Question

2. A dam 20 m long retains H = 7 m of water, as shown in figure below. Find the total resultant hydrostatic force (F) acting on the dam and the location of the center of pressure (hcp) if a is 60° 111

Answer 1

Total hydro static force acting on inclined plane surface = $\rho g A h$

$\rho$ = 1000kg/m³

A, Submerged area of dam A = Inclined length * Length of dam

= H/sin$\alpha$ * 20 = 7/sin 60 *20 = 161.66 m²

h, depth of CG of area from water surface can be calculated as shown in figure.

M W 083m m AB = u ino: 8.083m .LG along inclined sunforce 8.083 = 4:04m from figure, Cas 30 = h ; h= 4:04 los 30 = 3.5m incl

F = 1000*9.81* 161.66* 3.5 = 5550596.1 N

Location of centre of pressue hcp = $\frac{I_{G}}{Ah}+h$

I_{G =} 20*8.083^3/12   = 880.17 mm⁴

  hcp = 880.17/(161.66*3.5) + 3.5 = 5.056 m