(^728) „„„„„ A Textbook of Engineering Mechanics
Example 35.9. An isosceles triangle of base 3 metres, and altitude 6 metres, is immersed
vertically in water, with its axis of symmetry horizontal as shown in Fig. 35.11.
Fig. 35.11.
If the head of water, on its axis is 9 metres, locate the centre of pressure both vertically and
laterally.
Solution. Given: Base of the triangle (b) = 3 m; Altitude of the triangle (h) = 6 m and
head of the water on its axis () 9m.x =
Vertical location of centre of pressure
We know that area of the triangular plate
A=^2
36
9m
22
bh ×

Since the plate is lying with its symmetrical axis, parallel to the water surface, therefore mo-
ment of inertia of the plate about the axis AD will be obtained as discussed below :

1. Split up the plate into two triangles viz. ABD about ADC.


2. Find out the moments of inertia of the two triangles, ABD and ADC, separately, about the
   line AD.


3. Add the moments of inertia of the two triangles, which will give the required moment of
   inertia of the triangle ABC about the axis AD.
   We know that the moment of inertia of triangle ABD about AD

=

3

6(1.5) 1.6875 m 4

12

=

Similarly, moment of inertia of triangle ADC about AD

= 1.6875 m^4

∴ Moment of inertia of the triangle ABC about AD,

IG= 1.6875 + 1.6875 = 3.375 m^4

We also know that depth of centre of pressure of the plate from the water surface,

h =

3.375

99.04m

99

IG x

Ax

+= +=

×^

Ans.

Horizontal location of centre of pressure

The centre of pressure, in horizontal direction will coincide with the centre of the triangle.

Therefore centre of pressure will be at a distance of

6

3

= 2 m from BC. Ans.