Chapter 7 : Moment of Inertia 115

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

parallel to Y-Y axis,

`3`

64

2

`20 (60)`

0.36 10 mm

G 12

I

×

==×

and distance of centre of gravity of rectangle (2) from Y-Y axis,

h 2 = 50 – 25 = 25 mm,

∴ Moment of inertia of rectangle (2) about Y-Y axis

=+ = ×+ × =×IahG 22226 0.36 10 [1200 (25) ]2 641.11 10 mm

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

IYY = (0.517 × 10^6 ) + (1.11 × 10^6 ) = 1.627 × 10^6 mm^4 Ans.

Example 7.13. Figure 7.17 shows the cross-section of a cast iron beam.

Fig. 7.17.

Determine the moments of inertia of the section about horizontal and vertical axes passing

through the centroid of the section.

Solution. As the section is symmetrical about its horizontal and vertical axes, therefore

centre of gravity of the section will lie at the centre of the rectangle. A little consideration will show

that when the two semicircles are placed together, it will form a circular hole with 50 mm radius or

100 mm diameter.

Moment of inertia of the section about horizontal axis passing through the centroid of the section.

We know that moment of inertia of the rectangular section about its horizontal axis passing

through its centre of gravity,

33

120 (150) 33.75 10 mm 64

12 12

`bd ×`

== = ×

and moment of inertia of the circular section about a horizontal axis passing through its centre of

gravity,

`( )^44 (50) 4.91 10 mm^64`

44

`r`

`ππ`

== =×

∴ Moment of inertia of the whole section about horizontal axis passing through the centroid

of the section,

IXX = (33.75 × 10^6 ) – (4.91 × 10^6 ) = 28.84 × 10^6 mm^4 Ans.