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Basic forces on a concrete dam

note: a kip is a kilo-pound, equal to 1000 pounds (lbs.)

  • forces of the reservoir water, Ww and Hw.
    • Ww: vertical component of force of water -- weight of water

      Ww = (unit weight of water) * (volume of water)
      the unit weight of water = 0.0624 kip/ft3 or 9810 N/m3

      The weight of water should be disregarded for simulations.

    • Hw: the horizontal component of the force of the water
      Water applies pressure (*) in the shape of a triangle. The deeper the water, the more horizontal pressure it exerts on the dam. So at the surface of the reservoir, the water is exerting no pressure and at the bottom of the reservoir, the water is exerting maximum pressure.

      The base of the triangle is equal to the unit weight of water. The total force exerted by the triangle equals the area of the triangle:

      Hw = 0.5*(the unit weight of water)*height2
      height = the height of the water

      The force acts at the center of gravity of the triangle -- one-third of the way up from the bottom. The maroon arrow shows where the total force of the water acts.

  • uplift force, U: if there is no drainage, the uplift force is the result of the water pressure under the dam pushing up on the dam

  • weight of concrete, Wc
    Wc = (the specific weight of concrete)*(volume of concrete)
    the specific weight of concrete = 0.150 kips/ft3 or 23.6 N/m3

In a two-dimensional simulation, the thickness is assumed to be one. So the volume is just equal to the area times 1. The weight is assumed to act at the center of gravity (for simulation purposes, just make a guess at where the center of the dam is and apply the weight there).

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