A trapezoidal channel 2 m wide at the bottom and 1 m deep has side slopes of 1:1. Determine: (i) the total pressure, and (ii) the centre of pressure on the vertical gate closing the channel when it is full of water.

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Pressure on a Trapezoidal Channel Gate

Problem Statement

A trapezoidal channel 2 m wide at the bottom and 1 m deep has side slopes of 1:1. Determine: (i) the total pressure, and (ii) the centre of pressure on the vertical gate closing the channel when it is full of water.

Given Data

  • Width at bottom, \( b_{bottom} = 2 \, \text{m}\)
  • Depth of channel, \( d = 1 \, \text{m}\)
  • Side slopes = 1:1 (for a depth of 1 m, the horizontal width of the slope is 1 m)
  • Top width, \( b_{top} = 2 + 1 + 1 = 4 \, \text{m}\)
  • Fluid is water, so density \(\rho = 1000 \, \text{kg/m}^3\)

Diagram

Cross-section of the trapezoidal channel gate.

Diagram of the trapezoidal channel gate

Solution (Decomposition Method)

(i) Total Pressure (Force)

First, calculate the total area by splitting the trapezoid into a central rectangle (\(A_1\)) and two side triangles (\(A_2\)).

$$ A_1 \text{ (rectangle)} = 2 \, \text{m} \times 1 \, \text{m} = 2 \, \text{m}^2 $$ $$ A_2 \text{ (total for two triangles)} = 2 \times (\frac{1}{2} \times 1 \, \text{m} \times 1 \, \text{m}) = 1 \, \text{m}^2 $$ $$ A_{total} = A_1 + A_2 = 2 + 1 = 3 \, \text{m}^2 $$

Next, find the depth of the Centre of Gravity (\(\bar{h}\)) for the composite shape.

$$ \bar{h}_1 \text{ (for rectangle)} = \frac{1}{2} = 0.5 \, \text{m} $$ $$ \bar{h}_2 \text{ (for triangles)} = \frac{1}{3} \times 1 = 0.333 \, \text{m} $$ $$ \bar{h} = \frac{A_1 \bar{h}_1 + A_2 \bar{h}_2}{A_1 + A_2} $$ $$ \bar{h} = \frac{(2 \times 0.5) + (1 \times 0.333)}{3} $$ $$ \bar{h} = \frac{1 + 0.333}{3} \approx 0.444 \, \text{m} $$

Now, calculate the total pressure (force):

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 3 \times 0.444 $$ $$ F \approx 13079.9 \, \text{N} $$

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

Find the moment of inertia (\(I_G\)) of the trapezoid about its centroid using the parallel axis theorem.

1. M.O.I. of the rectangle about the trapezoid's C.G.:

$$ I'_{rect} = (\frac{bd^3}{12}) + A_1 \times (\bar{h}_1 - \bar{h})^2 $$ $$ I'_{rect} = (\frac{2 \times 1^3}{12}) + 2 \times (0.5 - 0.444)^2 $$ $$ I'_{rect} = 0.1667 + 2 \times (0.056)^2 $$ $$ I'_{rect} = 0.1667 + 0.00627 \approx 0.1729 \, \text{m}^4 $$

2. M.O.I. of the two triangles about the trapezoid's C.G.:

$$ I'_{triangles} = 2 \times [(\frac{bh^3}{36}) + A_{tri} \times (\bar{h} - \bar{h}_2)^2] $$ $$ I'_{triangles} = 2 \times [(\frac{1 \times 1^3}{36}) + (\frac{1}{2} \times 1 \times 1) \times (0.444 - 0.333)^2] $$ $$ I'_{triangles} = 2 \times [0.0278 + 0.5 \times (0.111)^2] $$ $$ I'_{triangles} = 2 \times [0.0278 + 0.00616] \approx 0.0679 \, \text{m}^4 $$

3. Total M.O.I. and Centre of Pressure:

$$ I_G = I'_{rect} + I'_{triangles} $$ $$ I_G = 0.1729 + 0.0679 = 0.2408 \, \text{m}^4 $$ $$ h^* = \frac{I_G}{A\bar{h}} + \bar{h} $$ $$ h^* = \frac{0.2408}{3 \times 0.444} + 0.444 $$ $$ h^* \approx 0.1808 + 0.444 \approx 0.625 \, \text{m} $$

Alternate Solution (Formula Method)

Using standard formulas for a trapezoid (where \(a\) is bottom width, \(b\) is top width):

$$ \bar{h} = (\frac{2a + b}{a + b}) \times \frac{d}{3} = (\frac{2 \times 2 + 4}{2 + 4}) \times \frac{1}{3} $$ $$ \bar{h} = (\frac{8}{6}) \times \frac{1}{3} \approx 0.444 \, \text{m} $$

Moment of Inertia \(I_G\) from formula:

$$ I_G = \frac{(a^2 + 4ab + b^2)}{36(a+b)} \times d^3 $$ $$ I_G = \frac{(2^2 + 4(2)(4) + 4^2)}{36(2+4)} \times 1^3 $$ $$ I_G = \frac{(4 + 32 + 16)}{36 \times 6} \times 1 $$ $$ I_G = \frac{52}{216} \approx 0.2407 \, \text{m}^4 $$

Centre of Pressure \(h^*\):

$$ h^* = \frac{I_G}{A\bar{h}} + \bar{h} $$ $$ h^* = \frac{0.2407}{3 \times 0.444} + 0.444 $$ $$ h^* \approx 0.1808 + 0.444 \approx 0.625 \, \text{m} $$
Final Results:

Total Pressure (Force): \( F \approx 13080 \, \text{N} \)

Centre of Pressure: \( h^* \approx 0.625 \, \text{m} \) below the free surface

Explanation of Concepts

Total Pressure (Hydrostatic Force): This is the net force exerted by the fluid on the surface of the gate. It is calculated as the pressure at the centroid (C.G.) of the area multiplied by the total area (\(F = P_{CG} \times A = (\rho g \bar{h}) A\)).

Centre of Pressure: This is the point on the immersed surface where the total hydrostatic force effectively acts. Because fluid pressure increases with depth, this point is always located below the geometric centroid of the surface. Its precise location is found by considering the moment of inertia of the surface area.

Physical Meaning

The total force of approximately 13,080 N (or about 1.3 metric tons) is the load that the gate and its supports must be designed to withstand.

The centre of pressure is at a depth of 0.625 m, which is below the centre of gravity (0.444 m). This lower point of action for the force is crucial for structural design. If a single support were to resist this force, it would need to be placed at this depth to prevent the gate from rotating. In gates with multiple supports or hinges, this location determines the distribution of the load among them.

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