4. LL PL LAND A Water Content: 20% Pore number (e): 0.95 Table 1. Soil Properties LAND B 59% 24% 44% 0.87 a. What types of laboratory tests can be used to determine the permeability of soil B in Table 1.

4. LL PL LAND A Water Content: 20% Pore number (e): 0.95 Table 1. Soil Properties LAND B 59% 24% 44% 0.87 a. What types of laboratory tests can be used to determine the permeability of soil B in Table 1.

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 Compaction Test

Soil compaction is the major area of geotechnical engineering. Soil is composed of multiple pores present on its surface, while the soil is mainly composed of gravel, sand, rocks, etc. The arrangement of these rocks and gravels is such that they build up voids (spaces) between them, which allows spaces for the roots of the plants and crops to spread and absorb the required amount of nutrients. Soil compaction is a major problem primarily caused by human activities, micro-organisms, and other multiple factors.

Question

4.
LL
PL
LAND A
Water Content: 20%
Pore number (e): 0.95
Table 1. Soil Properties
LAND B
59%
24%
44%
0.87
a. What types of laboratory tests can be used to determine the permeability of soil B in Table
1.

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B. ( A Soil Specimen has the following characteristics: % passing No.4 sieve = 85 % passing No.200 sieve = 11 D60 = 2 mm D30 = 0.35 mm D10 = 70 μm L.L = 36% P.L = 31% D8 = 2 μm 1. Classify the specimen according to the Unified Soil Classification System (USCS). Assign the group name and the group symbol. 2. Determine the soil activity. 3. Determine the soil Liquidity index and consistency index if we = 26%. J
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
Question 14 Classify the following soils using the Unified soil classification system. Give the group symbol. No. 4 = 100 No. 200 = 77 LL = = 63 PL 25 %D A) CL В ML MH CH
Given: wet soil sample weight 45.6 lbs volume of soil before drying 0.40 cubic ft dry sample wt. 37.8 lbs specific gravity of solids 2.65 MUST include a Phase Diagram - draw & label. Find 1 2 3 545 water content unit weight of moist soil void ratio (hint, need Vw and Vs) porosity degree of saturation
Sieve Analysis- Percent Finer Plasticity Index Liquid Limit Soil No.10 No.40 No.200 92 68 30 42 18 90 48 31 5. 1. AASHTO CIlassification for oil A 2. Group Index for Soil A 3. Group Index for Soil B 22
Subject: Soil Mechanics A. Classify the Soil Using USCS (with explanation/ justification) Given data: D10 = ? D30= 0.0023mm D60= 0.032mm Liquid Limit= 49% Plastic Limit= 24% CHOICES: a. CL (lean clay with gravel) b. CL ( lean clay with sand) c. ML (silt with gravel) d. ML (Silt with sand)
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.
Sr B. ( A Soil Specimen has the following characteristics: % passing No.4 sieve = 85 % passing No.200 sieve = 11 D60 = 2 mm D30=0.35 mm D₁0 = 70 μm D8=2 μm L.L = 36% P.L = 31% 1. Classify the specimen according to the Unified Soil Classification System (USCS). Assign the group name and the group symbol. 2. Determine the soil activity. 3. Determine the soil Liquidity index and consistency index if wc = 26%.
Classify the following soil by the AASHTO classification system. Give the group index. Sieve Analysis – Percent Finer Liquid Plasticity Index Soil No. 10 No. 40 No. 200 Limit D 92 68 30 42 18
Using 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)
18. The results for sieve analysis is shown in the following table: Sieve No. 4 10 20 40 60 100 200 % Passing 69.3 59.1 48.3 38.5 28.4 19.8 4.5 Liquid Limit Plastic Limit Non-Plastic (NP) Non-Plastic (NP) a. Determine the classification of soil according to AASHTO. b. Determine the classification of soil according to USCS.
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.