With reference to the sketch and tables, find the safety factors for the four cases posed using the method of slices. Include all important calculations. If you use a spreadsheet, include suitable data prints. Complete a table for each of the conditions following:

1 With water table (WT) 1 and no relieving benches

2 With WT 2 and no relieving benches

3 With WT 2 and a 140?60 ft (42.7?18.3 m) relieving bench at the crest

4 With WT 1, a 140?60 ft relieving bench at the crest, and a relieving Bench 2 at the toe required to give a factor of safety of 1.05. Compute the length L of Bench 2 in this case.

Factor of safety = ________________________

Notes: Units are feet-ft (m).

Water table 1 (WT₁) is 30 ft (9.14 m) below the ground surface.

Water table 2 (WT₂) is 100 ft (30.5 m) below the ground surface.

Areas ft² (m²):

Parts of relief bench 1: slices:

A?= 118 (10.96) ? ? ? ? ??1 = 470 (43.7)

B?= 1,682 (156.3) ? ? ? ?2 = 1,937 (180)

C?= 4,800 (445.9) ? ? ? ?3 = 5,451 (506.4)

D?= 1,800 (167.2) ? ? ??4 = 12,246 (1,137.7)

Notes

a. A blank means zero (0). Drift is cohesion less, clay and interfaces are frictionless. The difference between dry and wet specific weight is neglected in this problem.

Transcribed Image Text:

Not to scale Ground surface 140 WTIAelief benchl. 45° Young drift 60 -Slope face Old drift 40 WT 2 R= 160 Relief bench 2 Medium clay Interface I 38 Varved clay 22 Interface 2 Wn P w'n tan o Rc L cL Section Weight a Ws Properties": Cohesion c psf (kPa) Friction angle o (°) Specific weight y pef (kN/m³) Property material T10 (17.4) 125 (19.8) Young drift Old drift 36 33 800 (38.3) 2,300 (I10.2) 2,800 (134.1) 700 (33.5) Interface I Medium clay Varved clay Interface I Iron formation 125 (19.8) 105 (16.6)