2. A sample of soil has 60% passing the No. 4 sieve and 42% passing the No. 200 sieve. The liqui limit (LL) for the soil is 28%, and the plasticity index (PI) is 4.5%. Classify this soil according to the USCS.
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- Question 11 The results of the particle-size analysis of a soil are as follows: Percent passing the No. 10 sieve = 80 • Percent passing the No. 40 sieve = 40 Percent passing the No. 200 sieve = 20 %3D The liquid limit and plasticity index of the minus No. 40 fraction of the soil are 30 and 5, respectively. Classify the soil by the AASHTO system. A A-2-4 B А-2-6 C) A-3 D A-2-5QA:If the specific gravity of a soil 2.69 and dry unit weight 1.915 gm/cm2, the sample of this soil is dried from its state at plastic limit, the volume change is 25% of its own volume at plastic limit, Similarly, when the same sample is dried from its state of liquid limit, the volume change is 45% of its own volume at liquid limit. determine the plasticity index if the value of liquid limit = 35%. (Soil mechanic)QA:If the specific gravity of a soil 2.69 and dry unit weight 1.915 gm/cm2, the sample of this soil is dried from its state at plastic limit, the volume change is 25% of its own volume at plastic limit, Similarly, when the same sample is dried from its state of liquid limit, the volume change is 45% of its own volume at liquid limit. determine the plasticity index if the value of liquid limit = 35%.
- Classify 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 = 0Using Unified Soil Classification System, determine the following: (a) Coefficient of Gradation (b) Group Symbol. Passing No. 4 = 47 % Passing No. 200 = 10 % D10 = 3.45 mm %3! D30 = 6.29 mm De0 = 13.72 mm Plasticity for the minus No. 40 fraction: Liquid Limit = 28 % Plastic Limit = 17 % O GC-GM GP-GC O GC O Coef. Of Gradation = 0.84 O GP O Coef. Of Gradation = 3.97Using AASHTO Classification System, determine the following: (a) Group Index (b) Soil Classification Passing No. 10 = 55 % Passing No. 40 = 47 % Passing No. 200 = 23 % Plasticity for the minus No. 40 fraction: Liquid Limit = 22 % Plastic Limit = 17 % O GI = - 0.40 say 0 O GI = 0.60 say 1 O GI = 0 O A-2-4 (1) O A-2-7 (0) O A-1-b (0) O A-2-6 (1)
- PROBLEM No. 6 Atterberg Limit From the laboratory results of the Atterberg Limit Test, the Liquid Limit of the soil is 65% and the Plastic Limit of the soil is 40%. a. Determine the Plasticity Index. b. If the in-situ moisture content of the soil is 85%, determine the Liquidity Index of the soil. c. Determine the state of consistency of the in-situ soil. d. Determine the Consistency Index of the soil.QA:If the specific gravity of a soil 2.69 and dry unit weight 1.915 gm/cm, the sample of this soil is dried from its state at plastic limit, the volume change is 25% of its own volume at plastic limit, Similarly, when the same sample is dried from its state of liquid limit, the volume change is 45% of its own volume at liquid limit. determine the plasticity index if the value of liquid limit=35%. %3DA sample of inorganic soil has 100% passing the No. 4 sieve and 75% passing the No. 200 sieve. The liquid limit for the soil was found to be 60 %, while the plasticity index was 24%. Classify this soil according to the USCS.
- The liquid limit (LL) of a soil has been found as 78% and the plastic limit ( PL ) as 37% . Calculate the following: The plasticity index (PL) The liquidity index (LI) of the sample soil if water content ( W ) has been found as 56%The following soil as specified by the AASHTO System has the following results: Percent finer than No. 10 sieve . 90 . 76 . 34 Percent finer than No. 40 sieve Percent finer than No. 200 sieve Liquid limit = 37 Plasticity index = 12 Calculate the partial group index for the soil that belongs to groups A-2-6. 2.38 2.85 0.85 0.38The 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-ML