Question

Question 4: (25 points, (a): 10 pts, (b): 10 pts, (c):5pts)). For the below given slope, calculate the factor of safety against rotational failure; a) For static condition b) For seismic condition (kn = 0.15) c) Comment on the adequacy of the calculated safety factors for static and seismic conditions. 300 500 R=20m X = 16m Co = 40 kPa x = 12m Co = 70 kPa W = 2000 KN FS = Mresisting - Cu,1*11*RA + Cu.2 * L2 * R2 Mdriving Mdriving

Answer 1

Reyr- solya D 50 Given information 3ol] R=20m Radius Razom Xy = 16m Weight W=2000kN. (6=40kpa X6=12m Cu = 70kpa W= 2000 KN Calculation for overtwining moments case ar static Weight Wa 2000 KN moment arm Xb 12m moment of static = Waxb) 2000 (12) 24000 KN-m. Case Ho - Pseudo static Total W= kg W 0.15 X 2000 = 300 KN, moment arm Xv=16m moment of pseudostatic = (kaw) (x4) - 300x16 - 48 00 KN-M. Therefore, Total moment momat of static t prendostatic moment で 24000 + 4800 28.800 KN-m.

Calculation of resisting moments. (EMR) EMR = CILIR + C₂ La R. c = 40 kpa + 70kpa Given in the figure, R = 20m 2Ãx20x30 4 = 10.47m 360 L2= 25x20x50 - 17.45m 360 Total Resisting moment EMR - = (40) (10.47 (20) + (70) (17:45) (20) 8376 + 24430 {{MR = 32806 KN-M} @ factor of saftey for static condition Eme Static moments 32806 24000 1.366 (b) factor of saftey for seismic condition EMR Static momentst Prendo moments 32806 32806 28,800 24000 + 4800 (Fs) = 1.139

- The factor of saftey for static condition is less than 1.5. Therefore it's uncate F-o.s for seismic condition is less than the Fois for static condition.