Chapter 35 : Hydrostatics „„„„„ 735

∴Total pressure on the plate,

P=P 1 – P 2 = 369.5 – 19.2 = 350.3 kN^ Ans.

Centre of pressure

Let h = Depth of centre of pressure of the plate from the water

surface.

We know that the moment of inertia of the main circular section about its centre of gravity,

IG 1 =

(4)^444 m

64

π

=π

Similarly, IG 2 =

(1)^44 m

64 64

ππ

=

∴Depth of centre of pressure of the main plate from the water surface,

h 1 =

2 2

1

1

11

sin 4 sin 30º

3

43

IG x

Ax

θ π

+= +

π× =

(0.5)^2

3

3

+ = 3.08 m

Similarly, h 2 =

2 2

2

2

22

sin sin 30º

(^64) 2.5
0.25 2.5
IG x
Ax
π
θ
+= +
π×

(0.5)^2
2.5
40

• = 2.5 m
  Now taking moments about the water surface and equating the same,
  350.3×h = (369.5 × 3.08) – (19.2 × 2.5) = 1090
  ∴ h =
  1090
  3.11 m
  350.3
  =^ Ans.
  35.14.PRESSURE DIAGRAMS
  A pressure diagram may be defined as a graphical representation of the variation in the inten-
  sity of pressure over a surface. Such diagrams are very useful for finding out the total pressure and the
  centre of pressure of a liquid on the vertical surface (i.e., wall or dam etc.). A vertical surface may be
  subjected to the following types of pressures :

1. Pressure due to one kind of liquid on one side,


2. Pressure due to one kind of liquid, over another, on one side, and


3. Pressure due to liquids on both the sides.
   Now we shall discuss the above three cases, one by one.

35.15.PRESSURE DIAGRAM DUE TO ONE KIND OF LIQUID ON ONE SIDE

Consider a vertical wall subjected to pressure due to one
kind of liquid, on one of its sides as shown in Fig. 35.16.

Let H = Height of liquid
w = Specific weight of the liquid and
l = Length of wall.
We know that the intensity of pressure on the wall will be
zero at the liquid surface, and will increase by a straight line law
to (wH) at the bottom.

Therefore the pressure diagram will be a triangle ABC as
shown in Fig. 35.19.

Fig. 35.19. Pressure diagram due

to one kind of liquid.