Question

1. A soil sample is taken from a small test hole in a soil borrow pit. The volume of the test hole is 5 ft. The wet mass of the soil is 18 kg and the dry mass is 21 kg. The specific gravity of the soil solids is 2.65. ( m 35. 3147ft)(1 kge 2.205 lb) ( 1 lb= 0.2248 N) Determine: a. Wet and dry density (Mg/m) b. Wet and dry unit weight (pcf) C. Water content (%) d. Void ratio (not above 1) e. Degree of saturation (%). Classify soils A, B, and C by the AASHTO classification system, with the calculated GI. Be sure to provide docume Appendix of this exam. 2. ntation as to how this is determined, this can be written in the table provided in the Description Percent finer than No. 10 sieve Percent finer than No. 40 sieve Percent finer than No. 200 sieve 100 92 86 70 38 83 48 20 20 15 94 80 63 40 21 Liquid limit Plastic limit

Answer 1

V = 0.5 ft^3 = 0.0141584 m^3 w = 18 kg: w_s = 21 kg g_s = 2.65 note: w_s = 18 kg + wd w_wet = w-= 21 kg/wet density(bulk) = 21.0 times 10^3 /0.0141584 1 mg = 10^3 kg = 1.4832 times 10^6 mg/m^3 dry = 1.8 times 10^3/0.0141584 = 1.2713 times 10^6 mg/m^3 (b) wet density = 21 times 2.205/92.61 lb/ft^3 or 92.61 pcf dry density = 18 times 2.05/0.5 79 = .38 pcf (c) w% = (21 - 18/18) times 100 = 16.67% d = g_s times w/1 + e e = 2.65 times 62.45 - /79.38 1.0848 tildeequal 1.0 s_r = w.g_/e = 0.1666 times 2.65/1 times 100 = 0.44 times 100 = 44% \r\n N_10 rightarrow 94 N_40 rightarrow 80 N_200 63 w_L = 40 w_p = 21 I_p = 4. - 21 = 19 here more than 35% passing N_200 hence it is materials for a - 5 and a
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