A circular opening, 3 m diameter, in a vertical side of a tank is closed by a disc of 3 m diameter which can rotate about a horizontal diameter.

Force and Torque on a Vertical Circular Gate

Problem Statement

A circular opening, 3 m diameter, in a vertical side of a tank is closed by a disc of 3 m diameter which can rotate about a horizontal diameter. Calculate: (i) the force on the disc, and (ii) the torque required to maintain the disc in equilibrium in the vertical position when the head of water above the horizontal diameter is 4 m.

Given Data

Diameter of disc, $d = 3 \, \text{m}$
Area of disc, $A = \frac{\pi}{4}d^2 = \frac{\pi}{4}(3)^2 = 7.0685 \, \text{m}^2$
Depth of centre of gravity (horizontal diameter), $\bar{h} = 4.0 \, \text{m}$
Density of water, $\rho = 1000 \, \text{kg/m}^3$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Diagram

Visual representation of the submerged circular disc.

Solution

(i) Force on the Disc

The total force on the disc is given by the hydrostatic force equation:

$$ F = \rho g A \bar{h} $$

Substitute the given values:

$$ F = 1000 \times 9.81 \times 7.0685 \times 4.0 $$ $$ F = 277368 \, \text{N} = 277.37 \, \text{kN} $$

(ii) Torque required to maintain equilibrium

To find the torque, we first need the location of the centre of pressure ($h^*$), where the force $F$ acts.

$$ h^* = \frac{I_G}{A\bar{h}} + \bar{h} $$

For a circular section, the moment of inertia $I_G$ can be simplified in the formula as:

$$ h^* = \frac{\frac{\pi d^4}{64}}{\frac{\pi d^2}{4} \bar{h}} + \bar{h} = \frac{d^2}{16\bar{h}} + \bar{h} $$

Calculate the depth of the centre of pressure:

$$ h^* = \frac{(3)^2}{16 \times 4.0} + 4.0 = \frac{9}{64} + 4.0 $$ $$ h^* = 0.140625 + 4.0 = 4.140625 \, \text{m} $$

The torque is the force $F$ multiplied by the distance between the pivot (the horizontal diameter at $\bar{h}$) and the centre of pressure (at $h^*$).

$$ \text{Torque} = F \times (h^* - \bar{h}) $$

Substitute the values to find the torque:

$$ \text{Torque} = 277368 \times (4.140625 - 4.0) $$ $$ \text{Torque} = 277368 \times 0.140625 $$ $$ \text{Torque} \approx 38988 \, \text{N-m} = 38.99 \, \text{kN-m} $$

Final Results:

(i) Force on the disc: $ F \approx 277.37 \, \text{kN} $

(ii) Torque required: $ \text{Torque} \approx 38.99 \, \text{kN-m} $

Explanation of Concepts

Force on the Disc: This is the total hydrostatic force exerted by the water on the entire surface of the disc. It's calculated as if the entire pressure was uniform and equal to the pressure at the disc's geometric center (centroid).

Centre of Pressure ($h^*$): This is the point where the total hydrostatic force effectively acts. Because pressure increases with depth, this point is always below the centroid for a vertical surface.

Torque: Torque is a measure of the turning effect of a force. Since the total force $F$ acts at the centre of pressure and not at the pivot point (the horizontal diameter), it creates a moment that would cause the disc to rotate downwards. An equal and opposite torque must be applied to keep the disc in a stable, vertical position.

Physical Meaning

The results show a total force of nearly 277.4 kN (over 28 tons of force) acting on the disc. However, this force is not applied evenly.

The calculated centre of pressure is at 4.14 m, which is about 14 cm below the pivot point at 4 m. This small distance might seem insignificant, but when multiplied by the large hydrostatic force, it results in a substantial torque of almost 39 kN-m. This is the torque that a locking mechanism or motor would need to provide to hold the gate closed against the water pressure. Without this counteracting torque, the disc would swing open from the bottom.