The face of the dam adjacent to the water is vertical and is 90 ft. wide at the top, 60 ft. wide at the bottom, and 20 ft. high in the form of an isosceles trapezoid. Find the total force on the face of the dam due to the water pressure.
The face of the dam adjacent to the water is vertical and is 90 ft. wide at the top, 60 ft. wide at the bottom, and 20 ft. high in the form of an isosceles trapezoid. Find the total force on the face of the dam due to the water pressure.
Change to SI unit:
Top width=90ft=27.43m
Bottom width=60ft=18.28m
Height of dam(h)=20ft=6.09m
Slope due to line of different sections constant width=(18.28-27.43)/6.09=-1.50
Width equation in form of depth
w=27.43-1.5h
Force due to submerged water= integrate over the range of 0 to 6.09 of PdA
P=1000*9.81*h
dA=wdh
dA=(27.43-1.5h)dh
PdA=1000*9.81*h*(27.43-1.5h)dh
PdA=9810(27.43h^2/2-1.5h^3/3)
F=9810{(27.43*6.09^2/2-1.5*(6.09^3/3)}
F=9810(508.66-112.93)
F=9810(508.66-112.93)
F=3882079.27N
F=3882.07927kN
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