Find the total pressure and position of the centre of pressure on a triangular plate of base 2 m and height 3 m which is immersed in water in such a way that the plane of the plate makes an angle of 60° with the free surface of the water. The base of the plate is parallel to the water surface and at a depth of 2.5 m from the water surface.

Pressure on a Triangular Plate

Problem Statement

Given Data

Base of plate, $ b = 2 \, \text{m} $
Height of plate, $ h_{plate} = 3 \, \text{m} $
Inclination Angle, $ \theta = 60^\circ $
Depth of base from surface = $ 2.5 \, \text{m} $

Diagram

Configuration of the submerged triangular plate.

Solution

1. Area and Centroid Calculation

$$ \text{Area, } A = \frac{1}{2} \times b \times h_{plate} $$ $$ A = \frac{1}{2} \times 2 \times 3 $$ $$ A = 3 \, \text{m}^2 $$

The centroid of a triangle is located at 1/3 of its height from the base. We need to find the vertical depth of this point ($\bar{h}$).

$$ \text{Distance from base to C.G. (along plate height), } AG = \frac{1}{3} h_{plate} $$ $$ AG = \frac{1}{3} \times 3 $$ $$ AG = 1 \, \text{m} $$

$$ \text{Vertical depth of C.G., } \bar{h} = (\text{Depth of base}) + (AG \sin\theta) $$ $$ \bar{h} = 2.5 + (1 \times \sin 60^\circ) $$ $$ \bar{h} = 2.5 + (1 \times 0.866) $$ $$ \bar{h} = 3.366 \, \text{m} $$

2. Total Pressure Force (F)

The total force is calculated using the depth of the centroid.

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 3 \times 3.366 $$ $$ F = 99061.38 \, \text{N} $$

3. Centre of Pressure ($h^*$)

The position of the centre of pressure is found using the parallel axis theorem.

$$ \text{Moment of Inertia about C.G., } I_G = \frac{b h_{plate}^3}{36} $$ $$ I_G = \frac{2 \times 3^3}{36} $$ $$ I_G = \frac{54}{36} $$ $$ I_G = 1.5 \, \text{m}^4 $$

$$ h^* = \frac{I_G \sin^2\theta}{A\bar{h}} + \bar{h} $$ $$ h^* = \frac{1.5 \times (\sin 60^\circ)^2}{3 \times 3.366} + 3.366 $$ $$ h^* = \frac{1.5 \times (0.866)^2}{10.098} + 3.366 $$ $$ h^* = \frac{1.5 \times 0.75}{10.098} + 3.366 $$ $$ h^* = \frac{1.125}{10.098} + 3.366 $$ $$ h^* = 0.1114 + 3.366 $$ $$ h^* \approx 3.477 \, \text{m} $$

Final Results:

Total Pressure Force: $ F \approx 99061 \, \text{N} $ or $ 99.06 \, \text{kN} $

Centre of Pressure: $ h^* \approx 3.477 \, \text{m} $ below the free surface

Explanation of Concepts

Centroid vs. Centre of Pressure: The centroid ($\bar{h}$) is the geometric center of the plate's area, and it's used to calculate the average pressure and thus the total force. The Centre of Pressure ($h^*$) is the point where this total force effectively acts. Because water pressure increases with depth, this point is always located deeper than the centroid.

Moment of Inertia ($I_G$): This property describes the plate's resistance to rotation. It is crucial for finding the exact location of the centre of pressure. The formula for $I_G$ depends on the shape of the submerged area (in this case, a triangle).