(^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.