For a soil specimen, the following are given:% passing Sieve No. 4 = 92%% passing Sieve No. 10 = 81%% passing Sieve No. 40 = 78%% passing Sieve No. 200 = 65%LL = 48%PI = 32%Classify the soil by the AASHTO system.O A-2-7 (15)O A-7-6 (18)O A-7-6 (32)O A-2-5 (6)

Look at the % passing through No 200 sieve=65>35% silt and clay minerals

Answered: For a soil specimen, the following are…

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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A 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 AASHTO system.Justify your answer. a. A-5 (6) b. A-6 (5) c. A-7-6 (20) d. A-4 (5)
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-2-5
A 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 AASHTO system.Justify your answer. provide fbd a.A-6 (5) b.A-5 (6) c.A-4 (5) d.A-7-6 (20)
The following data are the results of a sieve analysis. Detemine the percentage retained on sieve #200. Mass of soil Sieve no. Opening retained (mm) on each sieve 4 4.75 10 2 18.5 20 0.85 53.2 40 0.425 90.5 60 0.25 81.8 100 0.15 92.2 200 0.075 58.5 Pan 26.5 Select one: а. 94 b. 58 c. none of the choices d. 27 е. 38
The soil tested at the laboratory was found that the particle passing sieve No.200 is 89%, retained on sieve No. 4 is 3%, Liquid limit of 70%, organic matter of 25%, and Plasticity Index of 40%. Determine the type of soil * PLASTICITY CHART 60 50 CH A LINE: Pl 0,73(LL-20)| 30 CL MH&OH 20 CL ML ML&OL 10 20 30 40 50 60 70 80 90 100 LIQUID LIMIT (LL) (%) O Organic clay O Clay with high plasticity O Organic Silt O Sandy Clay PLASTICITY INDEX (PI) (%)
Do the following: a) Calculate the percent retained and percent passing each sieve, and graph the results on a semi-log graph (you may use the attached chart, or graph in Excel and include with your submission). Then, characterize the gradation of the material. b) Calculate the values of D10, D30, D60, Cu, and Cc c) If the LL is 48 and the PL is 26, determine the USCS classification of the soil, and state its suitability for the purpose stated in the scenario.
mpt=541762&cmid%3D103601 Marks 25 Question 1 A sample of AL Batina soil tested in the laboratory. The gradation analysis results are as shown in table Q1. The Atterberg limit tests indicate that the liquid limit is LL = 30% and the plastic limit is PL=27%. Table Q1 Sieve No.# Sieve size (mm) Finer (%) 4 4.75 90 8. 2.36 84 16 1.3 60 30 0.6 30 50 0.2 10 200 0.075 8 i. Determine the uniformity coefficient Marks] ii. Determine curvature coefficient [5 [5 Marks] iii. Classify AL Batina soil according to Unified Soil Classification System (USCS). [10 Marks] iv. Determine the group symbol and group name Marks) [5
2. The following are the results of a sieve analysis. U.S. Mass of soil sieve no. retained (g) 4 10 18.5 20 53,2 40 90.5 60 81.8 100 92.2 200 58.5 Pan 26.5 a. Determine the percent finer than each sieve and plot a grain-size distri- bution curve. b. Determine Du, Dy, and D, for each soil. c. Calculate the uniformity coefficient C. d. Calculate the coefficient of gradation C.
42.) The results of the sieve analysis done on a soil are shown below. Sieve No. Diameter Soil A 4 4.760 96 6 2.380 91 10 2.000 89 20 0.840 85 40 0.420 81 60 0.250 78 100 0.149 67 200 0.074 64 LL 40 PL 30 MH CH ML None of the above CL
43.) The results of the sieve analysis done on a soil are shown below. Sieve No. Diameter Soil A 4 4.760 100 6 2.380 98 10 2.000 91 20 0.840 85 40 0.420 84 60 0.250 79 100 0.149 61 200 0.074 55 LL 60 PL 40 ML None of the above. CL CH MH
B. Solve the following problems. Write your answer in a long bond paper. Avoid Erasures. 1. The results of the sieve analysis are shown below: a) What is the percentage of the soil retained in No. 200 sieve? b) What is the effective grain size of the soil in mm? c) Determine the uniformity coefficient. Sieve No. 8 Diameter (mm) 4.76 2.38 Mass retained (g) 25 80
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)

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