Engineering Mechanics

Chapter 35 : Hydrostatics „„„„„ 737

The pressure P 1 on the surface AD, due to liquid 1, may be found out as usual from the area of
triangle ADE. The pressure on the surface DB will consist of pressure P 2 due to superimposed liquid 1,
as well as pressure P 3 due to liquid 2.

This pressure will be given by the area of the trapezium BCED (i.e. area of rectangle BFED
due to superimposed liquid i.e. P 2 = w 1 H 1 × H 2 and the area of triangle FCE due to liquid 2).
The total pressure P will be sum of these three pressures (i.e. P = P 1 + P 2 + P 3 ). The line of action
of the total pressure may be found out by equating the moments of P, P 1 , P 2 and P 3 about A.

Note: From the geometry of the figure, we find that

P 1 =

2

11

2

wH

Similarly, P 2 =w 1 H 1 × H 2 and P 3 =

2

22

2

wH

Example 35.15. Find the magnitude and line of action of the resultant force exerted upon
the side of a box tank, which is 10 m square and 1 metre deep. The box tank is half filled with a liquid
having specific gravity of 2, while the remainder is filled with a liquid having a specific gravity of 1.

Solution. Given: Side of the square tank = 10 m and depth of tank = 1 m.

Magnitude of the resultant force (i.e., pressure)

The pressure diagram on one side of tank is shown in Fig. 35.19. We know that the
pressure DE

=w 1 H 1 = (9.8 × 1) × 0.5 = 4.9 kN/m^2

Similarly, pressure FC =w 2 H 2 = (9.8 × 2) 0.5 = 9.8 kN/m^2

∴ Pressure per metre length of the tank

P= Area of figure ADBCE

= Area of triangle ADE + Area of rectangle BFED

+ Area of triangle FCE

=

1

(4.9 0.5) (0.5 4.9)

2

×+× 2

1

(9.8 0.5) kN/m

2

+×

= 6.125 kN/m^2

and magnitude of the total pressure on 10 m long wall,

= 10 × 6.125 = 61.25 kN Ans.

Line of action of the resultant force (i.e., pressure)

Fig. 35.22.

Let h = Depth of the line of action of the resultant pressure from A.