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.

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/m) 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)