Question

Wind blowing over the water surface in the reservoir developed a wave of height 2 m. Then the wave force in kN/m on the upper portion of gravity dam is

• A. 100
• B. 50
• C. 80
• D. 40

Hint:

Wave forces on gravity dams are critical for stability analysis. The wave pressure depends on the wave height and water density, and the force distribution is typically triangular, centered around the still water level. Simplified formulas like \( F = 2 \gamma_w h_w^2 \) are often used in preliminary design.

Answer 1

The Correct Option is C

Step 1: Identify the given data.
The wave height is given as 2 m. We need to find the wave force per unit length (in kN/m) on the upper portion of a gravity dam. The options are:
(1) 100
(2) 50
(3) 80
(4) 40
The unit weight of water is not specified, so we assume the standard value of \( \gamma_w = 10 \, \text{kN/m}^3 \) (commonly used in such problems, equivalent to 1000 kg/m³ with \( g = 10 \, \text{m/s}^2 \)).
Step 2: Recall the formula for wave force on a gravity dam.
For gravity dams, the wave force due to wind-induced waves is often calculated using empirical formulas, such as the one provided in IS:6512 (Indian Standard for Design of Gravity Dams) or other hydraulic engineering standards. A commonly used formula for the wave pressure (in kN/m²) on the upstream face of a dam is:
\[ p = 2 \gamma_w h_w \] Where:
\( p \): Wave pressure (kN/m²)
\( \gamma_w \): Unit weight of water (kN/m³)
\( h_w \): Wave height (m)
This pressure acts over the height of the wave-affected zone, typically taken as \( 1.25 h_w \) above the still water level (SWL) and \( 0.75 h_w \) below the SWL, so the total height of the pressure distribution is often approximated as \( 2 h_w \). However, the force is per unit length (kN/m), so we need the force over this height per unit width. The total wave force per unit length (kN/m) can be approximated using a simplified formula for gravity dams:
\[ F = 2 \gamma_w h_w^2 \] This accounts for the pressure distribution over the wave height, integrated to give the force per unit length.
Step 3: Substitute the values and calculate the wave force.
Given \( h_w = 2 \, \text{m} \) and \( \gamma_w = 10 \, \text{kN/m}^3 \):
\[ F = 2 \gamma_w h_w^2 = 2 \times 10 \times (2)^2 = 2 \times 10 \times 4 = 80 \, \text{kN/m} \]
Step 4: Verify the calculation.
The pressure at the still water level is \( p = 2 \times 10 \times 2 = 40 \, \text{kN/m}^2 \). The pressure distribution is triangular (from 0 at the wave crest to a maximum at the SWL and decreasing below). The effective height of the pressure distribution is often taken as \( h_w \), and the force per unit length is the pressure times the height:
\[ F = \frac{1}{2} \times p \times h_w \times 2 = \frac{1}{2} \times 40 \times 2 \times 2 = 80 \, \text{kN/m} \] This matches the simplified formula and confirms the result.
Step 5: Select the correct option.
The wave force is 80 kN/m, which matches option (3).
\[ \boxed{80 \, \text{kN/m}} \]