Question

QUESTION 3 (20 marks) A. Following are the results of a standard penetration test in sand. Determine the corrected standard penetration numbers, (N1), at various depths. Note that the water table was not observed within a depth of 35ft below the ground surface. Assume that the average unit weight of sand is 110 lb/ft! (5 marks) 5 Depth, Z NA 8 10 7 15 13 20 15 25 12 B. For the soil in Question 3. A (above) estimate the soil average friction angle from Z = 0 to Z = 7.5m. (5 marks) C. Table 3 displays the results of sieve analysis performed on four different soils. Classify these soils according to the USDA textural classification system. (4 marks) Table 3 Soll Particle-size distribution А B с D Gravel Sand Silt Clay 10 20 41 29 21 12 35 32 0 18 14 58 12 22 26 40 D. Refer to Figure 3. Given: B = 300, y = 155 kN/m", -20%, and c=15 kN/m². • Determine the height, H, that will have a factor of safety, FS, of 2 against sliding along the soil-rock interface (3 marks) • Calculate the height, H, for critical equilibrium (3 marks) Figure 3

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Answer 1

unit 3 weight of sand = 11016/ft 3 weight of water = 62.4 lb/ft 3 Depth N60 5 8 10 Z 15 13 20 15 25 12 At d = 50 ft o overbunden pressure (6) Yand xd -26 110X5 62.4x5 238 lb/ft² (n = Normalizing face for = 0.77 logget cn = 0.77 logio (Alе tо реек) 2000 2000 238 0.711 Corrected standard penetreation test o (N1) 60 = .cn Noo = 0.71X 8 - 5.688

I Lege دهم (8) o. 4800 con N60 At d = isft 10XLS - 62.4x15 2000 con 0.47 logio tour (N16) 60 = 30-344 0.3zaxis 5.16 # = الى 20 ft 6!= 40 x 20- 62.4x20 952 2000 en- 8.77 logio (N1)60 = 0 248 x 20