Engineering Mechanics

Chapter 6 : Centre of Gravity „„„„„ 85

Example 6.5. A uniform lamina shown in Fig. 6.14 consists of a rectangle, a circle and a
triangle.

Fig. 6.14.

Determine the centre of gravity of the lamina. All dimensions are in mm.

Solution. As the section is not symmetrical about any axis, therefore we have to find out the

values of both x and yfor the lamina.

Let left edge of circular portion and bottom face rectangular portion be the axes of reference.

(i) Rectangular portion

a 1 = 100 × 50 = 5000 mm^2

1

100

25 75 mm

2

x=+ =

and 1

50

25 mm

2

y ==

(ii) Semicircular portion

22 2

ar 2 22 (25) 982 mm

ππ

=× = =

2

4425

25 – 25 – 14.4 mm

33

r

x

×

== =

ππ

and 2

50

25 mm

2

y ==

(iii) Triangular portion

2

3

50 50

1250 mm

2

a

×

==

x 3 = 25 + 50 + 25 = 100 mm

and 3

50

50 66.7 mm

3

y =+ =
We know that distance between centre of gravity of the section and left edge of the circular
portion,

11 2 2 3 3

123

ax a x a x

x

aa a

++

=

++

(5000 75) (982 14.4) (1250 100)

5000 982 1250

×+ × + ×

=

++

= 71.1 mm Ans.