Answer Happy

For the slope shown below, determine factor of safety for each of the following cases. Assume the reinforcement only acts in the horizontal direction. Do not consider tension crack in your analysis Radius = 55 ft Center of Circle (15 ft. 40 ft) a = 123° El = 30 ft Weight = 215,600 lbs/ft coordinates = (30.65, 12) El = 10 ft T = Horizontal Reinforcement Force = 800 lbs/ft El = 0 Ft (0, 0) Clay C = 600 psf, 0 = 0° y = 100 pcf a) Calculate the factor of safety for static case using the forces and dimensions shown in the figure. b) Repeat your factor of safety calculation this time for a pseudo-static case. For this site, you are given the peak ground acceleration (PGA) of 0.4 g. Use Kn = PGA/2 and assume the same failure surface is applicable for the pseudo-static case. Only consider the horizontal component of seismic loading. c) If you were asked to NOT use the same slip surface and re-search to find the most critical failure surface for the pseudo-static case, what would you expect in terms of the shape of failure surface for the pseudo-static case versus the static case? No calculation is needed for this part and just explain the differences between the critical slip surfaces from the two cases.