Given the potential one-foot thick slope failure along the circular arc (R = 300 ft, 91.44 m)
Question:
Given the potential one-foot thick slope failure along the circular arc (R = 300 ft, 91.44 m) shown in the sketch and the data in Table 2.14, find the safety factor possible with drawdown of the water table below the failure surface. 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?.
Transcribed Image Text:
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 (0.575) 2.778 (1.245) 4.489 (18.29) (9.14) (0.440) 2.453 (15.85) (1.100) 4.101 (0.370) 1.304 (0.584) 1.826 28 52 30 (9.14) 30 24 75 (2.011) 6.413 (2.873) 10.90 (9.14) 30 (22.86) (1.837) (0.818) 1.443 6.248 13 06 (27.43) (2.800) 97 (9.14) 56 (0.646) 0.570 10.885 (17.18) (29.57) (4.877) 97 (18.29) (29.57) (4.999) 82 (0.255) 1.976 (0.885) 4.185 (1.875) 3.629 (4.58) 11.33 (5.076) 10.29 (4.610) 5.985 11.158 10 60 24 75 9.400 (22.86) (24.99) (4.2110) 4.716 (18.29) (18.29) (2.113) 1.965 (17.18) (4.57) (0.880) 8. 38 60 60 (2.681) 2.779 (1.626) 1.965 15 45 6. 56 (1.245) 1.283 (0.880) İ.172 (0.525) 0.799 (0.358) 10 83 66 (0.575) (25.30)
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