Question 3.P.18: Determine the magnitude and direction of the force on the el......
Determine the magnitude and direction of the force on the elliptical tank portion AB as shown in Fig P. 3.18.
The horizontal force on the elliptical portion equals the force on the projected area, considering 1 m width, \bar{h} = 5 m, A = 4 × 1
γ A \bar{h} = 9810 × 4 × 5 = 196200 N acts at 1.333 m from bottom
Vertical force on the elliptical portion equals the weight of water above this area. Area of ellipse = π bh/4 where b and h are minor and major axis. b = 6, h = 8. A = 12π. The area here is 1/4 th of the ellipse. Hence, A = 3 π m^{2} (ellipse portion)
Rectangular portion above
= 3 × 3 = 9 m². Volume = 1 × (3 π + 9) m^{3}
Weight = 9810 (3 π + 9)
= 180747 N = Total vertical force
The centre of gravity of the quarter of elliptical portion
= (4b/3π), \\ = 4 × 3/3 π = 4/π m from the major axis (as b = 3 m)
Centre of gravity of the rectangle = 1.5 m from the wall Taking moments and solving, the location x of vertical force
= [9 × 1.5 + 3 π (3 – 4/π)] / [9 + 3 π] = 1.616 m from wall
Resultant force = (196200^{2} + 180747^{2})^{0.5} = 266766 N
To determine the line of action, let this line cut OA at a distance of h below O at P.
Then, taking moments about P, (2.666 – h) 192600 = (3 – 1.616) 180747
∴ h = 1.392 m, the line of action passes through a point P, 1.392 m below O at an angle of 47.35° from vertical.