A circular plate 3.0 m diameter is immersed in water in such a way that the plane of the plate makes an angle of 60° with the free surface of water. Determine the total pressure and position of centre of pressure when the upper edge of the plate is 2 m below the free water surface.

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Inclined Circular Plate Problem

Problem Statement

A circular plate 3.0 m diameter is immersed in water in such a way that the plane of the plate makes an angle of 60° with the free surface of water. Determine the total pressure and position of centre of pressure when the upper edge of the plate is 2 m below the free water surface.

Given Data

  • Diameter of plate, \( D = 3 \, \text{m} \)
  • Angle with free surface, \( \theta = 60^\circ \)
  • Depth of upper edge = 2 m
  • Fluid is water, \( \rho = 1000 \, \text{kg/m}^3 \)

Solution

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

First, calculate the area of the circular plate.

$$ A = \frac{\pi}{4} D^2 $$ $$ A = \frac{\pi}{4} (3)^2 $$ $$ A \approx 7.0686 \, \text{m}^2 $$

Next, find the vertical depth of the plate's centroid (\(\bar{h}\)). The centroid is at a radius (\(D/2\)) distance along the incline from the upper edge.

$$ \bar{h} = (\text{Depth of upper edge}) + \left( \frac{D}{2} \sin\theta \right) $$ $$ \bar{h} = 2 + \left( \frac{3}{2} \sin 60^\circ \right) $$ $$ \bar{h} = 2 + (1.5 \times 0.866) $$ $$ \bar{h} = 2 + 1.299 $$ $$ \bar{h} = 3.299 \, \text{m} $$

Now, calculate the total force on the plate.

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 7.0686 \times 3.299 $$ $$ F \approx 228694 \, \text{N} $$

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

The vertical depth of the centre of pressure for an inclined plane is given by:

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

First, calculate the moment of inertia (\(I_G\)) for the circular plate.

$$ I_G = \frac{\pi D^4}{64} $$ $$ I_G = \frac{\pi \times (3)^4}{64} $$ $$ I_G = \frac{81 \pi}{64} \approx 3.976 \, \text{m}^4 $$

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

$$ h^* = \frac{3.976 \times (\sin 60^\circ)^2}{7.0686 \times 3.299} + 3.299 $$ $$ h^* = \frac{3.976 \times (0.866)^2}{23.319} + 3.299 $$ $$ h^* = \frac{3.976 \times 0.75}{23.319} + 3.299 $$ $$ h^* = \frac{2.982}{23.319} + 3.299 $$ $$ h^* \approx 0.1279 + 3.299 $$ $$ h^* \approx 3.427 \, \text{m} $$
Final Results:

Total Pressure (Force): \( F \approx 228.69 \, \text{kN} \).

Position of Centre of Pressure: \( h^* \approx 3.427 \, \text{m} \) (vertical depth from free surface).

Explanation of Concepts

Inclined Surfaces: When a surface is inclined, we must use trigonometry to find the true vertical depth of its centroid. The force calculation (\(F = \rho g A \bar{h}\)) always uses the vertical depth (\(\bar{h}\)), not the distance along the incline.

Centre of Pressure on an Inclined Plane: The formula for the centre of pressure must also account for the inclination. The term \(I_G \sin^2 \theta\) in the numerator correctly projects the moment of inertia into the vertical plane. The result, \(h^*\), gives the true vertical depth of the centre of pressure from the free surface. It is always deeper than the centroid's vertical depth (\(\bar{h}\)).

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