# (^92) A Textbook of Engineering Mechanics

and 1

120 300 2 200

56 mm

3300200

y

⎛⎞+×

=×⎜⎟=

⎝⎠+

(ii) Semicircle

2 222

11

(90) 4050 mm

22

ar=×π=×π× = π

and 2

4 4 90 120

mm

33

r

y

×

== =

πππ

We know that distance between centre of gravity of the section and AB,

11 2 2

12

120

(30 000 56) 4050

30 000 4050

×− π×⎛⎞

− ⎜⎟⎝⎠π

# −−π

ay a y

y

aa

mm

= 69.1 mm Ans.

Example 6.14. A semicircular area is removed from a trapezium as shown in Fig.6.24

(dimensions in mm)

Fig. 6.24.

Determine the centroid of the remaining area (shown hatched).

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

values of xand yfor the area. Split up the area into three parts as shown in Fig. 6.25. Let left face and

base of the trapezium be the axes of reference.

(i) Rectangle

a 1 = 80 × 30 = 2400 mm^2

1

80

40 mm

2

x==

and 1

30

15 mm

2

y==

(ii) Triangle

2

2

80 30

1200 mm

2

a

×

# 2

80 2

53.3 mm

3

x

×

and 2

30

30 40 mm

3

y =+ =

(iii) Semicircle

22 2

ar 3 22 (20) 628.3 mm

ππ

=× = =

Fig. 6.25.