Problem Statement
A dam has a parabolic shape \( y = y_0 \left( \frac{x}{x_0} \right)^2 \) with \( x_0 = 6 \, \text{m} \) and \( y_0 = 9 \, \text{m} \). The fluid is water. Compute the horizontal, vertical, and the resultant thrust exerted by the water per metre length of the dam.
Given Data
- Parabolic shape: \( y = 9 \left( \frac{x}{6} \right)^2 = \frac{x^2}{4} \implies x = 2\sqrt{y} \)
- Total height of water, \( y_0 = 9 \, \text{m} \)
- Width of dam, \( b = 1 \, \text{m} \)
- Density of Water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to Gravity, \( g = 9.81 \, \text{m/s}^2 \)
Diagram of Parabolic Dam
Solution
1. Horizontal Thrust (\(F_x\))
The horizontal force is the pressure on the vertically projected area (OB).
The depth of the centroid of this projected area is:
The horizontal thrust is:
2. Vertical Thrust (\(F_y\))
The vertical force is the weight of the water in the volume OAB.
We find the area of OAB by integrating the curve's equation from y=0 to y=9.
The vertical thrust is the weight of this volume of water:
3. Resultant Thrust (\(F\)) and Direction (\(\theta\))
The magnitude of the resultant thrust is:
The direction of the resultant thrust is:
The resultant thrust is \( F \approx 531574 \, \text{N} \) or \( 531.57 \, \text{kN} \).
The direction of the resultant is \( \theta \approx 41.63^\circ \) with the horizontal.
Explanation of Concepts
Force on a Parabolic Surface: For surfaces defined by a mathematical equation like a parabola, we use integral calculus to find the exact area under the curve. This area represents the cross-section of the fluid supported by the dam. The weight of this fluid volume constitutes the vertical component of the hydrostatic force. The horizontal force is calculated more simply by considering the force on the dam's vertical projection.
