Problem 1

h

B

A

h₁

50 Points

A triangular concrete dam separates two different reservoirs, as indicated in the figure. The water

level to the left of the dam is at h and is just not overflowing the dam. The water level to the

right of the dam is at hy. The dam is made out of concrete, but is able to pivot around point

A. The pressure underneath the dam is a direct function of its weight: at point A, the pressure is

PA-yhr, while the pressure increases linearly to PB = yh at point B./nPart a: Draw a free-body diagram with all forces represented.

Part b: Calculate the resultant force, and the point of application (= center of pressure) at the

left-hand wall of the dam.

Part c: Calculate the resultant force, and the point of application (= center of pressure) at the

slanted right-hand wall of the dam.

Part d: Calculate the resultant force, and the point of application (= center of pressure) at the

bottom wall of the dam.

Part e: Calculate the weight of the dam, and the point of application (= center of gravity) of the

dam, if the density of concrete is pe=2,500kg/m³3.

Part f: Perform a sum of moments around the pivot point A, to obtain a symbolic expression for

the length of the wall L (in terms of all other parameters) when the concrete dam would just start

to topple over.

Part g: Calculate the length L for the wall under the conditions of part e if the height h = 30 m

and h = 10 m, and the density of water is pu = 1,000kg/m³3. Hint: note that the equation

from part f has a small non-linear term that also depends on L, so you need to do some type of

approximation or interpolation for the angle of the triangle.