Answered: Classify the following soils using the…

Classify the following soils using the Unified soil classification system. Give the group symbol. No. 4 = 60 No. 200 = 40 LL = 26 PL = 4 A GW-GM В GW-GC C) GP-GM D) GM-GC

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:

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% Finer • Particle size analyses were carried out on two soils-Soil A and Soil B-and the particle size distribution curves are shown in Figure 100 90 80 70 60 50 40 30 Clay 20 10 0 0.001 + Silt 0.01 Soil A Sand Soil B 0.1 1 Particle size (mm) - log scale Gravel 10 Soil A LL 26 (oven-dried; assume same for not dried) B Nonplastic 1. Which of the two soils, in your opinion, is a better material for the core of a rolled earth dam? PL 18
4. The result of sieve analysis of three soils is given. It is required to classify these soils according to the AASHTO Classification system. Sieve No. 4 8 10 20 40 60 100 200 LL PL Diam. (mm) 4.760 2.380 2.000 0.840 0.420 0.250 0.149 0.074 SIEVE ANALYSIS Soll A 85 80 68 62 52 48 38 25 42 33 Percent Passing Soil B 98 90 85 78 62 55 50 45 45 37 Soil C 100 95 88 75 60 52 40 34 38 18
Classify the soil according to the soil classification system AASHTO: With the following data: Percentage passing: sieve No. 200 = 45% sieve No. 40 = 10% sieve No. 10 = 45% LP = 32 LL = 54
10.. , The following are the characteristics of two soils. ClassifY the soils according te the AASHTO system. Soil A: Percent passing No.4 sieve = 98 Percent passing No.1 0 sieve = 90 Percent passing No. 40 sieve = 76 Percent passing No. 200 sieve = 34 Liquid limit = 38 Plastic limit = 26 Soil B: Percent passing No.4 sieve = 100 Percent passing No.1 0 sieve - 98 Percent passing No. 40 sieve = 86 Percent passing No. 200 sieve = 58 Liquid limit = 49 Plastic Limit = 28
Example#2 Classify Soils A and B as given in Example 9-1 and obtain the names. Assume Soil B to be inorganic. Soil A: Percent passing No. 4 sieve = Percent passing No. 10 sieve = Percent passing No. 40 sieve = Percent passing No. 200 sieve = 98 %3D 90 %3D 76 %3D 34 Liquid limit = 38 %3D Plastic limit 26 %3D
Classify the foloowing soils according to the AASHTO classification system and give the group indices.
3.11 Classify the following soils by using the Unified soil classification system. Give the group symbols and the group names. Sleve analysls- Percent finer Liquld Ilmit Plasticity Index Soll No. 4 No. 200 C, C. 1 70 30 33 21 48 20 41 22 95 70 52 28 4 100 82 30 19 100 74 35 21 6. 87 26 18 38 69 88 78 38 8. 99 57 54 26 9. 71 11 32 16 4.8 2.9 10 100 2 NP 7.2 2.2 11 89 65 44 21 12 90 39 31 3.9 2.1 Ans.: 1. SC (clayey sand with gravel), 2. GC (Clayey Gravel with sand), 3. CH (sandy fat clay), 4. CL (Lean clay with sand), 5. CL (Lean clay with sand), 6. SC (CIlayey sand), 7. CH (sandy fat clay), 8. CH (sandy fat clay), 9. SP-SC (Poorly Graded sand with gravel and sand), 10. sW (Well-graded sand), 11.CL (Sandy lean clay), 12. SP-SC (Poorly Graded sand with gravel and sand) 3.12 For an inorganic soil, the following grain-size analysis is given. U.S. Percent Sleve No. passing . 4 100 10 90 20 64 40 38 80 18 200 13 For this soil, LL = 23 and PL = 19. Classify the soil by using (a) AASHTO soil…
3. A soil has the following particle size distribution Gravel = 20% Sand = 10% Silt = 30%mm Clay = 40% Classify the soil according to USDA textural classification.
Determine the soil classification Moisture content = 5.75% Limit liquid = 53.5 Plastic limit = 31.6 Granulometry Sieve % Pass No. 4 100 # 40 68 # 200 54.3 As in the plasticity graph the coordinates are difficult to read use IP = (wL-20), to obtain IP and compare.
A. Answer in True or False 1. A soil has a Cu =1, then it is considered as a poorly graded soll. 2. A soll has L.L-60% and P.L-35%, then it plots a point below the A-line in the plasticity chat 3. A soil described with a symbol (SP) has % passing No. 200 sievs more than 50% 4. Decreasing water content below shrinkage limit is accompanied with a decreasing in void r 5. Water molecules are considered polar and known as dipoles. 6. Dry Density is the smallest density. 7. Capillary rise is directly proportional to the pore channel diameter. 8. The unconfined compressive strength is twice the undrained cohesion. 9. Piping or boiling occurs when effective stress equals to zero. 10. It is assumed that the secondary compression begins when the primary consolidation endis
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A. Answer in True or False 1. A soil has a Cu =1, then it is considered as a poorly graded soil. 2. A soil has L.L360% and P.L-35%, then it plots a point below the A-line in the plasticity chart. 3. A soil described with a symbol (SP) has % passing No. 200 sieve more than 50%. 4. Decreasing water content below shrinkage limit is accompanied with a decreasing in void ratio. 5. Water molecules are considered polar and known as dipoles. 6. Dry Density is the smallest density. 7. Capillary rise is directly proportional to the pore channel diameter. 8. The unconfined compressive strength is twice the undrained cohesion. 9. Piping or boiling occurs when effective stress equals to zero. 10. It is assumed that the secondary compression begins when the primary consolidation ends.