4. A soil sample having a liquidity index of 0.85, a plasticity index of 16.50 and a plastic limit of 11.45. The soil sample has a specicfic gravity of 2.68 and a void ratio of 0.77. a. Compute the degree of saturation. b. Compute the air void ratio.
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- The volume of the solids is 65cc. The wet weight of the soil is 198g having a saturation at 48% and the dried weight is 176g. Determine the following: a. Total volume of the sample, cc. b. Volume of the voids, cc. c. Air void ratio, %.A sample soil is obtained and sieve analysis is performed to determined its particle distribution. Sieve Opening Mass Retained % Finer # 6 3.35 146 # 8 2.36 88 #16 1.18 125 # 40 0.425 56 #50 0.3 42 #140 0.106 127 # 200 0.075 126 Pan 29 Determine: 1. Effective Size (in mm and in 4 decimal places) 2. Uniformity Coefficient (in 2 decimal places) 3. Coefficient of Curvature (in 2 decimal places) 4. Sorting Coefficient (in 2 decimal places)b. Compute the plastic limit. C. Compute the natural water content. d. Compute the plasticity index. e. Compute the liquidity index. f. Compute the consistency index. PROBLEM NO. 4 From the table shows a summary of a liquid limit and plastic limit test. LIQUID LIMIT TEST Soil Wt. of wet Wt. of dry Water No. of Sample Soil (kg) Soil (kg) A. Content Blows 9.5 6.7 28 B 7.95 5.54 20 8.21 5.69 18 D. 12.85 8.83 13 PLASTIC LIMIT TEST Wt. of Dry Soil(kg) Soil Wt, of Wet Sample Soil(kg) 9.05 7.35 2 8.66 7.01 Natural Water Content Scil Wt. of Wt. of Sample Wet Soil Dry Soi! 1 9.69 7.0 9.47 6.86 a. Compute the liquid limit. b. Compute the plastic limit. C. Compute the natural water content. d. Compute the plasticity index. e. Compute the liquidity index GEOTECHNICAL ENGINEERING 1
- The results of the particle-size analysis of the soil are as follows: Percent passing No. 10 sieve = 100; Percent passing No. 40 sieve = 80; Percent passing No. 10 sieve = 58.The LL and PI of minus No. 40 fraction of the soil are 30 and 10 respectively. Classify the soil by USCS system a. CH b. SP-SC c. CL-ML d. CL-MLClassify the soil sample using AASHTO Classification System. Passing No. 10 = 89 % Passing No. 40 = 64 % Passing No. 200 = 48 % Plasticity for the minus No. 40 fraction: Liquid Limit = 36 % Plastic Limit = 27 % O GI = 2.0 O A-6 (2) O A-5 (0) O A-4 (2) O A-7-5 (0) O GI = 1.0 O GI = 05.5 The modified Proctor test (ASTM D 698, method C) was performed for a soil with G, = 2.70, and the results are as follows: Moist Weight in Mold, gf Water Content, % 6.5 3250 9.3 3826 12.6 4293 14.9 4362 17.2 4035 18.6 3685 (a) Plot the Ya versus w relation. (b) Determine Ya,max and wopt- (c) Calculate S and e at the maximum dry unit weight point. (d) What is Y, at Wopt? What is the range of water content if the relative compaction (R.C.) is required to be 95% of the modified Proctor Ya.max?
- 5.5 The modified Proctor test (ASTM D 698, method C) was performed for a soil with G, = 2.70, and the results are as follows: Moist Weight Content, % in Mold, gf Water 6.5 3250 9.3 3826 12.6 4293 14.9 4362 17.2 4035 18.6 3685 (a) Plot the Ya versus w relation. (b) Determine Ya,max and wopt- (c) Calculate S and e at the maximum dry unit weight point.Following are the results from the liquid and plastic limit test for a soil. No. of blows (N) Moisture Content % 16 39.5 20 36.1 28 30.0 a. Compute the liquid limit. b. Compute the plasticity index. C. Compute the liquidity index if the insitu moisture content is 26.5%. d. Compute the consistency index.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%.
- Answer ALL Questions Time: 2 Hours Q3: Classify the following soil by using the Unified soil classification system. Give the group symbols and the group names. Table 1. Sample 1 Grain Size Results (ASTM D422). Table 2. Atterberg Limits Results (ASTM D4318). Sleve No. Dia. (mm) % Passing Sample PL LL % in 12.7 100 G0 40 14 4.75 97 #10 2 94 #20 0.85 84 # 40 0.425 57 #60 0.25 32 140 0.106 15 #200 0.075 100 90 Atterberg Limit Results: LL = 60%, PL = 40% E 80 * 70- 60 - 50 40 30- a 20 10 1 30 a40 100 0.1200 10 80 0.01 Particle Diameter (mm) Answer ALL Questions Time: 2 Hours 70 CH or OH 60 A-line PI = 0.73(LL - 20) Or 50 40 30 CL MH OL 20 or CL-ML OH ML 10 - or OL 10 16 20 30 40 50 60 70 80 90 100 Liquid limit Figure 5.3 Plasticity chart Percent (%) Finer by Weight Plasticity index U-line PI = 0.(LL - 8) O Cengage Learning 2014The 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.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%. J