Question 3.P.21: A hemispherical bulge of 3 m diameter inwards is as shown in......

Force on the surface ABC is required. The projected area = π  D^{2}/4

(i) Horizontal force = 9810 × 3.5 × π × 3^{2}/4 = 242700  \mathrm{N} Depth of centre of pressure

= 3.5 + (π/64) (3^{4} × 4/3.5 × π × 3^{2}) = 3.66071  \mathrm{m}

Vertical force = weight of the liquid displaced

γ (1/2) ( 4 π  R^{3}/3) = 69343  \mathrm{N} and acts upwards

For the hemisphere, centre of action from surface = 3 R/8 = 0.5625  \mathrm{m} from wall.

Note : The vertical force on the surface AB is due to the liquid column above it and acts upwards. The vertical force on the surface BC is due to the liquid column above it and acts downwards. So the net force is due to the weight of the volume of liquid displaced and acts upwards.

The resultant is given by [242700^{2} + 69343^{2}]^{0.5} = 252412  \mathrm{N}

The direction is given by (angle with vertical) θ,

θ = \tan^{– 1}  (24700/69343) = 74.05°.

The angle with the horizontal will be 15.95°

Check whether the resultant passes through the centre by taking moment. It does.

\{0.5625 × 69343 – (0.16071 × 242700)\} ≅ 0

(ii) When water level comes up to the edge, horizontal force

= γ h A = 9810 × 1.5 × (π × 3^{2}/4) = 104014  \mathrm{N}

Horizontal force acts at 1.5 + (π  3^{4}/64) (1/1.5) (4/π × 3^{2}) = 1.875  \mathrm{m}

The vertical force remains the same.

Resultant = [104014^{2} + 69343^{2}]^{0.5} = 125009  \mathrm{N}

Does the resultant pass through the centre? Check.

Line of action, angle with vertical = \tan^{– 1}  (104014/69343) = 56.31°

(iii) When water comes to the centre, horizontal force

= 9810 × 0.75 × (π × 3^{2}/8) = 26004  \mathrm{N} \\ I_{G} = I_{b}  –  A  (\bar{h^{2}} ),   \bar{h} = 2D/3  π,  I_{b} = π  D^{4}/128

Centre of pressure = 0.75 + [0.55565/(0.6366 × π × 1.5^{2}/2)]

= 0.9973 m, down from B.

Vertical force = γ (1/4) (4 π R^{3}/3) = 34672  \mathrm{N} (upwards)

Resultant = 43340N, θ = 36.87° with vertical.