(^720) „„„„„ A Textbook of Engineering Mechanics
35.4.PRESSURE HEAD
We have already discussed in the last article, that when a liquid is contained in a vessel, it
exerts pressure on all the sides and bottom of the vessel. Now consider a vessel containing some
liquid as shown in Fig. 35.2. Now let a cylinder be made to stand in the liquid.
Let w= Specific weight* of the
liquid,
H= Height of liquid in the
cylinder, and
a= Cross-sectional area of the
cylinder.
We know that weight of the liquid in the cylinder,
=wHa
∴ Intensity of pressure at the bottom of the cylinder,
p= Weight
Area
wHa
wH
a

The above expression shows that the intensity of pressure at any point, in a liquid, is propor-
tional to the depth of the point below the liquid surface.
Note: The value of specific weight of water is generally taken as 9.8 kN/m^3
Example 35.1. Find the pressure at a point, which is 4 m below the free surface of water.
Take specific weight of water as 9.8 kN/m^3.
Solution. Given: Depth of the point below the free surface of water (H) = 4 m and specific
weight of water (w) = 9.8 kN/m^3.
We know that pressure at the point,
p=wH = 9.8 × 4 = 39.2 kN/m^2 Ans.
Example 35.2. Find the depth of oil of specific gravity 0.8 which will produce a pressure
of 50 kN/m^2.
Solution. Given: Specific gravity = 0.8 or specific weight (w) = 9.8 × 0.8 = 7.84 kN/m^3 and
intensity of pressure (p) = 50 kN/m^2.
Let H= Depth of oil.
We know that intensity of pressure of oil (p)
50 =wH = 7.84 H
∴ H=
50
6.38 m
7.84
= Ans.
35.5.TOTAL PRESSURE
The total pressure on an immersed surface, exerted by the liquid, may be defined as the gross
pressure acting on it. Mathematically, the total pressure may be found out by dividing the whole
immersed surface into a number of small strips. Now the total pressure,
P=p 1 a 1 + p 2 a 2 + p 3 a 3 + ...
where p 1 , p 2 , p 3 , ... = Intensities of pressure on different strips of the surface, and
a 1 , a 2 , a 3 , ... = Areas of corresponding strips.
Fig. 35.2. Pressure head.

• It is the weight per unit volume.