Calculate the horizontal and vertical components of the water pressure exerted on a tainter gate of radius 8 m as shown in the figure. Take width of gate unity (1 m).

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Tainter Gate Fluid Pressure Problem (R=8m)

Problem Statement

Calculate the horizontal and vertical components of the water pressure exerted on a tainter gate of radius 8 m as shown in the figure. Take width of gate unity (1 m).

Given Data

  • Radius of Gate, \( R = 8 \, \text{m} \)
  • Vertical height of water, \( BD = 4 \, \text{m} \)
  • Width of Gate, \( b = 1 \, \text{m} \) (unity)
  • Density of Water, \( \rho = 1000 \, \text{kg/m}^3 \)
  • Acceleration due to Gravity, \( g = 9.81 \, \text{m/s}^2 \)

Diagram of Tainter Gate

The tainter gate with an 8m radius submerged in water.

Diagram of the Tainter Gate (R=8m)

Solution

We calculate the horizontal and vertical components of the force separately.

1. Horizontal Force (\(F_x\))

The horizontal force is the total pressure on the projected vertical area (area BD).

$$ A_{proj} = \text{Projected Height (BD)} \times \text{Width} $$ $$ = 4.0 \, \text{m} \times 1 \, \text{m} $$ $$ = 4.0 \, \text{m}^2 $$

The depth of the centroid of this projected area from the free surface is:

$$ \bar{h} = \frac{\text{Projected Height}}{2} $$ $$ = \frac{4.0}{2} $$ $$ = 2.0 \, \text{m} $$

The horizontal force is calculated as:

$$ F_x = \rho g A_{proj} \bar{h} $$ $$ = 1000 \times 9.81 \times 4.0 \times 2.0 $$ $$ = 78480 \, \text{N} $$

2. Vertical Force (\(F_y\))

The vertical force equals the weight of the imaginary volume of water enclosed by the curved surface (portion CBDC).

First, we find the angle from the geometry: \( \sin \theta = \frac{BD}{BO} = \frac{4}{8} = 0.5 \), so \( \theta = 30^\circ \).

The area of portion CBDC is the area of the sector CBO minus the area of the triangle BOD.

$$ \text{Area of Sector CBO} = \frac{30^\circ}{360^\circ} \times \pi R^2 $$ $$ = \frac{1}{12} \times \pi (8)^2 $$ $$ \approx 16.755 \, \text{m}^2 $$
$$ \text{Area of } \triangle BOD = \frac{1}{2} \times BD \times DO $$ $$ = \frac{1}{2} \times 4 \times (8 \cos 30^\circ) $$ $$ = 2 \times (8 \times 0.866) $$ $$ = 13.856 \, \text{m}^2 $$
$$ \text{Area (CBDC)} = \text{Area of Sector} - \text{Area of } \triangle BOD $$ $$ = 16.755 - 13.856 $$ $$ = 2.899 \, \text{m}^2$$

The vertical force is the weight of the water in this volume:

$$ F_y = \rho g \times \text{Area (CBDC)} \times \text{Width} $$ $$ = 1000 \times 9.81 \times 2.899 \times 1 $$ $$ \approx 28439 \, \text{N} $$
Final Result:

The horizontal component of the force is \( F_x = 78480 \, \text{N} \) or \( 78.48 \, \text{kN} \).

The vertical component of the force is \( F_y \approx 28439 \, \text{N} \) or \( 28.44 \, \text{kN} \).

Explanation of Concepts

Horizontal Force on Curved Surface: The horizontal component of the hydrostatic force on any curved surface is always equal to the force on the vertical projection of that surface. This allows us to simplify the problem by analyzing a simple flat plate.

Vertical Force on Curved Surface: The vertical component is equal to the weight of the fluid volume directly above (or enclosed by) the curved surface, extending up to the free surface. This force acts through the centroid (center of gravity) of that fluid volume.

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