A circular plate 3 metre diameter is submerged in water as shown. Its greatest and least depths are below the surfaces being 2 metre and 1 metre respectively. Find: (i) the total pressure on front face of the plate, and (ii) the position of centre of pressure.

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Pressure on an Inclined Circular Plate

Problem Statement

A circular plate 3 metre diameter is submerged in water as shown. Its greatest and least depths are below the surfaces being 2 metre and 1 metre respectively. Find: (i) the total pressure on front face of the plate, and (ii) the position of centre of pressure.

Given Data

  • Diameter of plate, \( d = 3.0 \, \text{m}\)
  • Greatest depth (bottom edge), \( h_{max} = 2.0 \, \text{m}\)
  • Least depth (top edge), \( h_{min} = 1.0 \, \text{m}\)
  • Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)

Diagram

Illustration of the inclined circular plate submerged in water.

Diagram of the inclined circular plate in water

Solution

First, we find the sine of the inclination angle, \(\theta\).

$$ \sin\theta = \frac{h_{max} - h_{min}}{d} $$ $$ \sin\theta = \frac{2.0 - 1.0}{3.0} = \frac{1}{3} $$

Next, we find the vertical depth of the plate's centroid (\(\bar{h}\)).

$$ \bar{h} = h_{min} + (\frac{d}{2}) \sin\theta $$ $$ \bar{h} = 1.0 + (\frac{3.0}{2}) \times \frac{1}{3} $$ $$ \bar{h} = 1.0 + 0.5 = 1.5 \, \text{m} $$

(i) Total Pressure Force (\(F\))

The total pressure force is given by \( F = \rho g A \bar{h} \). First, we calculate the area \(A\):

$$ A = \frac{\pi}{4} d^2 $$ $$ A = \frac{\pi}{4} (3.0)^2 \approx 7.0686 \, \text{m}^2 $$

Now, calculate the total force:

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 7.0686 \times 1.5 $$ $$ F \approx 104013 \, \text{N} = 104.01 \, \text{kN} $$

(ii) Centre of Pressure (\(h^*\))

The vertical depth to the centre of pressure, \(h^*\), is given by:

$$ h^* = \frac{I_G \sin^2\theta}{A\bar{h}} + \bar{h} $$

The moment of inertia \(I_G\) for a circular area is:

$$ I_G = \frac{\pi d^4}{64} = \frac{\pi (3.0)^4}{64} \approx 3.976 \, \text{m}^4 $$

Now, substitute the values into the formula for \(h^*\):

$$ h^* = \frac{3.976 \times (\frac{1}{3})^2}{7.0686 \times 1.5} + 1.5 $$ $$ h^* \approx \frac{3.976 \times \frac{1}{9}}{10.603} + 1.5 $$ $$ h^* \approx \frac{0.4418}{10.603} + 1.5 $$ $$ h^* \approx 0.0417 + 1.5 $$ $$ h^* \approx 1.5417 \, \text{m} $$
Final Results:

Total Pressure Force: \( F \approx 104.01 \, \text{kN} \)

Position of Centre of Pressure: \( h^* \approx 1.542 \, \text{m} \) below the free surface

Explanation of Concepts

Total Pressure Force: This is the total force exerted by the water on the plate. It is calculated by finding the pressure at the plate's geometric center (centroid) and multiplying it by the plate's area. This provides a single resultant force that represents the distributed load of the water.

Centre of Pressure: This is the point on the plate where the total pressure force effectively acts. Because pressure increases with depth, this point is always located below the plate's centroid. Knowing its exact position is vital for analyzing the torque or turning moment on the plate, which is crucial for designing any supports or hinges.

Physical Meaning

The plate is subjected to a force of approximately 104,000 Newtons, which is equivalent to the weight of about 10.6 metric tons. The supports for the plate must be engineered to handle this significant load.

This force acts at a depth of 1.542 m, which is about 4.2 cm vertically below the plate's center (at 1.5 m). This small offset creates a rotational force (moment) that would tend to push the bottom of the plate more than the top. This effect must be accounted for in the structural design.

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