Chapter 7 : Moment of Inertia 113

We know that moment of inertia of rectangle (1) about an axis through its centre of gravity

and parallel to X-X axis,

3

34

1

`0 (20)`

40 10 mm

G 12

I

6×

==×

and distance between centre of gravity of rectangle (1) and X-X axis,

h 1 = 130 – 60.8 = 69.2 mm

∴ Moment of inertia of rectangle (1) about X-X axis,

23 2 34

=+ =× + ×IahG 111 (40 10 ) [1200 (69.2) ]= ×5786 10 mm

Similarly, moment of inertia of rectangle (2) about an axis through its centre of gravity and

parallel to X-X axis,

3

34

2

`20 (100)`

1666.7 10 mm

12

`IG`

`×`

==×

and distance between centre of gravity of rectangle (2) and X-X axis,

`h 2 = 70 – 60.8 = 9.2 mm`

∴ Moment of inertia of rectangle (2) about X-X axis,

(^) =+ =IahG 222 23 234(1666.7 10 )× + ×[2000 (9.2) ]= ×1836 10 mm

Now moment of inertia of rectangle (3) about an axis through its centre of gravity and parallel

to X-X axis,

3

34

3

100 (20)

66.7 10 mm

G 12

I

×

==×

and distance between centre of gravity of rectangle (3) and X-X axis,

h 3 = 60.8 – 10 = 50.8 mm

∴ Moment of inertia of rectangle (3) about X-X axis,

= IahG 333 +=×+ ×^23 (66.7 10 ) [2000 (50.8) ]^2 = 5228 × 10^3 mm^4

Now moment of inertia of the whole section about X-X axis,

IXX = (5786 × 10^3 ) + (1836 × 10^3 ) + (5228 × 10^3 ) = 12 850 × 10^3 mm^4 Ans.

Example 7.12. Find the moment of inertia about the centroidal X-X and Y-Y axes of the

angle section shown in Fig. 7.16.

Solution. First of all, let us find the centre of gravity of the section. As the section is not

symmetrical about any section, therefore we have to find out the values of xand y for the angle

section. Split up the section into two rectangles (1) and (2) as shown in Fig. 7.16.

Moment of inertia about centroidal X-X axis

Let bottom face of the angle section be the axis of reference.

Rectangle (1)

a 1 = 100 × 20 = 2000 mm^2

and (^1)

100

50 mm

2

y ==

Rectangle (2)

a 2 = (80 – 20) × 20 = 1200 mm^2

and 2

20

10 mm

2

y ==

Fig. 7.16.