22. 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% a. Classify the soil by the AASHTO system. b. Classify the soil by the USCS.

22. 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% a. Classify the soil by the AASHTO system. b. Classify the soil by the USCS.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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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:

Question

22. 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%
a. Classify the soil by the AASHTO system.
b. Classify the soil by the USCS.

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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 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)
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.
7. Following details regarding the particle size distribution and other index properties pertinent to the classification of soil. Choose three soils and classify each soil according to USCS and AASHTO classification systems. When necessary, assume Cu = 4.48 and C = 1.22. Your solution may be presented in similar form as what was included in your module 4. Sieve Analysis (Percent Finer) Soil No Liquid Limit Plastic Limit No. 10 No. 40 No. 200 1 98 80 50 38 29 2 100 92 80 56 23 3 100 88 65 37 22 4 85 55 45 28 20 5 92 75 62 43 28 6 48 28 6 N.A. NP 7 87 62 30 32 24 8 90 76 34 37 25 9 100 78 8 N.A. NP 10 92 74 32 44 35
The grain-size analysis is for a soil is given as follows: Sieve No. Passing Diameter% (mm) 4.75 2.00 0.850 0.425 0.250 0.150 0.075 # 4 #10 #20 # 40 #60 #100 # 200 94 63 21 10 5. 3. Given that the soil is non plastic, classify the soil by using the Unified Soil Classification System. What is the value of the uniformly coefficient Cu. What is the value of the coefficient of gradation Ce What is the value of the sorting coefficient So. O Classify the type of soil
7. Following details regarding the particle size distribution and other index properties pertinent to the classification of soil. Choose three soils and classify each soil according to USCS and AASHTO classification systems. When necessary, assume Cu = 4.48 and Cc = 1.22. Your solution may be presented in similar form as what was included in your module 4. Sieve Analysis (Percent Finer) Soil No Liquid Limit Plastic Limit No. 10 No. 40 No. 200 98 80 50 38 29 2 100 92 80 56 23 100 88 65 37 22 4 85 55 45 28 20 92 75 62 43 28 6 48 28 N.A. NP 7 87 62 30 32 24 8 90 76 34 37 25 100 78 8 N.A. NP 10 92 74 32 44 35
Question 42 The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Tables are provided below. Sieve No. Diameter (mm) Percent Passing No. 4 4.760 100 No. 8 2.380 88 No. 10 2.000 69 No. 20 0.840 61 No. 40 0.420 56 No. 60 0.250 43 No. 100 0.150 19 No. 200 0.075 9. Characteristics of No. 40 Fraction LL PL General classification Granular materials (35% or less of total sample passing No. 200) A-1 A-2 Group classification A-1-a A-1-b A-3 A-2-4 A-2-5 A-2-6 A-2-7 Sieve analysis (percent passing) No. 10 50 max. No. 40 No. 200 30 max. 15 max. 50 max. 25 max 51 min. 10 max. 35 max. 35 max. 35 max 35 max. Characteristics of fraction passing No. 40 Liquid limit. Plasticity index 40 max. 41 min. 40 max. 41 min. 6 max NP 10 max. 10 max. 11 min. 11 min. Usual types of significant constituent Stone fragments. gravel, and sand Fine Silty or clayey gravel and sand sand materials SUS F1 F2 F3 F4 FS F6 F7 F8 24 4 % & 5 R Y 因

0 10 16 20 30 40 Liquid limit 50 60 70 80 90 100 0 10- CL-ML 20 Plasticity index 30 J A-line PI = 0.73(LL - 20) 40- U-line PI = 0.9(LL - 8) 70 7 Table 5.2 Unified Soil Classification System (Based on Material Passing 76.2-mm Sieve) Criteria for assigning group symbols O Gravels More than 50% of coarve fraction retained on No. 4 sieve Sands 50% or more of coerse fraction passes No. 4 sieve Coarse-grained soils More than 50% of retained on No. 200 sieve Fine-grained soils 50% or more passes No. 200 sieve Silts and clays Liquid limit less than 50 Silts and clays Liquid limit 50 or more Clean Gravels Less than 5% fines Gravels with Fines More than 12% fines Clean Sands Less than 5% fines Sands with Fines More than 12% fines Inorganic Organic Inorganic Organic C, 4 and 1 c, s 3 C 3r PI 7 and plots on or above "A" line (Figure 5.3) C = 6 and 1 c, ≤ 3 C6 and/or C, 3º Pl 7 and plots on or above "A" line (Figure 5.3) Pl> 7 and plots on or above "A" line (Figure 5.3 PI < 4 or plots below "A"…
Question 44 The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Tables are provided below. Sieve No. Diameter (mm) Percent Passing No. 4 4.760 99 No. 8 2.380 91 No. 10 2.000 90 No. 20 0.840 85 No. 40 0.420 81 No. 60 0.250 79 No. 100 0.150 67 No. 200 0.075 60 Characteristics of No. 40 Fraction LL 39 PL 32
46.) The results of the sieve analysis done on a soil are shown below. Sieve No. Diameter Soil A 4 4.760 89 6 2.380 70 10 2.000 63 20 0.840 55 40 0.420 40 60 0.250 31 100 0.149 20 200 0.074 4 LL 60 PL 40 Determine its classification using USCS SP GW SW GP None of the above
PROBLEM No. 3 The table below shows the laboratory results of the sieve analysis of Soil A and B. What is the classification of each soil according to the Unified Soil Classification System? Give the Group Symbol and Group Name. Percent Passing Sieve No. Diameter (mm) Soil A Soil B 4 4.760 100 90 8 2.380 99 64 10 2.000 98 58 20 0.850 92 35 40 0.425 70 22 60 0.250 46 15 100 0.150 25 10 200 0.074 4 D10 0.10 0.15 D30 0.17 0.66 D60 0.32 2.10
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