 3  .........
dw1  dw 2  dw 3  .........
y
 ydw
 dw
=
 (ydw)
 dw
and
z
z1dw1  z 2dw 2  z3dw 3  ...........
dw1  dw 2  dw 3  ...........
z
 zdw
 dw
=
 (zdw)
 dw
If the density of the body is uniform, then, weight W = volume × weight density
W  V 
dW  dV 
x
 xdW   x  dV   x  dV   xdV
 dW  dV
 dV  dV
y
 ydW   y  dV   y  dV   ydV
 dW  dV
 dV  dV
z
 zdW   z  dV   z  dV   zdV
 dW  dV
 dV  dV
When the thickness of the body is very small and is uniform through out the area A,
then, volume = Area × thickness.
V=A×t
dV = dA × t
x
 xdV
 dV
=
 (xdAt)
 (dAt)
=
 (xdA)
 dA
y
 ydV   ydAt   (ydA)
 dV  dAt  dA
z
 zdV   zdAt   (zdA)
 dV  dAt  dA
When the cross sectional area is uniform, such as a wire, volume = Area of cross section
x length
V=A×L
dV = A × dl
Similarly,
x
 xdV   xAdl   (xdl)
 dV  Adl  dl
y
 ydV   yAdl   (ydl)
 dV  Adl  dl
z
 zdV   zAdl   (zdl)
 dV  Adl  dl
Centroid of wires
When a line segment or wire has an axis of symmetry, the centroid will be on that axis.
When a line segment or wire has two axes of symmetry, the point of intersection of the axes of
symmetry will be the centroid of the wire.
Determination of position of centroid of a wire involves taking moments of elemental
lengths with respect to some axes. Therefore some frame of reference or co-ordinate axes are to
be selected. The vertical line through the left extreme point or line can be taken as the Y-axis and
the horizontal line through the lower most point or line can be taken as the X-axis. Distance of
centroid from the X-axis is denoted by y and that from the Y-axis is denoted by x . Although the
position of centroid does not depend upon the reference axes, the value of the x and y depends
on the position of the reference axes. The axes passing through the centroid are called centroidal
axes. x and y will be zero, if the reference axes chosen pass through the centroid.
Example 2.1
Find the centroid of a horizontal line of AB of length L.
Solution:
Consider an elemental length dl at a distance x from the Y axis.
L
x
 xdl
 dl

 xdl
0
L
 dl
0
L
=
x
 x2 
 
 2 0
L2
0
2

L0
 L0L
L
;y  0
2
Example 2.2
Locate the centroid of a circular arc when the arc is kept symmetrical with respect to Xaxis.
Solution
Consider an elemental length of wire AB at an angle  from the X axis substanding an
angle of d at the centre. The length of elemental arc dL = R.d .
x  R cos  and
y  R.sin 

2
 R cos Rd
x
 xdl   2
 dl

2
 Rd

2

R 2 sin  2  
=
R  

2 

2
2
 
 
R sin  sin(  ) 
2
2 
= 


 ( )
2
2
x
2R sin


2
Since the arc is symmetrical with respect to X axis,
y0
Example 2.3
Locate the C.G. of the semicircular arc of radius R.
Solution.
Since the arc is symmetrical with respect to the Y axis, x  0

ydl 0

y


dl

 Rd
R sin  Rd
0
R 2 cos 0

=
R  0

= R
= R
y
cos   cos 0

 1  1

2R

Example 2.4
Locate the centroid of a quarter circular arc of radius R.
Elemental length dl = R d
x  R cos 
y  R sin 

2
 xdl  0
x
 dl
R cos Rd

2
 Rd
0

=R
sin 02

2
0
 
x
2R

R
1  0
 
 2  0 

2
 ydl  0
y
 dl
R sin Rd

2
 Rd
0
=R

2
0
  cos 

2
0
 
= R
y
0  1

2
2R

Example 2.5
Locate the centroid of the line ABCD as shown in fig. 2.10.
Solution.
x
 xdl   xdl  x1l1  x 2l2  x3l3
l1  l2  l3
 dl  dl
l1 = 100 mm, l2 = 150 mm, l3 = 100 mm
l1  l2  l3 = 100 + 150 + 100 = 350 mm
x1 
100
cos 60  25mm
2
x 2  100 cos 60 
150
 125mm
2
x 3  100 cos 60  150 
x
100
cos 30  243.3mm
2
25 100  125 150  243.3  100
100  150  100
= 130.23 mm
y
y1l1  y1l2  y3l3
l1  l2  l3
y1 
100
sin 60  43.30
2
y2  100sin 60  86.60
y3  100sin 60 
y
43.3 100  86.6 150  111.6  100
100  150  100
y  81.37mm
Centroid of areas
100
sin 30  111.60
2
When a plane area has an axis of symmetry, the centroid will be on that axis. When
there are two axes of symmetry, the point of intersection of the axes of symmetry will be the
centroid of the area.
In the previous section it was proved that the distance of centroid of an area from Y axis
 xdA . where dA is an elemental area and x is the distance of this elemental
is given by, x 
 dA
area from Y axis. Similarly y 
 ydA . where y is the distance of elemental area dA from X axis.
 dA
Centroid of rectangle
To find y
Consider a horizontal strip of thickness by dy at a distance y from the X axis as shown
in fig. 2.22.
Elemental area dA = b × dy
d
 ydA
y
 dA

 ydA
0
d
d

 dA
0
 ybdy
0
d
 bdy
0
d

 y2 
b 
 2 0
b  y
d
0
b 2
d
d
2

bd
2
y
d
2
To find x
Consider a vertical strip of thickness dx at a distance x from the Y axis as shown in fig.
2.23.
Elemental area, dA = d × dx
b
xdA 0

x
 b
dA

 d  dx
xd  dx
0
b
=
 x2 
d 
 2 0
x
d  x 0
b
b2
b
 2 
db
2
d
b
2
When the reference axes, X-X and YY axis, pass through the centre of rectangle,
x  0 and y  0 as shown in fig. 2.24.
Centroid of a right angled triangle
Consider right angled triangle of sides b and h. Consider an elemental strip of thickness
dy at a distance y from the X axis. Area of element, dA = b1 × dy. Using property of similar
b
(h  y)
triangles, 1 
b
h
b1 
b
(h  y)
h
 dA 
b
(h  y)dy
h
h
y
b
 y  h (h  y)dy
 y dA  0
h
b
 dA
 (h  y)dy
0
h

h
0 (hy  y )dy 
h
0 (h  y)dy
2
 hy 2 y3 
 

3 0
 2

y2 
hy



2 0

1 1
h3   
2 3
 
 1
h 2 1  
 2
y
h
h
h 2 h3

2
3

2
h
h2 
2
h3
 62
h
2
h
3
Consider an elemental strip of thickness dx at a distance x from the Y axis as shown in
fig.2.26.
Area of the element, dA = h1dx.
Using property of similar triangles,
h1 (b  x)

h
b
h
(b  x)
b
h1 
 dA 
x
 xdA
 dA
b

h
(b  x) dx
b
h
0 x b (b  x)dx
b
0
h
(b  x)dx
b

 bx 2 x 3 
 

3 0
 2

x2 
bx



2 0

b
b
 b3 b3 
3
   b
2 3

 6
 2 b2  b2
b  
2
2

x
b
3
Thus, the distance of centroid of a triangular lamina from its base is one third of its
altitude.
Example 2.10
Locate the centroid of the triangular lamina as shown in fig. 2.27.
Solution
1
2
x  b b  b
3
3
1
2
yh h  h
3
3
Centroid of Isosceles triangle
x
b
2
y
1
h
3
Centroid of a Circle
When the reference axes are selected as shown in fig. 2.29,
xy
d
2
Example 2.11
Locate the centroid of one quarter of circular area of radius R.
Consider an element as shown in fig. 2.30.
dA =
r d × dr
x
=
r cos 
y
=
r sin 
x
 xdA
 dA

2R

  r cos  r d dr
0 0

2R
  r dr d
0 0

2 3 R
r

  3 
0
cos d
0

2 2 R
r
  2 
0
0

R3
sin 02
 3

R2
 02
2
d
and
 

sin  sin 0 

2R  2
  4R

3
3
 
 2  0 

2R
 y dA  0 0
y

 dA
2R
  r d dr
r sin  r d dr
0 0

2  3 R
r

  3 
0
sin  d
0

2  2 R
r
  2 
0
d
0

R3
  cos 02
3


R2

 02
2
2R
0  1
 3

2
y
4R
3
Example 2.12
Locate the centroid of one quadrant of a circle of radius r as shown in fig. 2.32.
Solution
xr
4R
3
yr
4R
3
Example 2.13
Locate the centroid of sector of a circle of radius R with an included angle of .
Consider an elemental area as shown in fig. 2.33.
Area of the element, dA = r d × dr
x = r cos 
Since the area is symmetrical with respect to X axis, y  0
 xdA
 dA
x

2 R
  r cos  r d dr


0
2

2 R
  r d dr

0
2

2

R
 r3 
  3  cos  d

0

2

2
R
 r2 
  2  d

0

2


R3
sin  2

3
2

2


2
2
R
 
2
 
 
sin  ( sin ) 

2R  2
2 

 
3

 2  ( 2 ) 

4R sin
3

2
Centroid of composite areas
In many engineering applications some composite areas are considered. These
composite areas may consist of triangles, rectangles and other simple figures. Hence it is
necessary to find the centroid of such composite areas. Composite area can be divided into a
number of areas like rectangle, triangle, semicircle for which areas and location of centroids of
each area are known. The distance of centroid of composite area from Y axis is given by
x
 x a   (x a)  a1x1  a 2x 2  ..... , where a1, a2 etc are area of each section and x1,
a1  a 2  .....
 a a
x2 etc are the distance of centroid of a1, a2 etc from the Y axis. Similarly,
y
 ya   (ya)  a1y1  a 2 y2  .....
a1  a 2  .....
 a a
Example 2.17
Locate the centroid of the ‘T’ section shown in fig. 2.37.
Solution
Since the section is symmetrical with respect to the Y axis, x  0 .
y
a1y1  a 2 y 2
a1  a 2
a1 = 300 × 20 = 6000 cm2
a2 = 200 × 20 = 4000 cm2
y1 
300
 150cm
2
y 2  300 
 y
20
 310cm
2
6000  150  4000  310
 214 mm
6000  4000
Example 2.18
Locate the centroid of the area shown in fig. 2.39.
Solution
a1 = 14 × 2 = 28 cm2
a2 = 20 × 2 = 40 cm2
a3 = 8 × 2 = 16 cm2
x1 
14
 7 cm
2
x2 
2
 1cm
2
x3 
8
 4cm
2
y1 
2
 1cm
2
y3  2  20 
x

20
 12cm
2
2
 23cm
2
a1x1  a 2 x 2  a 3x 3
a1  a 2  a 3
28  7  40  1  16  4
= 3.57 cm
28  40  16
y

y2  2 
a1y1  a 2 y 2  a 3 y3
a1  a 2  a 3
28  1  40  12  16  23
= 10.43 cm
28  40  16
Example 2.20
Determine the centre of gravity of the thin homogeneous plane shown in fig. 2.43.
Solution
1
 100  150  7500mm 2
2
a1 
a2 = 200 × 150 = 30000 mm2
a3 
r 2 
  502  3927mm 2
2
2
1


x1  100   100  = 66.67 mm
3


x 2  100 
200
 200mm
2
x 3  100  50  50  200mm
1


y1  150   150   100mm
3


y2 
150
 75mm
2
y3 
4r 4  50

 21.22mm
3 3  
x

7500  66.67  30000  200  3927  200
= 170.21 mm
7500  30000  3927
y

a1x1  a 2 x 2  a 3x 3
a1  a 2  a 3
a1y1  a 2 y 2  a 3 y3
a1  a 2  a 3
7500  100  30000  75  3927  21.22
= 86.88 mm
7500  30000  3927
Example 2.21
Locate the centroid of the area shown in fig. 2.45.
Solution
a1 
r 2  60
1002  60
 
 5235.99mm 2
360
360
a2 
1
 100cos 60  100sin 60  2165.06mm 2
2
OG1 

60
4  100  sin
2 
2

3
3
 60
180
4r sin
x1  OG1 cos

= 63.66 × cos 30 = 55.13 mm
2
y1  OG1 sin

= 63.66 × sin 30 = 31.83 mm
2
1
x 2   100cos 60  16.67mm
3
1

 2
y 2  100  sin 60   100sin 60    100sin 60  57.74mm
3

 3
x
55.13  5235.99  16.67  2165.06
 43.88 mm
5235.99  2165.06
y
31.83  5235.99  57.74  2165.06
 39.41 mm
5235.99  2165.06
2.5. Moment of inertia of lamina
Moment of inertia or an lamina is a purely mathematical term which gives a quantitative
estimate of the relative distribution of area with respect to some reference axis. If r is the distance
of an elemental area, dA, from a reference axis AB, then the some of the terms,  r 2dA , to
cover the entire area is called moment of inertia of the area about the reference axis AB and is
denoted by I AB .
I AB   r 2dA   r 2dA
The product r  dA is the first moment of elemental area about the reference axis. The
second moment of elemental area about the reference axis is r 2  dA . Thus moment of inertia of
an area is the integral of second moment of an elemental area about the reference axis. The term
second moment of area is more proper than moment of inertia because the term moment of
inertia should be used when the integral of mass is taken. Generally the moment of inertia of area
is called area moment of inertia or second moment of area and moment of inertia of physical
body is called mass moment of inertia. The axis passing through the centroid is called centroidal
axis and the moment of inertia about the centroidal axis is denoted by IG .
2.6. Radius of gyration
Consider an area A which has a moment of inertia I with respect to a reference axis AB.
Let us assume that this area is compressed to a thin strip parallel to the axis AB. For this strip to
have the same moment of inertia I, with respect to the same reference axis AB, the strip should
I
be placed at a distance k from the axis AB such that I  Ak 2 . k 
is called radius of gyration
A
of the area with respect to the given axis AB.
2.7. Perpendicular axis theorem
If I xx and I yy are the moment of inertia of an area A about mutually perpendicular axis
XX and YY, in the plane of the area, then the moment of inertia of the area about the ZZ axis
which is perpendicular to XX and YY axis and passing through the point of intersection of XX
and YY axis is given by I ZZ  IXX  IYY
Proof:
Consider a plane area A. Let XX and YY be the two mutually perpendicular axes in the
plane of the area, intersecting at O. Let ZZ be an axis through O and perpendicular to the plane
of area A. Consider an elemental area dA at distance y from the XX axis, x from the YY axis and
r from the ZZ axis, then
Izz   r 2dA   (x 2  y 2 )dA .
=  x 2dA   y 2dA
= IXX  IYY  IYY +IXX
I ZZ  IXX  IYY .
2.8. Parallel axis theorem
It is a transfer theorem which is used to transfer moment of inertia from one axis to
another axis. These two axes should be parallel to each other and one of these axes must be a
centroidal axis.
It states that, IG is the moment of inertia of a plane lamina of area A, about its
centroidal axis in the plane of the lamina, then the moment of inertia about any axis AB which is
parallel to the centroidal axis and at a distance “h” from the centroidal axis is given by
IAB  IG  Ah 2
Proof:
Consider an elemental area dA at a distance y from the centroidal axis. The first
moment of elemental area about the axis AB as shown in fig.2.87 is dA (y + h). Second moment
of elemental area about the axis AB is dA (y  h) 2 . The second moment of the area about the
axis AB is  dA (y  h) 2
IAB   dA (y  h) 2
=  dA (y 2  h 2  2 hy)
=  y 2dA   h 2dA   2 h y dA
= IG  h 2  dA  2h  ydA
= IG  h 2A  2h (Ay)
IAB  IG  Ah 2
y  0, because it is the distance of centroid G from the axis from which y is measured.
In fig. 2.87, y is measured from the centroidal axis itself.
Example 2.39
Calculate the moment of inertia of a rectangular cross section about the centroidal axes
and also about its base AB.
Solution.
Consider an elemental of thickness dy at a distance y from the X axis.
Area of the element, dA = b.dy.
Second moment of this elemental area about the X axis, d IXX  y2d A
= y 2 b.dy
IXX 
d
2
 y bdy
2

d
2
d
3 2
y
= b.  
 3  d
2
=
b d 3
d 
( )  (  )3 

3 2
2 
=
b  d3 d3 
  
3 8
8
=
I XX 
b d3
3 4
b d3
12
Consider an element of thickness dx at a distance x from the Y axis.
Area of the element dA = d.dx.
Second moment of this elemental area about the Y axis,
d IYY  x 2dA  x 2 d.dx
I YY 
b
2
 x d dx
2

b
2
b
 x3  2
= d 
 3  b
2
=
d
3
 b 3 b 3 
( 2 )  ( 2 ) 


=
d
3
 b3 b3 
  
8
8
d b3 d b3

=
3 4
12
I YY 
d b3
12
Moment of inertia of a rectangular lamina about its centroidal axis =
1
 length of side
12
parallel to the axis × (length of other side)3 .
Moment of inertia about the base can be calculated using parallel axis theorem
IAB  IG  Ah 2
bd 3
d
 bd  ( ) 2
=
12
2
bd 3 bd 3

=
12
4
= I AB 
bd3
3
Moment of inertia of a rectangular lamina about its base is
bd 3
3
Example 2.40
Determine the moment of inertia of a triangle about its centroidal axis, parallel to its
base.
Consider an element of thickness dy at a distance y from the vertex R. Let the length of
this element be b.
Area of element, dA = b dy.
Second moment of this elemental area about the axis through the vertex R,
dIR  y2 (bdy)
H
I R   b y 2 dy
0
b y
B
 , b   y
B H
H
H
IR 
B
 H yy
2
dy
0
H
B  y4 
=  
H  4 0
B H4
=
H 4
BH3
IR 
4
Using parallel axis theorem,
IR  IG  Ah 2 ,
IG  IR  Ah 2
=
BH 3 1
2
  B  H  ( H) 2
4
2
3
=
BH 3 2
  BH3
4
9
IG 
BH3
36
Again using parallel axis theorem,
IPQ  IG  Ah 2
=
BH3 1
H
  B  H  ( )2
36
2
3
=
BH3 BH3

36
18
I PQ 
BH 3
12
Moment of inertia of a triangular lamina about its base is
BH 3
.
12
Moment of inertia of a triangular lamina about the centroidal axis, parallel to the base is
3
BH
.
36
Moment of inertia of a triangular lamina about an axis through the vertex and parallel to
BH 3
.
the base is
4
Example 2.41
Determine the moment of inertia of a circular lamina about its centroidal axes.
Solution.
Consider an elemental area as shown in fig. 2.91. The area of a element, dA  r d  dr .
Second moment of this elemental area about XX axis, dIXX  y2dA
I XX   y 2 r d dr
2 R
=
  (r sin )
2
r d dr
0 0
2 R
=
 r
3
sin 2  dr d
0 0
2
=

0
R
 r4 
2
  sin  d
 4 0
R4
4
2
R4
=
4
2
=
 sin
2
 d
0
1  cos 2 
 d
2

 
0
2
=
R 4  sin 2 

4  2 
2  0
=
R4
 2  0  (0  0) 
8
4
D
 
4
R
 D4
2
   
=
4
4
64
D 4
, where D is the diameter of the circle. Because of symmetry of
64
D 4
 IXX 
64
I XX 
circular area, I YY
Example 2.42
Determine the moment of inertia of a semicircular lamina about its centroidal axes.
Consider an elemental area as shown in fig. 2.92.
Area of the element dA  r d dr
Second moment of this elemental area about the base AB, d I AB   y 2 dA
y  r sin 
R
IAB    (r sin ) 2 r dr d
00
R
=
r
00
3
sin 2dr d

=
R4
sin 2  d

4 0

R 4 1  cos 2
=
(
)d
4 0
2

=
R4 
sin 2 
  2

4 2 
2  0
=
R4
(  0)  (0  0) 
8
=
R 4   D 
  
8
82
4
D 4
=
128
Moment of inertia of a semicircular lamina about its base is
D 4
.
128
Using parallel axis theorem.
IAB  IG  Ah 2
IXX  IAB  Ah 2
=
R 4 R 2 4R 2

( )
8
2
3
=
R 4 R 4  8

8
9
 8 
= R4   
 8 9 
IXX  0.11R 4
Moment of inertia of a semicircular lamina about its centroidal axis, parallel to the base
is 0.11R .
4
Moment of inertia about the centroidal axis, perpendicular to the base.
Consider an elemental area as shown in fig. 2.93.
Area of the element, dA  r d dr
x  r cos 
Second moment of elemental area about YY axis, d IYY  x 2dA .
R
IYY    (r cos ) 2 r d dr
00

=
R4
cos 2  d

4 0

R4  1
1

=
  sin 2

4 2
4
0
=
R4 1

  0

4 2

4
D
 
4
R
D4
2
   
=
8
8
128
Moment of inertia of a semicircular lamina about the centroidal axis, perpendicular to
D 4
the base is
.
128
Example 2.43
Determine the moment of inertia of one quarter of a circle about its centroidal axes.
Consider an elemental area as shown in fig. 2.94. Area of the element, dA  r d dr .
y  r sin 
Second moment of elemental area about the axis OA,
IOA   y 2dA

2R
IOA    (r sin ) 2 r d dr
00

4 2
=
R
4
 sin
2
 d 
0

4 2
R
4

2
=
R 1
1
  sin 2

4 2
4
0
=
R4  1 

  0

4 2 2

4
1  cos 2 
 d
2

0
 
R 4 D4

16
256
Because of symmetry, IOA  IOB 
D4
256
Using parallel axis theorem,
IOA  IXX  Ah 2
IXX  IOA  Ah 2
R 4 R 2  4R 
=

 
16
4
 3 
2
4

= R4   
16 9 
= 0.055R 4
Because of symmetry, IXX  IYY  0.055R 4 .
Example 2.44
Find the moment of inertia of the angle shown in fig. 2.95 about X and Y axes shown.
Solution.
A1 = 100 × 10 = 1000 mm2
A2 = 115 × 8 = 920 mm2
x1 
100
 50mm
2
x2 
8
 4mm
2
y1 
10
 5mm
2
y 2  10 
115
 67.5mm
2

 
IX  IG1XX  A1y12  IG 2XX  A2 y2 2

1
 1

   100  103  1000  52     8  1153  920  67.52 
 12
  12

= 52.39 × 105 mm4

 
IY  IG1YY  A1x12  IG 2YY  A 2 x 22


1
 10  1003  1000  502
12
= 33.53 × 105 mm4
Example 2.45
Find the moment of inertia of an unequal angle iron section of 250mm×20mm about its
centroidal axes.
Solution.
A1 = 125 × 20 = 2500 mm2
A2 = 230 × 20 = 4600 mm2
x1 
125
 62.5mm
2
x2 
20
 10mm
2
y1 
20
 10mm
2
y 2  20 
230
 135mm
2
x
A1x1  A 2 x 2 2500  62.5  4600  10
= 28.49 mm

A1  A 2
2500  4600
y
A1y1  A 2 y 2 2500  10  4600  135
= 91 mm

A1  A 2
2500  4600
IG1XX 
1
 125  203  83333.33mm 4
12
IG 2XX 
1
 20  2303  20278333.33mm 4
12
IG1YY 
1
 20  1253  3255208.33mm 4
12
IG 2YY 
1
 230  203  153333.33mm 4
12

 
IG XX  IG1XX  A1h12  IG2XX  A2h 22

h1  GG1  y  y1  91  10  81mm
h 2  GG 2  y2  y  135  91  44mm

 
IG XX  83333.33  2500  812  2027833.33  4600  44 2

= 16485833.33 + 29183933.33
= 45669766.66 mm4

 
IG YY  IG1YY  A1h12  IG2YY  A2h 22

h1 is the horizontal distance GG1 and h2 is the horizontal distance GG2.
h1  x1  x  62.5  28.49  34.01mm
h 2  x  x 2  28.49  10  18.49mm

 
IG YY  3255208.33  2500  34.012  153333.33  4600  18.492

= 7872890.37 mm4
Example 2.46
Determine the moment of inertia of the T-section shown in fig. 2.99 about the centroidal
axes.
Solution.
A1 = 100 × 20 = 2000 mm2
A2 = 80 × 20 = 1600 mm2
y1 
100
 50mm
2
y 2  100 
y
20
 110mm
2
A1y1  A 2 y 2 2000  50  1600  110
= 76.67 mm

A1  A 2
2000  1600
Moment of inertia of section (i) about its own horizontal centroidal axis is IG1XX.
IG1XX 
20  1003
 1666666.67mm
12
Moment of inertia of section (2) about its own horizontal centroidal axis is
IG 2XX 

80  203
53333.33mm 4
12
 
IG XX  IG1XX  A1h12  IG 2XX  A2h 22

h1 and h2 are the vertical distances of G1 and G2 from G.
h1  y  y1  76.67  50  26.67
h 2  y2  y  110  76.67  33.33
IGXX  1666666.67  (2000  26.672 )  53333.33  (1600  33.332 )
= 4.92 × 106 mm4
Since G1, G2 and G are on the same YY axis, IG YY  IG1YY  IG 2YY
IG1YY is the M.I of section (1) about its own vertical centroidal axis, and IG 2YY is the
M.I of section (2) about its own vertical centroidal axis.
IG YY 
100  203 20  803

 9.2  105 mm 4
12
12
Example 2.52
Calculate the moment of inertia of the shaded area, as shown in fig. 2.112, with respect
to the centroidal axes.
Solution.
1
 60  80  2400cm 2
2
a1 
a 2  r 2  176.715cm2
x1 
2
 60  40cm
3
x 2  30cm
y1 
2
 80  53.333cm
3
y2  50cm
x
a1x1  a 2 x 2
2400  40  176.715  30

2400  176.715
a1  a 2
y
a1y1  a 2 y 2
2400  53.33  176.715  50
= 53.598 cm

2400  176.715
a1  a 2

 
IG XX  IG1XX  A1h12  IG 2XX  A2h 2 2

2
   154

 60  803 1
15 
2
 
  60  80  (0.265)   
      (3.598) 2 


2
2
 36
  64

= (853333.333 + 168.54) – (2485.049 + 2287.677) = 853304.5 cm4

 
IG YY  IG1YY  A1h12  IG 2YY  A2h 2 2

2

 80  603 1
    154
15 
 
  60  80  (0.795) 2   
      (10.795) 2 


2
2
 36
  64

= 480000 + 1516.86 – 2485.049 – 20592.957
IGYY  458438.854cm4
Example 2.53
Find the moment of inertia Ix of the section shown in fig. 2.114.
Solution
Consider an elemental strip of thickness dy at a distance y from the X axis.
Area of the element, a  x dy  ky 2dy
Second moment of this elemental area about X axis,
dIx  ay 2  ky 2dy x y 2  ky 4dy
a
 y5 
k
 IX   dy dy  k    a 5
 5 0 5
0
s
4
x  ky 2
At A, x = b and y = a
b  ka 2
k
b
a2
 IX 
b
a 3b
k 5
a = 2 a5 
5
5
5a
2.9 Mass Moment of Inertia
Mass is the quantitative measure of the resistance to change the motion of a body. The
inertia of a body is the property by virtue of which it resists any change in its state of rest or of
uniform motion. Translatory inertia is defined as mass and rotational inertia is known as moment
of inertia.
Consider two bodies having the same mass but of different shape as shown in fig. 2.116.
Resistance to rotation about the axis is different for the two bodies, even though they
have the same mass. The resistance to rotation about the axis AB depends on the expression
2
 dm x , where dm is an elemental mass of the body at a distance x from the AB axis. Thus the
moment of inertia of a body with respect to a given axis is defined by the integral, I   dm x 2 .
The unit of mass moment of inertia is kg-m2. The radius of gyration of the body is expressed on
I
, where m is the mass of body. The moment of inertia of a body about an axis at a
k
m
distance d and parallel to centroidal axis is equal to the sum of moment of inertia about
centroidal axis and product of mass and square of distance between the parallel axes.