Based on the sieve analysis a soil has % finer for the following sieves: No. 10- 86% ; No. 40- 56% and No. 200- 8%. No Plasticity. Classify the soil by AASHT0 classification and give also the group index.

Based on the sieve analysis a soil has % finer for the following sieves: No. 10- 86% ; No. 40- 56% and No. 200- 8%. No Plasticity. Classify the soil by AASHT0 classification and give also the group index.

Principles of Geotechnical Engineering (MindTap Course List)

9th Edition

ISBN:9781305970939

Author:Braja M. Das, Khaled Sobhan

Publisher:Braja M. Das, Khaled Sobhan

Chapter5: Classification Of Soil

Section: Chapter Questions

Problem 5.10P

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Concept explainers

Soil classification and characteristics

Soil classification is a method of categorizing soil based on its properties, types, and uses. The motive of soil classification is to arrange various types of soils and divide them into groups for various engineering and agricultural purposes. Soil classification is carried out by various agencies as per their purposes. From an engineering perspective, soil classification is done to find out the suitability of soil for various construction purposes like the construction of dams, foundations, highways, infrastructures, and so on. For engineering purposes, the soil is classified by the following systems:

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Based on the sieve analysis a soil has % finer for the following sieves: No. 10- 100% ; No. 40- 78% and No. 200- 82%. The liquid limit is 32 and Plasticity index is 12. Classify the soil by AASHT0 classification and give also the group index.
For a given sieve analysis: No.4 - % finer is 94, for No 200 - %finer 3. Cu=4.48 Cc=1.22. Classify the soil using USCS. Justify your answer.
For a given sieve analyis: No. 4 - % finer 100%, for No. 200 -%finer 86. LL=55, PL= 28. Classify the soil using USCS. Justify your answer.
2. Determine the Unified Soil Classification System (USCS) classification for the following inorganic soil. [Ans. CH]. A. Grain Size Distribution Test. Percent Sieve No. 4 10 20 40 60 100 200 Passing (%) 100 100 99 95 93 88 86 B. Plastic Limit Test Tare Number Wt. Wet Soil + Tare (g) Wt. Dry Soil + Tare (g) Wt. of Tare (g) C. Liquid Limit Test Determination Number of Blows Water Content (%) 1 2 6.54 7.24 5.40 5.95 1.17 1 18 69 1.17 2 27 64 3 34 61
Based on the sieve analysis a soil has % finer for the following sieves: No. 10- 80% ; No. 40- 68% and No. 200- 48%. The liquid limit is 30 and Plasticity index is 11. Classify the soil by AASHT0 classification and give also the group index.a. A-6 (2) b. A-7-5(8)c. A-6 (9) d. A-7-6(8)
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-ML
A soil sample has 98% passing the No. 4 sieve and 74% passing the No. 200 sleve. The LL for the soils 58% and the PI is 30%. Classify the soil according to the USCS. Show step by step solutions.
Based on the sieve analysis a soil has % finer for the following sieves: No. 10- 90% ; No. 40- 74% and No. 200- 58%. The liquid limit is 38 and Plasticity index is 12. Classify the soil by AASHT0 classification and give also the group index. a. A-7-5(8) b. A-6 (5) c. A-6 (9) d. A-7-6(8)
Classify the given soils using the Unified Soil Classification System (USCS). Give the group symbols and group names. (Show the path of the classification.) Sieve Analysis (percent finer) Soil No. No. 4 No. 200 LL PL Comments A 94 3 NP Cy = 4.48 and C = 1.22 - В 100 77 63 25 B.
Use the unified soil classification system to classify the soil with 65% of the soil sample retained on No 4 Sieve, 30% retained on No 200 Sieve, Cu=3 and Cc-0.9.Also the LL = 28% and PL = 24.5% THE GROUP NAME IS
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 0.38
Problem 1 The following grain size distribution test results were obtained from sieve analysis test. Percent Passing U.S. Sieve Number 4 100 10 90 20 64 40 38 08 200 18 13 Also, the Atterberg test results revealed the following results LL = 23 and PL = 19 Classify the soil by using a) AASHTO soil classification system b) USCS soil classification system Give group names and group symbols. Show the GSDC and sketch of finding necessary information from GSDC. Problem 2