Chapter 7 : Moment of Inertia „„„„„ 105

4

44

0

2()()

42 32

r

ZZ

x

Ird

⎡⎤ ππ

=π⎢⎥= =

⎣⎦

... substituting

2

d

r

⎛⎞

⎜⎟=

⎝⎠

We know from the Theorem of Perpendicular Axis that

IXX + IYY = IZZ

∴ *

(^1) () (^44) ()
2 2 32 64
ZZ
XX YY
I
II d d
ππ
== =× =
Example 7.3. Find the moment of inertia of a circular section of 50 mm diameter about an
axis passing through its centre.
Solution. Given: Diameter (d) = 50 mm
We know that moment of inertia of the circular section about an axis passing through its
centre,
()^4434 (50) 307 10 mm
XX 64 64
Id
ππ
==×=× Ans.
7.11.MOMENT OF INERTIA OF A HOLLOW CIRCULAR SECTION
Consider a hollow circular section as shown in Fig.7.6,
whose moment of inertia is required to be found out.
Let D = Diameter of the main circle, and
d = Diameter of the cut out circle.
We know that the moment of inertia of the main circle
about X-X axis
()^4
64
D
π

and moment of inertia of the cut-out circle about X-X axis
()^4
64
d
π

∴ Moment of inertia of the hollow circular section about X-X axis,
IXX = Moment of inertia of main circle – Moment of inertia of cut out circle,
()– ()^4444 ( – )
64 64 64
Dd Dd
πππ

Similarly,
(–)^44
YY 64
IDd
π

Note : This relation holds good only if the centre of the main circular section as well as that
of the cut out circular section coincide with each other.

• This may also be obtained by Routh’s rule as discussed below
  XX 4
  I =AS
  (for circular section)
  where area, Ad 4 2
  =×π
  and sum of the square of semi axis Y-Y and Z-Z,
  (^22)
  0
  24
  S=+=⎛⎞dd
  ⎜⎟⎝⎠
  ∴
  2
  2
  (^44) () 4
  XX 4464
  d d
  IdAS
  ⎡⎤π××
  ⎢⎥⎣⎦ π
  == =
  Fig. 7.6. Hollow circular section.