Question 3.P.23: A square section tank of 3 m side and 2 m length as shown in......
A square section tank of 3 m side and 2 m length as shown in Fig. P. 3.23 has the top of one side wall in the shape of a cylinder as indicated. The tank is filled with water as indicated. Determine the horizontal and vertical forces on the curved surface. Also locate the line of action of the resultant force. The water is under a gauge pressure of 20,000 N/m².
The horizontal force is the force due to water pressure on the projected area. It can be split up into two components (i) due to the water column and (ii) due to the pressure on the fluid
The horizontal force = γ \bar{h} A + P A
= 9810 × 0.75 × (1.5 × 2) + 20,000 × (1.5 × 2) = 22072.5 + 60,000
= 82072.5 N (to the right)
The first component acts at the centre of pressure and the second at the centre of gravity.
Centre of pressure due to fluid pressure
= 0.75 + (1/12) (2 × 1.5³/ 0.75 × 2 × 1.5) = 1.0 m (from top).
Location of the net force is determined by taking moments about the top.
= {(22072.5 × 1) + (60000 × 0.75)} / (22072.5 + 60,000)
= 0.8172 m from top.
The vertical force can also be considered as the result of two action (i) the weight of displaced volume and (ii) the pressure on the projected area.
= 9810 [(120 × π × 1² × 2/360) + (\cos 60 × \sin 60 × 2/2)] + \sin 60 × 2 × 20000
= 24794 + 34641 = 59435 N (upwards)
The resultant = (82072.5² + 59435²)^{0.5} = 101333 \mathrm{N}
The resultant acts at an angle (with vertical), \tan^{– 1} (82072.5 / 59434.9) = 54.09°
Note : The problem can also be solved by considering an additional head of fluid equal to the gauge pressure.