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.