Hydrostatic force on gates

(a) Vertical rectangular gate

Given: $b = 2\ \text{m}$, $H = 3\ \text{m}$, top edge at 2 m depth.

Depth of centroid: $\bar h = 2 + 1.5 = 3.5\ \text{m}$. Area $A = 2 \times 3 = 6\ \text{m}^2$.

$$F = \rho g \bar h A = 1000 \times 9.81 \times 3.5 \times 6 = 206\,010\ \text{N} = 206.0\ \text{kN}$$

Centre of pressure:

$$\begin{aligned} I_G &= \frac{bH^3}{12} = \frac{2 \times 27}{12} = 4.5\ \text{m}^4 \\ h^* &= \bar h + \frac{I_G}{A\bar h} = 3.5 + \frac{4.5}{6 \times 3.5} = 3.714\ \text{m} \end{aligned}$$

So the centre of pressure is $3.714 - 2 = 1.714\ \text{m}$ below the hinge.

Taking moments about the hinge, with force $P$ at the bottom edge (3 m below the hinge):

$$\begin{aligned} P \times 3 &= F \times 1.714 \\ P &= \frac{206.01 \times 1.714}{3} = 117.7\ \text{kN} \end{aligned}$$

(b) Quadrant gate

Horizontal component equals the force on the vertical projection ($2\ \text{m} \times 3\ \text{m}$), whose centroid is at 1 m depth:

$$F_H = \rho g \bar h A_{proj} = 1000 \times 9.81 \times 1 \times 6 = 58\,860\ \text{N} = 58.86\ \text{kN}$$

Vertical component equals the weight of water above the curved surface (quarter cylinder of water):

$$F_V = \rho g \left(\frac{\pi R^2}{4} \times w\right) = 1000 \times 9.81 \times (3.1416 \times 3) = 92\,457\ \text{N} = 92.46\ \text{kN}$$

Resultant:

$$\begin{aligned} F_R &= \sqrt{F_H^2 + F_V^2} = \sqrt{58.86^2 + 92.46^2} = 109.6\ \text{kN} \\ \theta &= \tan^{-1}\frac{F_V}{F_H} = \tan^{-1}\frac{92.46}{58.86} = 57.5^\circ \text{ below the horizontal} \end{aligned}$$

Because all pressure forces on a circular arc act normal to the surface, the resultant passes through the centre O.

Answer: (a) $F = 206\ \text{kN}$, $h^* = 3.714\ \text{m}$, $P = 117.7\ \text{kN}$; (b) $F_H = 58.86\ \text{kN}$, $F_V = 92.46\ \text{kN}$, $F_R = 109.6\ \text{kN}$ at $57.5^\circ$ to the horizontal