A vertical sluice gate is used to cover an opening in a dam. The opening is 2 m wide and 1.2 m high. On the upstream of the gate, a liquid of sp. gr. 1.45 lies up to a height of 1.5 m above the top of the gate. On the downstream side, water is available up to a height touching the top of the gate.

Force on a Vertical Sluice Gate

Problem Statement

A vertical sluice gate is used to cover an opening in a dam. The opening is 2 m wide and 1.2 m high. On the upstream of the gate, a liquid of sp. gr. 1.45 lies up to a height of 1.5 m above the top of the gate. On the downstream side, water is available up to a height touching the top of the gate. Find the resultant force acting on the gate and the position of the centre of pressure. Find also the force acting horizontally at the top of the gate which is capable of opening it, assuming the gate is hinged at the bottom.

Given Data

Width of gate, $b = 2 \, \text{m}$
Height of gate, $d = 1.2 \, \text{m}$
Area of gate, $A = b \times d = 2 \times 1.2 = 2.4 \, \text{m}^2$
Upstream liquid sp. gr., $S_1 = 1.45$, Density, $\rho_1 = 1450 \, \text{kg/m}^3$
Downstream liquid is water, Density, $\rho_2 = 1000 \, \text{kg/m}^3$

Diagram

Forces acting on the upstream and downstream sides of the sluice gate.

Solution

(i) Forces on the Gate

First, calculate the force on the upstream side ($F_1$). The depth to the centroid is:

$$ \bar{h}_1 = 1.5 \, \text{m} + \frac{1.2 \, \text{m}}{2} = 2.1 \, \text{m} $$ $$ F_1 = \rho_1 g A \bar{h}_1 $$ $$ F_1 = 1450 \times 9.81 \times 2.4 \times 2.1 $$ $$ F_1 = 71691 \, \text{N} $$

Next, calculate the force on the downstream side ($F_2$). The depth to the centroid is:

$$ \bar{h}_2 = \frac{1.2 \, \text{m}}{2} = 0.6 \, \text{m} $$ $$ F_2 = \rho_2 g A \bar{h}_2 $$ $$ F_2 = 1000 \times 9.81 \times 2.4 \times 0.6 $$ $$ F_2 = 14126 \, \text{N} $$

The resultant force is the difference between the two:

$$ F_{Res} = F_1 - F_2 $$ $$ F_{Res} = 71691 - 14126 $$ $$ F_{Res} = 57565 \, \text{N} $$

(ii) Position of Centre of Pressure

Calculate the centre of pressure for the upstream force ($h_1^*$). The moment of inertia is $I_G = \frac{bd^3}{12} = \frac{2 \times 1.2^3}{12} = 0.288 \, \text{m}^4$.

$$ h_1^* = \frac{I_G}{A\bar{h}_1} + \bar{h}_1 $$ $$ h_1^* = \frac{0.288}{2.4 \times 2.1} + 2.1 $$ $$ h_1^* = 0.0571 + 2.1 $$ $$ h_1^* \approx 2.1571 \, \text{m} $$

The distance of $F_1$ from the bottom hinge is:

$$ y_1 = (1.5 + 1.2) - h_1^* $$ $$ y_1 = 2.7 - 2.1571 $$ $$ y_1 = 0.5429 \, \text{m} $$

Calculate the centre of pressure for the downstream force ($h_2^*$):

$$ h_2^* = \frac{I_G}{A\bar{h}_2} + \bar{h}_2 $$ $$ h_2^* = \frac{0.288}{2.4 \times 0.6} + 0.6 $$ $$ h_2^* = 0.2 + 0.6 $$ $$ h_2^* = 0.8 \, \text{m} $$

The distance of $F_2$ from the bottom hinge is:

$$ y_2 = 1.2 - h_2^* $$ $$ y_2 = 1.2 - 0.8 $$ $$ y_2 = 0.4 \, \text{m} $$

The position of the resultant force ($y_{Res}$) above the hinge is found by taking moments about the hinge:

$$ y_{Res} = \frac{F_1 y_1 - F_2 y_2}{F_{Res}} $$ $$ y_{Res} = \frac{71691 \times 0.5429 - 14126 \times 0.4}{57565} $$ $$ y_{Res} = \frac{38921 - 5650.4}{57565} $$ $$ y_{Res} \approx 0.578 \, \text{m} $$

(iii) Force to Open the Gate

Let $P$ be the force at the top of the gate. To open the gate, this force must overcome the net moment from the fluid forces about the hinge. We take moments about the hinge:

$$ P \times 1.2 + F_2 \times y_2 = F_1 \times y_1 $$ $$ P = \frac{F_1 y_1 - F_2 y_2}{1.2} $$ $$ P = \frac{71691 \times 0.5429 - 14126 \times 0.4}{1.2} $$ $$ P = \frac{33270.6}{1.2} $$ $$ P \approx 27725.5 \, \text{N} $$

Final Results:

Resultant Force on Gate: $ F_{Res} \approx 57.57 \, \text{kN} $

Position of Centre of Pressure: $ 0.578 \, \text{m} $ above the hinge

Force to Open Gate: $ P \approx 27.73 \, \text{kN} $

Explanation of Concepts

Net Force: The gate experiences force from fluids on both sides. The upstream force is larger due to the higher fluid density and greater depth. The resultant force is the net effect of these two opposing forces.

Centre of Pressure: Each force ($F_1$ and $F_2$) acts at its own centre of pressure, which is lower than the geometric centre of the gate because pressure increases with depth. The "position of centre of pressure" for the entire system is the point where the single resultant force would need to act to produce the same net moment about the hinge.

Principle of Moments: To find the force $P$ needed to open the gate, we use the principle of moments. The gate is in equilibrium just before it opens. By summing the moments (Force × perpendicular distance from pivot) about the hinge and setting them to zero, we can solve for the unknown opening force $P$.

Physical Meaning

The results show a net force of over 57,000 N (about 5.8 metric tons) pushing the gate in the downstream direction. This is the force that the hinge and the top support must collectively withstand to keep the gate in place.

The final calculation shows that a force of over 27,000 N (about 2.8 metric tons) is required at the top of the gate to start opening it. This force is needed to counteract the net turning effect (moment) created by the fluid forces. Because the upstream force ($F_1$) is both larger and acts with a larger lever arm about the hinge than the downstream force ($F_2$), it creates a strong closing moment that must be overcome.