A soil sample has a unit weight of 1.9 g/cc and w=12%. If Gs of solids is 2.65, determine the void ratio in percent. CHOICES: a.56.8 b.1.7 c.36
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- SITUATION 3: A 100 cc proctor mold was used to collect a wet soil sample. Its weight was found to be 200 gms. Its oven dried weight was recorded as 189 grams. G = 2.7 7. The moisture content is most likely: a. 0.058201 c. 0.069841 8. Its void ratio is: a. 0.771429 c. 0.514286 b. 0.064021 d. 0.046561 b. 0.471429 d. 0.428571A dry sand is placed in a container having a volume of 0.30cu.ft. The dry weight of sample is 31lb. Water is carefully added to the container so as not to disturb the condition of the sand. When the container is filled, the combined weight of soil llus water is 38.2 lb. a. Compute the void ratio in the container. b. Compute the specific gravity of the soil particle. c. Determine the dry unit weight.3-8 A 1-ft3 of undisturbed soil sample has a dry weight of 107 lbs. The specific gravity of the soil solids is 2.70. Need Vs VT = 1 To solve for Vs = use the following equation Ws=Vs*Gs*yw (solve for Vs) a) Calculate the void ratio of the sample. b) Calculate the porosity of the sample. DC
- U B. ( A Soil Specimen has the following characteristics: % passing No.4 sieve = 85 % passing No.200 sieve = 11 D60 = 2 mm D30 = 0.35 mm D₁0 = 70 μm L.L = 36% P.L = 31% D8 = 2 µm 1. Classify the specimen according to the Unified Soil Classification System (USCS). Assign the group name and the group symbol. 2. Determine the soil activity. 3. Determine the soil Liquidity index and consistency index if we = 26%.Permeability tests were performed on a soil sample, under different void ratio and different temperatures and the following results were obtained. Test No. 1 2 Void ratio (e) 0.65 1.02 Temperature °℃ 25° 40⁰ Estimate the coefficient of permeability at a temperature of 20°C for a voids ratio of 0.80. Given the following physical properties of water: At 20°C, n = 10.09 × 10+ and p„ = 0.998 g/cm³ At 25°C, n = 8.95 × 104 g sec/cm² and p k(cm/s) 0.4x10 1.9x10-¹ At 40° C, n = 6.54 × 10-¹ g sec/cm² and p $0.997 g/cm³ = = 0.992 g/cm³The field wt of soil sample is 1900 kg/m³and the unit wt of the soil particle is 2660 kg/m³. A. Compute the dry unit weight if the moisture content is 11.5%. B. Compute the void ratio. C. Compute the degree of saturation.
- Organic soils are typically characterized by high void ratio, low specific gravity, and high compressibility. Following are the results of a consolidation test on a sample of organic soil obtained from southwest Florida. Pressure, o' (kN/m²) Change In dlal reading, AH (mm) 0.284 0.150 12 24 0.315 48 0.564 100 0.823 200 2.25 400 5.34 Given that the initial height of the specimen= 20.6 mm, mass of dry specimen = 12 g, area of specimen = 31.67 cm², and Gs = 2.49, d. Plot the e-log o' curve. Determine the pre-consolidation Calculate the compression index, Cc. е. pressure. f.A soil sample has a porosity of 40 percent. The specific gravity of solids is 2.70 . What is void ratio? A) 0.467 B) 0.567 C) 0.667 D) 0.743Find the value of Cc of the following soil data sample Weight Percentage Cumulative Percentage Sieve Weight retained (gm) percentage retained (C) retained finer S.NO Select one: O a. 10.85 size (gm) N = (100 - C) {W,/ W,} x 100 1 75mm 100 O b. 7.36 63mm 100 O c. 18.34 50mm 100 O d. 2,96 4 37.5mm 454 22.7 22.7 77.30 5 28mm 285 14.25 36.95 63.05 6 20mm 207 10.35 47.30 52.70 7 14mm 318 15.90 63.20 36.80 8 10mm 206 10.30 73.50 26.50 6.3mm 118 5.90 79.40 20.60 10 5.0mm 102 5.10 84.50 15.50 11 3.35mm 64 3.20 87.70 12.30 12 2.0mm 88 4.40 92.10 7.90 13 1.18mm 35 1.75 93.85 6.15 14 600µ 42 2.10 95.15 4.85 15 300µ 33 1.65 97.60 2.40 16 150µ 18 0.90 98.50 1.50 17 75µ 22 1.10 99.6 0.40 18 Pan 8 0.40 100 In the pan we have = 100 W,= 2000 2 -99.6 = 0.4 % of material 2. 3.
- Sr B. ( A Soil Specimen has the following characteristics: % passing No.4 sieve = 85 % passing No.200 sieve = 11 D60 = 2 mm D30=0.35 mm D₁0 = 70 μm D8=2 μm L.L = 36% P.L = 31% 1. Classify the specimen according to the Unified Soil Classification System (USCS). Assign the group name and the group symbol. 2. Determine the soil activity. 3. Determine the soil Liquidity index and consistency index if wc = 26%.The results of a laboratory consolidation test on a clay specimen are the following. Pressure, ir (lb/ft²) 500 1.000 2.000 4.000 8,000 16.000 Total height of specimen at end of consolidation (in.) 0.6947 0.6850 0.6705 0.6520 0.6358 0.6252 Given the initial height of specimen = 0.748 in.. G specimen = 95.2 g. and area of specimen = 4.91 in.²: a. Plot the e-log or curve = 2.68. mass of dry b. Determine the preconsolidation pressure c. Calculate the compression index, C.Q # 1. Organic soils are typically characterized by high void ratio, low specific gravity, and high compressibility. Following are the results of a consolidation test on a sample of organic soil. Pressure, o (kN/m2) Change in dial reading, AH (mm) 0.284 12 0.150 24 0.315 48 0.564 100 0.823 200 2.25 400 5.34 Given that the initial height of the specimen = 20.6 mm, mass of dry specimen = 12 g, area of specimen = 31.67 cm2 and G, = 2.49. Plot the e, log o' curve. b. Determine the preconsolidation pressure. a.