Question: Given the potential one-foot thick slope failure along the circular arc shown in the sketch and the data in Table 2.14, find the safety factor

Given the potential one-foot thick slope failure along the circular arc shown in the sketch and the data in Table 2.14, find the safety factor for slice 2 and for slice 7, then show algebraically the safety factor for the slope. Slope height is 120 ft (36.6 m), slope angle is 29?, and specific weight is about 95pcf (15kN/m³). Cohesion is 367 psf (17.6 kPa) and the angle of internal friction of the material is 16?.

P bf !05 N' Ibf 105 (MN) (MN) Par. slice Weight Ibf

P bf !05 N' Ibf 105 (MN) (MN) Par. slice Weight Ibf 105 a() Lft (MN) Wn bf 105 (MN) W, Ibf 105 (MN) H ft (m) (m) 1.283 40 60 30 0.983 0.825 (0.370) 1.304 (0.584) 1.826 (0.575) 2.778 (1.245) 4.489 (18.29) (9.14) (0.440) 2.453 52 (15.85) (1.100) 75 28 30 (9.14) 30 24 4.101 (2.011) 6.413 (2.873) 10.90 (4.58) 11.33 (5.076) 10.29 (0.818) 1.443 (9.14) 30 (22.86) (1.837) 6.248 13 06 (27.43) (2.800) 97 (9.14) 56 (0.646) 0.570 (0.255) 1.976 (0.885) 4.185 (1.875) 3.629 10.885 (17.18) (29.57) (4.877) 97 (18.29) (29.57) (4.999) 82 10 60 11.158 24 75 9.400 (4.610) 5.985 (22.86) (24.99) (4.2110) 60 4.716 38 60 (2.681) 2.779 (18.29) (18.29) (2.113) 15 (17.18) (4.57) (0.880) (1.626) 1.965 (0.880) 1.172 (0.525) 1.965 45 6. 56 (1.245) 1.283 0.799 10 66 83 (0.575) (25.30) (0.358)

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Given Circular are failure data in Fig 233 and Table 214 Find FS2 FS7 show total FS WT 1 2 3 4 1 6 5 ... View full answer

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