Find the magnitude and direction of the resultant water pressure acting on a curved face of a dam which is shaped according to the relation Y=X2/6 . The height of water retained by the dam is 12 m. Take the width of dam as unity.

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Parabolic Dam Pressure Problem

Problem Statement

Find the magnitude and direction of the resultant water pressure acting on a curved face of a dam which is shaped according to the relation \( y = x^2/6 \) as shown in the figure below. The height of water retained by the dam is 12 m. Take the width of dam as unity.

Given Data

  • Equation of dam curve: \( y = x^2/6 \)
  • Height of water, \( h = 12 \, \text{m} \)
  • Width of dam, \( b = 1 \, \text{m} \) (unity)
  • Fluid is water, \( \rho = 1000 \, \text{kg/m}^3 \)

Diagram of the Dam

Diagram of a dam with a parabolic face

Solution

The resultant force is found by calculating the horizontal and vertical components of the force separately.

1. Horizontal Force (\(F_x\))

The horizontal force is the total pressure on the vertical projection of the curved surface. This projection is a rectangle of height 12 m.

$$ A_x = h \times b $$ $$ A_x = 12 \times 1 = 12 \, \text{m}^2 $$

The depth of the centroid of this projected area is:

$$ \bar{h}_x = \frac{h}{2} $$ $$ \bar{h}_x = \frac{12}{2} = 6 \, \text{m} $$

Now, calculate the horizontal force component.

$$ F_x = \rho g A_x \bar{h}_x $$ $$ F_x = 1000 \times 9.81 \times 12 \times 6 $$ $$ F_x = 706320 \, \text{N} $$

2. Vertical Force (\(F_y\))

The vertical force is the weight of the water in the volume directly above the curved surface. We find this by integrating the area under the parabolic curve.

First, rearrange the equation for \(x\):

$$ y = \frac{x^2}{6} \implies x^2 = 6y \implies x = \sqrt{6y} $$

Now, integrate to find the area under the curve from y = 0 to y = 12.

$$ \text{Area} = \int_{0}^{12} x \,dy = \int_{0}^{12} \sqrt{6y} \,dy $$ $$ \text{Area} = \sqrt{6} \int_{0}^{12} y^{1/2} \,dy $$ $$ \text{Area} = \sqrt{6} \left[ \frac{y^{3/2}}{3/2} \right]_{0}^{12} = \sqrt{6} \times \frac{2}{3} [12^{3/2}] $$ $$ \text{Area} = \sqrt{6} \times \frac{2}{3} \times (41.569) \approx 67.88 \, \text{m}^2 $$
$$ V = \text{Area} \times b = 67.88 \times 1 = 67.88 \, \text{m}^3 $$

Now, calculate the vertical force component.

$$ F_y = \rho g V $$ $$ F_y = 1000 \times 9.81 \times 67.88 $$ $$ F_y \approx 665903 \, \text{N} $$

3. Resultant Force (\(F\))

The resultant force is the vector sum of the components.

$$ F = \sqrt{F_x^2 + F_y^2} $$ $$ F = \sqrt{(706320)^2 + (665903)^2} $$ $$ F \approx \sqrt{4.989 \times 10^{11} + 4.434 \times 10^{11}} $$ $$ F = \sqrt{9.423 \times 10^{11}} $$ $$ F \approx 970721 \, \text{N} $$

4. Angle of Action (\(\theta\))

The angle the resultant force makes with the horizontal is found using trigonometry.

$$ \tan \theta = \frac{F_y}{F_x} $$ $$ \tan \theta = \frac{665903}{706320} $$ $$ \tan \theta \approx 0.9428 $$ $$ \theta = \tan^{-1}(0.9428) $$ $$ \theta \approx 43.31^\circ \text{ or } 43^\circ 19' $$
Final Results:

Resultant Force: \( F \approx 970.72 \, \text{kN} \).

Angle of action with horizontal: \( \theta \approx 43^\circ 19' \).

Explanation of Concepts

Horizontal Component (\(F_x\)): The horizontal force on the curved dam face is equal to the force that would be exerted on its vertical projection. This projection is a simple rectangle with a height equal to the water depth.

Vertical Component (\(F_y\)): The vertical force is equal to the weight of the water contained in the volume directly above the curved surface. To find this volume, we first need to calculate the area of the parabolic section, which is done by integrating the equation of the curve with respect to the vertical axis (y) from the bottom to the water surface.

Resultant Force: Since the horizontal and vertical forces are perpendicular, the total resultant force is found using the Pythagorean theorem, and its direction relative to the horizontal is determined using the arctangent of the ratio of the forces.

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