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
12IG×
==×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.