Presentation on theme: "Forces Acting on Dams To design a dam, various forces must be considered to insure the safety of the dam. The most important forces are: 1. Weight of dam."— Presentation transcript:
Forces Acting on Dams To design a dam, various forces must be considered to insure the safety of the dam. The most important forces are: 1. Weight of dam 2. Water Pressure 3. Uplift 4. Wave pressure 5. Earthquake forces
Weight of Dam Weight of Dam The weight of the dam is calculated as follows: Where γ m is the specific weight of the dam’s material.
Water Pressure Water Pressure Water pressure acts perpendicular to the surface of the dam and is calculated per unit width as follows: Where γw is the specific weight of water and h is the height of water. PwPw h/3 h
Uplift Uplift Dams are subjected to uplift force under its base. Uplift acts upward. Where B is the width of the base of the dam. h U B B/3
Wave Pressure Wave Pressure The upper part of the dam (above the water level) is subjected to the impact of waves. The maximum wave pressure per unit width is: Where h w is the wave height. PwPw h w hwhw
Earthquake Forces Dams are subjected to vibration during earthquakes. Vibration affects both the body of the dam and the water in the reservoir behind the dam. Vibration forces are function of both the intensity (Rechter Scale) and its duration. The most danger effect occurs when the vibration is perpendicular to the face of the dam.
Body Forces: Body Forces: Body force acts horizontally at the center of gravity and is calculated as: Where α is the earthquake coefficient and W is the weight of the dam. Is taken 0.2 for practical reasons. W1W1 0.1 W 1 W2W2 0.1 W 2
Water Force: Water Force: Water vibration produces a force on the dam acting horizontally; Where C e is another coefficient (0.82) and h is the height of the water 2/5 h hP ew
Example: Example: Calculate the forces on given dam if; γ m = 2.5 t/m 3, γ w =1 t/m 3, hw = 1.5 m 6 m 24 m 30 m 10 m 33 m
1. Weight of water 1. Weight of water W 1 =2.5 * 6 * 40 = 600 t W 2 =2.5 *0.5*18*30 = 675 t = 675 t 2. Water Pressure 2. Water Pressure P = 0.5 * 1.0 * (33) 2 = t = t w1w1 w2w2 P
3. Uplift 3. Uplift U = 0.5 * 1.0 * 33 * 24 = 396 t = 396 t 4. Wave Pressure 4. Wave Pressure P w = 2.4 * 1.0 * 1.5 = 3.6 t = 3.6 t U PwPw
5. Earthquake Forces 5. Earthquake Forces a. Body Forces P em1 = 0.1 * 600 = 60 t = 60 t P em2 = 0.1 * 675 = 67.5 t = 67.5 t b. water Force P ew =(2/3)*0.82*0.1*(33)2 = t = t P em1 P em2 P ew