3 m water 8 m fconcrete 2200kgm 6 in The dam shown in the f

3 m water 8 m f)concrete = 2200kgm\" 6 in The dam shown in the figure is built of concrete having a density of 2200 kg/ m2 and rests on a solid foundation. Determine the minimum coefficient of friction between the dam and the foundation required to keep the dam from sliding at a water depth of 6.1 m. Assume that the dam is 100 m into the page.

Solution

Solution:-

Given

Density of concrete =2200kg/m³ =22 kN/m³

Height of dam = 8 m

b = 6 – (3 +1.9) =1.1 meter

Depth of water (d) = 6.1 meter

Water load produced by water weight(F_wv) = 1/2 *1.1 *6.1 *9.8 = 32.879 kN/m

Where density of water = 9.8 kN/m³

Hydrostatic force (F_wh) = 1/2 *6.1 *6.1 *9.8 =182.329 kN/m

Uplift force produced by ground water (F_u) = ½ *6.1 *6 *9.8 =179.34 kN/m

Weight of dam

W₁ = 1/2 *3*8*22 =264 kN/m

W₂ = 3 *8 *22 = 528 kN/m

Factor of safety = (f *V)/(H)

V =W₁ +W₂ +F_wv - F_u

       = 264 + 528 + 32.879 – 179.34 = 645.53 kN/m

H = F_wh = 182.329 kN/m

Let factor of safety = 1.5

Friction coefficient(f) = 1.5 * 182.329/(645.53)

                                      = 0.429        Answer