TECHGURU CLASSES for SSC-JE/RAILWAYS/ORDINANCE
CHAPTER - 6 : MEMENT OF INERTIA
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CHAPTER - 6 : MEMENT OF INERTIA
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CHAPTER - 6 : MEMENT OF INERTIA
6.1
AREA MOMENT OF INERTIA
Like centroid, area moment of inertia is also a geometrical property of the plane
sections. Following article will make understanding and significance of area
moment of inertia more clear.
6.1.1
Significance and definition of area moment of inertia
Consider the state of stresses* appearing on a beam section as a result of bending
couples applied on it. It will be seen later in chapter-10 that bending stress at a point
depends upon the distance of that point from an axis called neutral axis. Thus stress
is  = ky at a point shown in the figure 6.1.1(b). force at that point due to stress  *
is  dA = (ky) dA.
Moment of the force about neutral axis is (  dA) y = ky2dA. Summation of such
moments for the whole area can be represented as
The quantity
2
 ky dA  k  y
2
dA.
2
 y dA has nothing to do with the magnitude or nature of external
forces or couples.
2
 y dA
is a pure geometrical quantity and hence depends upon
the geometrical variations only.
X
.
N.A
M
M
X
X


y
X
Figure 6.1.1(a)
*
Figure 6.1.1(b) : State of stresses on a section of
the beam
Stress can be understood, for instance, as force per unit area thus stress F/A =  = dF/dA. For more
details on bending stress, see chapter 10
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y
Many times, in problems of fluid mechanics
and in the analysis of internal stresses in solid
mechanics, such a quantity appears. Quantity
x
2
 y dA is called area moment of inertia*.
dA
r
y
An area lying in xy plane is shown in figure6.1.1(c). Area moment of inertia about x, y and
z axes are given as follows.
I x  y 2 dA, I y  x 2 dA and I z  r 2 dA

A


A
O
z
x
Figure 6.1.1(c)
A
Following points are important about area moment of inertia :
(i)
Moment of inertia is a geometrical property of the section.
(ii) Moments of inertia of an area about different axes are distinct.
(iii) Moment of inertia of a plane area about an axis perpendicular to it is called
polar moment of inertia. (e.g. for the plane area shown in figure 6.1.1(c), Iz is
polar moment of inertia)
(iv) First moment of area 'dA' about x axis (or y) is y dA (or x dA), Moment of first
moment of area is y (ydA) = y2 dA. Therefore area moment of inertia is also
called second moment of area.
(v) Area moment of inertia is always a positive quantity.
6.1.2
Radius of Gyration
Consider again, the plane area A shown in figure 6.1.1(c). Consider a long narrow
strip of the same area A and parallel to x-axis.
If the distance KX of the strip from x-axis is such that area moment of inertia of the
long narrow strip (i.e. AkX2) is equal to the area moment of inertia of plane area A
of figure 6.1.1(c) about x-axis, then kX is called radius of gyration about x-axis
[Refer: figure-6.1.2(a)]
Thus
A K2x = Ix.

Kx =
Ix
A
Similarly, a long narrow strip of area 'A' can be chosen parallel to and at a distance
of Ky from y-axis. If the distance Ky of the strip is such that area moment of inertia of
*
Area moment of inertia is also called second moment of area because it contains square of the
distance multiplied to area.
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this strip (i.e. Ak2y) is equal to the area moment of inertia of plane area A [of figure
6.1.1 (c)] about y axis, then Ky is called radius of gyration about y-axis. [Refer:
figure-6.1.2(b)]
A Ky2 = Iy  Ky =
Thus
Iy
A
y
y
A
A
KX
O
O
x
z
x
z
Figure 6.1.2 (a)
y
y
KY
A
A
kz
O
O
x
z
x
z
Figure 6.1.2 (b)
Figure 6.1.2 (c)
Now, consider a narrow ring of area A and radius Kz such that polar moment of
inertia of the ring (i.e. AK2z) is equal to polar moment of inertia (Iz) of the plane Area
A in figure 6.1.1 (c). Kz is called radius of gyration about z axis. [Refer figure:
6.1.2(c)].
Thus
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6.2
THEOREMS
6.2.1
Perpendicular axes theorem
Reconsider, for instance, the area shown in figure. 6.1.1(c). Polar moment of inertia
Iz is expressed as
2
dA  (x 2  y 2 ) dA
2
dA  y 2 dA  I y  I x
Iz =
r
Iz =
x


Iz = Ix + Iy

Above expression is the mathematical form of an important theorem of moment of
inertia called perpendicular axes theorem which is stated as under.
"According to perpendicular axes theorem, polar moment of inertia of a plane figure
is equal to the sum of moments of inertia about x and y axes, x, y axes being
orthogonal, passing through the pole and lying in the plane of area".
6.2.2
Parallel axes theorem
Consider the plane area A as shown in figure 6.3.2 (a). Area moment of inertia of the
plane area about x axis is given as :
Ix =
y
2
dA
Consider an axis passing through centroid and
dA
parallel to x-axis as shown in figure. Perpendicular
distance between the two axes is 'k'. It is clear from
y
y
the figure that distance of differential area 'dA' from
X-axis, y = y + k
Ix =
 (y  k)
2
dA =
k
y
2
dA  k 2 dA  2k y dA


In above expression,

y
2
x
C
x
Figure 6.3.2 (a)
dA represents the area moment of inertia of the plane area about centroidal
axis X , i.e. I X .
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
 dA

 y dA is the first moment of area about centroidal
is the total area of the plane figure, i.e. 'A' and
X – axis. But first moment
of area about centroidal axis is always zero.
Ix = I x + Ak2
Therefore
.......6.2.2(i)
Above expression provides a method to transfer the moment of inertia between two
axes one of which passes through the centroid and both are parallel. Expression is
a mathematical form of an important theorem of moment of inertia called parallel
axes theorem. Parallel axes theorem may be stated as follows.
"Moment of inertia of a plane area about an axis (say x) equals the sum of moment
of inertia of the area about centroidal axis parallel to the axis x and area times
square of the distance between two axes".
Note that if moment of inertia is to be transferred
from one axis (say x1) to another (x2), neither of
x2
which passes through the centroid, then first, it is
k2
transferred from x1 to x -axis then from x to x2-axis
as shown and explained below.
I x = Ix1 –
Ak12
= I x = Ix2 –
x
G
k1
Ak22
x1
Figure 6.2.2(b)
Ix2 = Ix1 – Ak12 + A k22
6.3
MOMENT OF INERTIA OF PLANE SECTIONS
Using definition of moment of inertia as well as parallel and perpendicular axes
theorems, expressions of moment of inertia of various standard sections viz
rectangular, triangular etc will be derived in following articles. Following steps are
useful for deriving moment of inertia of a plane section.
Step (i)
Choose a suitable differential element. Find its area 'dA'.
Step (ii)
Find second moment of area 'dA' about an axis [about which moment of
inertia is required]. Call second moment of such differential area as dI.
Step (iii) Use suitable limits to integrate 'dI' to cover the whole area. Thus find I.
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6.3.1
Rectangular Section
Consider a rectangular area having width 'b' and depth 'd'
units. Now moment of inertia of this area about centroidal
axis x is found as below.
d
2
dy
y
x
Step (i)
Choose a rectangular strip of depth 'dy' as a
differential element. It is clear from the figure
d
2
that area of the strip dA = ‘b dy’
Step (ii)
x
b
Now the second moment of area 'dA' is y2 dA
2
Figure 6.3.1(a)
2
about x -axis thus dI x = y dA = by dy
Step (iii) Integrate dI x from y = -d/2 to +d/2 to cover the
whole area
d/2
d/2
Ix =

 y 
b d3
– d3 

–



3  8
 
   – d/2
by 2 dy = b 
-d/2
Ix=
y
y
b d 3
bd 3

24

Similarly,
Iy =

IX 
b/2
bd 3
12
bd 3

x
Figure 6.3.1(b)
Moment of inertia of rectangular area about its base can be derived with the help of
parallel axes theorem.

bd 3
d2
bd 3
d
 b. d

IX = I x + A   =

4

2

Ix =
bd 3

Similarly, moment of inertia of rectangular area about axis – y is given as
Iy =
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
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y
Example 6.1: Determine the moment of inertia
of the rectangular area shown in figure about
60 cm
k
Solution: Moment of inertia of the plane
rectangular area about its centroidal axis
50 cm
parallel to x is given as
Ix
x
10 cm
x'–axis.
x
20 cm
x
bd 3
60 (10) 3


= 5000 cm3

12
Figure: Ex-6.1(a)
Using parallel axes theorem to transfer it to the x' axis,
I x' = I x + A k2 = 5000 + (60 × 10) (35)2
Ix' = 740,000 cm4
6.3.2
Ans.
10 


K  50   – [20]

2

  35 cm
Triangular Section
A triangular area is shown in figure 6.3.2(a). 'b' and
C
'h' are base and height of triangle respectively.
M
Our aim is to determine moment of inertia of the plane
triangular area about (a) its base and (b) centroidal
axis parallel to the base. First, we will determine
dy
h
y
A
moment of inertia about base AB.
b
B
x
Figure 6.3.2(a)
(a) Moment of inertia of triangle about base AB.
Step (i):
N
Select a narrow strip (MN) as a differental area. In limit, we can
treat it as a rectangular area because its dimensions are very small.
Area dA of the differential strip is given as
dA = (MN) dy

 By similar triangles ABC and MNC,

h–y
 MN


AB
h


h–y
 AB.
 MN  
 h 

 h – y 

 (AB)  dy
dA = 
 h 

h–y
 b dy
dA = 
 h 
Step (ii): Determine second moment of area 'dA' about line AB i.e. x–axis.
dIx = y2 dA =
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h
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Step (iii): Integrate dIx from y = 0 to y = h to obtain moment of inertia Ix of the
whole area about base AB, i.e. x–axis.
yh
Ix =

dI x 
y0
b
h
yh
 (hy
2
– y 3 ) dy
y0
yh

y4 
b  hy 3
b
Ix 
–



h  3
4 y0
h

Ix 
b
h
 h4
h4

–
 3
4





 h4 
bh 3
  
  

 
Now, we will use parallel axes theorem to transfer Ix to a centroidal axis
parallel to the base as follows.
(b) Moment of inertia of triangle about centroidal axis parallel to base : We know
that centroid of a triangle lies at
h
Ix = I x + A  
3
h
from its base. Using parallel axis theorem,
3

x
h/3
2
6.3.3
3

h 
bh
 bh 
I x = Ix – A      –  2 


 

Ix =
2
h

 





x
Figure 6.3.2(b)
bh 3

Circular Section
y
(a) Moment of inertia of a plane circular area : A plane
circular area is shown in figure 6.3.3(a). Diameter of
O
the circular area is 'd'. First, we will determine
x
moment of inertia of the area about an axis
perpendicular to figure and passing through centre.
Then, we will use perpendicular axes theorem to
determine moment of inertia about diameteral axis.
Step (i):
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d
Figure 6.3.3(a)
Select an annular ring at a radius 'a' and radial width 'da'
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Area dA of the annular ring is given as;
da
dA = 2  ada.
a
Step (ii): Second moment of area 'dA' about pole 'O' is
determined as ;
r
O
d Iz = a2 dA = 2  a3 da.
Step (iii): Integrate dIz from a = 0 to a = r to obtain the polar
moment of inertia (Iz) of the whole area.
a r
Iz =

a r
a r
dI z 
a 0
Figure 6.3.3(b)
a4 
2  a 3 da  2  
 4  a 0
a 0


Iz 
2 4
r 4
(r ) 
4
2

Iz 
 d
d 4
  
2 2
32
4
Step (iv): Apply perpendicular axes theorem to obtain moment of inertia about
diameteral axis. Ix + Iy = Iz but Ix and Iy are equal by symmetry, therefore,


2I x  I z
 Ix 
Ix 
d 4 / 32
4
Ix 
r 4
4

Iz
2
d 4
64

Ix 
 d4
64
Now, we will determine moment of inertia of semicircular and quarter
circular area in following sections.
(b) Moment of inertia of a semicircular area about its
y
(diameteral) base.
da
A semicircular area of diameter d (radius 'r') is shown in
figure 6.3.3 (c). First, polar moment of inertia of
semicircular area is determined.
a
O
d
x
Figure 6.3.3(c)
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Step (i):
Select a semi annular ring (as shown in figure) as a differential area. Area
'dA' of the ring is given as,
dA =  a da
Step (ii): Second moment of area of differential area 'dA' is given as ;
dIz = a2 dA =  a3 da.
Step (iii): Integrate dIz from a = 0 to a = r to obtain
a r
a r
a r
 a4 
a da    
Iz = dI z  
 4  a 0
a 0
a 0


Ix 
3

 (r 4 )
d 4

4
64
Step (iv): Apply perpendicular axes theorem to obtain moment of inertia (Ix) about
diameteral axis, Ix + Iy = Iz. By symmetry, Ix and Iy are equal, 2Ix = Iz
(c)

Ix 
or
Ix 
Iz
d 4

2
128

Ix 
d 4
128
r 4
8
Moment of inertia of a semicircular area about centroidal axis parallel to the
base.
As we know, centroid of a semicircular area lies at
r
from the base [see figure
3
6.3.3(d)]. We can transfer the moment of inertia of the area from base to
centroidal axis by using parallel axes theroem.
 r 

Ix = I x  A 
 3 
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

  r  
r 4
16

I x  I x – A 

–
 
8
92
 9 

 8 
Ix  r4  –

  9 
or
1 
 
Ix  d4 
–

  1 
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 r 2  2


   r


c
x
4r/3
x
d
Figure 6.3.3(d)
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(d): Moment of Inertia of a quarter circular area about
any of its two bases.
y
A quarter circular area is shown in figure 6.3.3(e).
 d
Its Radius is ‘r’    . Moment of inertia of the area
 2
about x-axis is determined as follows. [By
symmetry, Ix and Iy must be equal]
Step (i)
da
a
O
d
2
x
Figure 6.3.3(e)
Select a quarter annular ring as a differential
area ‘dA’. [see figure 6.3.3(e)]
Area dA is given as :
dA =
Step (ii)

ada
2
Second moment of area of differential area dA about pole O is obtained as
dI z = a2 dA =
 3
a da
2
Step (iii) Integrate dIz from a = 0 to a = r to obtain polar moment of inertia of the
whole area.
ar
Thus,



a 3 da  [a 4 ]r0
2 a0
8
Iz =

Iz =
r 4 d 4

8
128
dI z 

Step (iv) Apply perpendicular axes theorem to obtain moment of inertia about
x-axis.
Ix + Iy = Iz  2Ix =

(e)
Ix =
r 4
8
r 4 d 4

16 256
Moment of inertia a quarter circular area about a centroidal axis parallel to
its base.
We will use parallel axis theorem to transfer the moment of inertia from its base
to centroidal axis. Distance between the two axes is
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CHAPTER - 6 : MEMENT OF INERTIA
4r
Ix = I x  A  
 3 

 r 
Ix  Ix –  
 4
2
y
2
 16r 2  r 4
4 4
 2 
 9  16 – 9 r


b
a

4  4
 
Ix   –
r
16
9


x
Figure-Ex-6.2(a)
Example 6.2 : Determine moment of inertia of area of
elliptical quadrant about x and y axes as shown in
figure-Ex-6.2(a).
y
Solution : Moment of inertia of elliptical quadrant
about x-axis.
M
b
N
dy
x
y
x
Step (i): Select a narrow strip of differential area ‘dA’
parallel to x-axis as shown in figure-Ex6.2.(b).
a
Figure-Ex-6.2(b)
In limit narrow strip MN can be treated as a rectangular area. As shown
in figure, its distance from x-axis is ‘y’ and width is ‘dy’.
Differential area ‘dA’ of the strip is given as
dA = x dy

dA =
a
b 2 – y 2 dy
b
 for ellipse, point N must satisfy


2
 x = a 1 – y2
b

x2 y 2

=1
a 2 b2
Step (ii): Second moment of area of differential area ‘dA’ about x axis is ‘y2 dA’

dIx = y2 dA =
a 2
y
b
b 2 – y 2 dy
Step (iii): Integrate dIx from y = 0 to y = b to obtain moment of inertia of the whole
area,
yb
a
dI x 
Ix =
b
y0

yb
y
2
b 2 – y 2 dy
y0
b

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– y
b2 
2
2
2
–1 y 
(b 2 – y 2 ) 3 
Ix = 
 y b – y  b sin

8 
b  0
 4
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


b2 
b2
     

0

0

b2
–
0

(0  0)  





Ix = 


8 
8
 2  
 


Ix =
a  b 4  ab 3


b  16 
16
Determining moment of inertia of elliptical quadrant about y-axis is left to the
reader. It is advised to take a narrow strip parallel to y-axis as a differential area and
proceed in the similar way as before. Thus it can be found that Iy =
a 3 b
16
Following table summarizes area moments of inertia of standard sections.
Table 6.1 : Area moment of inertia of Plane figures
S.No.
Figure
Area moments of inertia
y
1.
3
I x  bd /3, I y  db 3 /3
y
I x  bd 3 /12, I y  db 3 /12
d
x
x
b
Rectangular area
2.
I x  bh 3 /12
x1
I x  bh 3 /36
x
h
b
Triangular area
3.
I x1  bh 3 /4
x
Ix  Iy 
y
x
d
Iz 
z
d 4
32
Circular area
4.
d 4 r 4

128
8
1  4  8  4
 
Ix  
–
d  –
r
 128 18 
 8 9 
Ix 
x
x
d
= 2r
Semicircular area
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5.
d 4 r 4

256 16
4  4
 
Ix   –
r
 16 9  
y
Ix 
x
x
Quarter circular area
ab 3
a 3 b
, Iy 
16
16
4  3
 
Ix   –
 ab
 16 9  
y
6.
Ix 
y
x
b
a
Area of elliptical quadrant
x
4  3
 
Iy   –
a b
 16 9 
Try Yourself
Q.6.1:
Ans.
Q.6.2:
Ans.
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Determine the moment of inertia of the area of elliptical quadrant shown in figure of
example 6.2 about centroidal axes parallel to the base x–y axes.
4  3

 
I x   16 – 9  ab





4
 3
I   –
a b
y
 16 9 

Determine area moment of inertia of the
subparabolic area shown in figure about x and y
axes.

ab 3
I

 x
21

3
I  a b
y

5
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b
x
a
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6.3.4
Composite Section
A Section made up of a no of standard sections
is called composite section. A typical composite
section is shown in figure 6.3.4(a). It comprises
of a rectangular and a triangular area minus a
semicircular area.
The moments of inertia of a composite area
about a particular axis is algebric sum of
moment of inertia of its components.
Therefore–
y
20 cm
20 cm
10 cm
x
15 cm
20 cm 10 cm
Figure 6.3.4(a)
Ix = I1x + I2x – I3x for the composite area considered for instance. I1x, I2x and I3x are
moment of inertia, about x–axis, of area 1 (rectangle), area 2 (triangle) and area 3
(semicircle).
40 cm
20 cm
1
2
3
x
20 cm
55 cm
10 cm
Figure 6.3.4(b)
I1x = Moment of inertia of rectangular area about x–axis i.e. its base
=
( ) 
= 1,173,333.33 cm4

I2x = Moment of inertia of triangular area about x–axis i.e. its base
=
( ) 
= 6666.67 cm4

and I3x = Moment of inertia of semicircular area about x–axis
 r4
 () 

=
= 3926.99 cm4
8

Now, moment of inertia of composite area about x–axis
Ix = I1x + I2x – I3x = 1176073.01 cm4
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Following examples (6.3 – 6.10) involve calculation of moment of inertia of
composite areas.
Example 6.3: Determine the moments of inertia of the shaded area shown in figure:
Ex-6.3(a) about the x and y-axes.
Solution: Moment of Inertia of the composite area about
x–axis.
y
The given composite area is made up of a rectangular area
minus a semicircular area.
Ix = I1x – I2x =

a (2a) 3
 (a ) 

= 4.94 a4
3

2a
a
Ans.
x
Figure: Ex-6.3(a)
Moment of inertia of the composite area about y–axis.
Iy = I1y – I2y =
a (2a) 3
– I 2y  A 2 a 2
3

16 4  4   a 2
= 3 a –  a –  2


y
 2
 a = 16 a 4 –  a 4  3.37a 4

3


Ans.
1
2a
2a
y
1
2a
2 a
2
x
2a
a
a
Figure: Ex-6.3(b)
Figure: Ex-6.3(c)
10
mm
Example 6.4: Determine the moment of inertia of the
L-section shown in figure: Ex-6.4(a) about centroidal
axes parallel to the legs.
Solution: In this type of problem, where it is required to
determine the moment of inertia of a composite area
about centroidal axis, we will first locate the centroid of
the composite figure. Following table helps locating the
centroid with respect ot x and y axes shown in figure
Ex-6.4(b).
125
mm
10 mm
85 mm
Figure: Ex-6.4(a)
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S.No. Area (A) (mm2)
1.
750
2.
1250
xc (mm)
10 +
yc (mm)
A xc (mm3)
Ayc (mm3)
5
35625
3750
125
= 62.5
2
6250
78125
 Axc = 41875
 Ayc = 81875
75
= 47.5
2
5
 A = 2000
Now, determine the coordinates of centroid as
y
y
x 

 Ax C
= 20.94 mm and y = 40.94 mm
A
Moment of inertia of composite area about
125
mm
(20.94, 40.94)
2
x
centroidal x axis [see figure: Ex-6.4(c)]
I x = I1 x + I2 x , Now, using parallel axes
theorem, we get
2
 I x = [I1(x1) + A1 ( y – y1) ] +
x
75 mm
10
mm
[I2(x2) + A2 (y2 – y )2]
Figure : Ex-6.4(b)
 75 (10) 3
2
=  12  750 (35.94)  


x2
(y2 – y)
2
x
(y – y1)
 10 (125)3

 1250 (21.56)2 

 12

x1
1
y
Ans.
y2
Moment of inertia of composite area about
centroidal y axis [see figure: Ex-6.4(d)]
I y = I1 y + I2 y , Now, using parallel axes
theorem, we get
x2 = 5
y
(x – x2)
2
(x1 – x)
 I y = [I1(y1) + A1 (x1 – x )2] + [I2(y2) + A2 ( x – x2)2]
y1
 10 (75)3
2
=  12  750 (26.56)  


1
x1 = 47.5
 125 (10) 3

 1250 (15.94) 2  = 1,208,658.87 mm4 Ans.

 12

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Figure : Ex-6.4(c)
= 3,183,658.87 mm4

10 mm
1
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EXERCISES
(I)
'Objective' type Questions
6.1
6.2
Area moment of inertia is also called :
(a)
First moment of area.
(b)
Second moment of area.
(c)
Moment of couple.
(d)
None of these.
Polar moment of inertia means;
(a)
Area moment of inertia of a plane figure about an axis passing
through its centroid.
(b)
Area moment of inertia of a plane figure about an axis passing
through its centroid and perpendicular to it.
(c)
Area moment of inertia of a plane figure about any axis which is not
in its plane.
(d)
6.3
None of these.
Area moments of inertia of a rectangular area (b × d) about an axis through
its base and an axis through centroid but parallel to the base are ;
6.4
6.5
(a)
 3
bd ,


bd3

(b)

bd3,


bd3

(c)
 2
bd ,


bd2

(d)
None of these
Polar moment of inertia of a plane circular figure is ;
(a)
 4
d

(b)
 d3

(c)
 4
d

(d)
 r4

Area moment of inertia of a plane circular figure about any diameteral
axis is;
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(a)
d4

(b)
 r4

(c)
r4

(d)
 d4

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6.6
6.7
Parallel axes theorem is also called ;
(a)
Transfer of axes theorem
(b)
Perpendicular axes theorem
(c)
First theorem of area moment of inertia
(d)
None of these.
Radii of Gyration of a plane rectangular area are;
(a)
(b)
(c)
(d)
6.8
6.9
6.10
6.11
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d
 3
d
 2
,
,
b
2 3
b
3 2
d
,
12
d
 3
,
1
,
2
b2  d2
6
,
1
,
3
b2  d2
2
b2  d2
12
b
,
12
,
b
2 3
,
1
,
2
b2  d2
3
Set of co-ordinate axes of a plane figure, about which area moments of
inertia of the figure are maximum and minimum, are called :
(a)
polar axes
(b)
principal axes
(c)
rectangular axes
(d)
None of the above.
Mass moment of inertia depends upon :
(a)
Distribution of area
(b)
distribution of forces
(c)
velocity
(d)
distribution of mass
Unit of mass moment of inertia is
(a)
kg – m2
(b)
g – cm2
(c)
Neither (a) nor (b)
(d)
Both (a) and (b).
Area moment of inertia and mass moment of inertia of a body :
(a)
are always same
(b)
May be same
(c)
are always different
(d)
None of these.
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6.12
6.13
Mass moment of inertia of a slender rod of mass m about a centroidal axis
perpendicular to its length l is
(a)
Ml 2
12
(b)
Ml 2
3
(c)
5 Ml 2
12
(d)
Ml 2
2
Mass moments of inertia of a thin rectangular plate about centroidal x and
y axes lying in its plane are - (Mass, breadth and depth of the plate are M, b
and d respectively)
(II)
(a)
Md 2
Mb 2
and
12

(b)
Md 2
Mb 2
and
3

(c)
2Md 2
2Mb 2
and
3

(d)
M (b 2  d 2 )
M b2  d2 )
and
12

Fill in the Blank Type Questions
6.14
Area moment of inertia of a semicircular area of diameter d about a
centroidal axis parallel to its base is -----------------
6.15
------------------- theorem states that area moment of inertia of a plane figure
about a polar axis is equal to the sum of moments of inertia of the same area
about any two orthogonal axes intersecting at pole and lying in the plane of
figure.
6.16
Area moment of inertia of a plane figure is always a ------------------quantity.
6.17
---------------------- shows the distribution of an area in a plane with y respect
to a reference axis.
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6.18
dA
For a plane area as shown in figure,
y
Ix = I x + AK2 is mathematical expression
of ------------------------------------ theorem.
A
x
k
x
6.19
A triangle has base b and height h. Its moment
of inertia about base is --------------------- and about a centroidal axis parallel
to base is ------------------------.
6.20
Mass moment of inertia of a right circular cylinder of mass M and radius r
about its axis is ----------------------.
(III)
"Matching" type question.
6.21
(i)
A plane rectangular area.
(p)
 8  4
 –
 r
  9 
(ii)
A plane triangular area.
(q)
bd 3
12
(r)
bh3/36
(iii)
6.22
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A semicircular area.
(iv)
A quarter circular area.
(s)
4  4

 –
r

9


(i)
Right circular cylinder.
(p)
M 2
r
12
(ii)
Homogeneous solid sphere.
(q)
3
Mr2
10
(iii)
Right circular cone.
(r)
M 2
r
2
(iv)
Rectangular parallelepiped
(s)
2
Mr2
5
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6.23
d
(i)
D
(p)
 BD 3 – bd 3 




12


D
(q)
BD 3 bd 3
–
12
3
(r)
BD 3 – 2bd 3
12
(s)
BD 3 – 8bd 3
12
b
B
(ii)
d
b
B
d=
b
(iii)
D
2
D
B
B
(iv)
6.24
6.25
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b
D
b
D
d=
2
Ix
A
(i)
Parallel axes theorem.
(p)
(ii)
Perpendicular axes theorem.
(q)
(iii)
Radius of gyration.
(r)
(iv)
Principal moments of inertia.
(s)
Ix = I x + AK2.
Iz = Ix + Iy.
(i)
Area moment of inertia
(p)
 x y dA
(ii)
Mass moment of inertia
(q)
y
Ix  Iy

2
(iii)
Product of inertia
(r)
r
2
(iv)
Polar moment of inertia
(s)
r
2
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(IV)
"True/false" type questions.
6.26
The moment of inertia of a body depends upon angular velocity of body.
6.27
bh 3
The moment of inertia of a triangle about it's base is
36
6.28
The moment of inertia of a square about it's side is a4/3.
6.29
The perpendicular axis theorem shows that I = IG + Ah2.
6.30
The mass moment of inertia of a sphere about an axis passing through its
centre is
(V)

MR2 .

Theoretical questions
6.31
6.32
6.33
Write short notes on –
(i)
Second moment of area
(ii)
Radius of gyration
(iii)
Polar moment of inertia
State and prove
(i)
Parallel axes theorem and
(ii)
Perpendicular axes theorem as applied for area moment of inertia.
Prove that area moment of inertia of a rectangular section about its
centroidal axes are
bd 3
db 3
and
respectively' b and d being the breadth


and depth of the section.
6.34
Derive an expression for area moment of inertia of a triangular section
about an axis passing through centroid and parallel to the base. Take base
and height of the triangle as 'b' and 'h' respectively.
6.35
Define and explain "mass moment of inertia" and "parallel axes theorem" for
it.
6.36
Derive an expression for mass moment of inertia of following homogeneous
bodies about centroidal axes.
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(i)
A cylinder
(ii)
A sphere
(iii)
A cone
(iv)
A parallelepiped.
LUCKNOW
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AGRA
ALLAHABAD
9793424360
9919751941
PATNA
9534284412
NOIDA
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BHOPAL
9838004479
JAIPUR
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25
TECHGURU CLASSES for SSC-JE / RAILWAYS / ORDINANCE
CHAPTER - 6 : MEMENT OF INERTIA
6.37
(VI)
Derive an expression for area moment of inerfia of
(i)
a circle
(ii)
a semicircle
(iii)
a quarter circle about centroidal axes.
Numerical Questions
6.38
y
The shaded area shown in figure is equal to 50 cm2.
Determine its centroidal moments of inertia I x and I y .
The polar moment of inertia of the area about point A
shown in figure is 22.5 × 102 cm4 and 2 I y = I x .
x
C
d=6 cm
A
Figure: 6.38
6.39
Determine the moment of inertia of the shown
area about line AB.
B
A
120
mm
200mm
Figure: 6.39
6.40
Determine moment of inertia of the plane area
3m
shown in figure about an axis through its centroid
and parallel to the base.
3m
3m
2cm
10cm
Figure: 6.40
ANSWERS
6.1
(b)
6.2
(b)
6.3
(a)
6.4
(a)
6.5
(a)
6.6
(a)
6.7
(d)
6.8
(b)
6.9
(d)
6.10
(d)
Classroom Study Course
U
L
C
K
N
O
W
Correspondance
Study Material
Classroom & Online Test Series
Foundation Batches also for 2nd & 3rd Year sturdents
Regular & Weekend Batches
Interview Guidance
26
TECHGURU CLASSES for SSC-JE/RAILWAYS/ORDINANCE
CHAPTER - 6 : MEMENT OF INERTIA
(b)
6.15
Perpendicular axes theorem
6.16
6.17
Radius of gyration
6.18
Parallel axes theorem
6.19
bh 3 bh 3
,


6.20
Mr 2

6.21
(q), (r), (p), (s)
6.22
(r), (s), (q), (p)
6.23
(r), (s), (p), (q)
4.24
(r), (s), (p), (q)
6.25
(q), (s), (p), (r)
6.26
False
6.27
False
6.28
True
6.29
False
6.30
True
6.38
I x = 300 cm4. I y = 150 cm4.
6.39
34.21 × 106mm4.
6.40
182.4 cm4.
NEW DELHI
8860637779
LUCKNOW
9919526958
6.12
(d)
6.13
AGRA
ALLAHABAD
9793424360
9919751941
(a)
PATNA
9534284412
6.14
 8  4
 –
r
  9 
6.11
positive
NOIDA
8860637779
BHOPAL
9838004479
JAIPUR
GORAKHPUR KANPUR
9793424360
9838004494 9838004497
27