05: The Figure below represents the data that was obtained from a liquid-limit test on a soil. 50 percent of its particles is passing sieve #200. Two determinations for the plastic limit gave water contents of 20.3 % and 20.8 %. Determine 1- Liquid Limit (LL). 2- Plasticity index. 3- Liquidity index if the water content is 27.4% 4- Void ratio at the liquid limit, if G, 2.7 5- Classify the soil based on AASHTO system. 30 10 20 40 60 80 100 Penetration, df (mm) Moisture contemt, )

05: The Figure below represents the data that was obtained from a liquid-limit test on a soil. 50 percent of its particles is passing sieve #200. Two determinations for the plastic limit gave water contents of 20.3 % and 20.8 %. Determine 1- Liquid Limit (LL). 2- Plasticity index. 3- Liquidity index if the water content is 27.4% 4- Void ratio at the liquid limit, if G, 2.7 5- Classify the soil based on AASHTO system. 30 10 20 40 60 80 100 Penetration, df (mm) Moisture contemt, )

Principles of Geotechnical Engineering (MindTap Course List)

9th Edition

ISBN:9781305970939

Author:Braja M. Das, Khaled Sobhan

Publisher:Braja M. Das, Khaled Sobhan

Chapter4: Plasticity And Structure Of Soil

Section: Chapter Questions

Problem 4.4P: Results from a liquid limit test conducted on a soil are given below. a. Determine the liquid limit...

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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
Q1-) A soil sample in plastic limit state weighs 2.535 N and becomes 1.980 N after oven drying. The plasticity index and the shrinkage limit of the soil are 11% and 15% respectively. Determine the quantity of water in liter to be added or taken out of the sample to change its state to: i- Solid state. ii- State at which the liquidity index equals to two.
PLS SHOW YOUR COMPLETE SOLUTIONGiven the laboratory results of the Atterberg Limits Test in the following figure, plot the water content versus the number of blows. a. Determine the nearest value to the Liquid Limit of soil. Answer: 47.1%b. Determine the nearest value to the Plastic Limit of soil. Answer: 38.34%c. Determine the nearest value to the Liquidity Index of soil. Answer: 1.55
3. A soll sample was determine in the laboratory to have a liquid limit of 41% and a plastic limit of 21.1%, if the water content is 30%; What is the plasticity index? What is the liquidity index? b. 17.44 b. 0.56 c. 19.90 c. 0.45 What is the characteristic of the soil? a. liquid b. moist c. semi plastic d. plastic 3.1. 3.2. 3.3. 4.2. 4.3. A. 16.38 A. 0.38 4. A sand soll 8.5 meters thick is composed of dry layer and a saturated layer due to the water table located at 3.5 meters below the ground level. The dry part of sand has Gs=2.66 and n=37.89%. below the water table, e=0.48 and Gs=2.67. 4.1. What is the effective stress developed at 3.5 m below the ground level? a. 56.74 KPa b. 64.84 KPa c. 47.16 KPa d. 73.08 KPa What is the effective stress at the bottom most part of the soil? a. 120.31 b. 76.80 c. 115.64 d. 137.50 At what height the ground water table will rise so that the effective stress at the bottom most of the soll be 104 KPa. a. 1,32 m b. 1.40 m c. 2.65 m d. 1.87 m 5. A sand…
A liquid limit test on a clay was performed with the following results. The natural water content of the clay is 38% and plastic limit is 21%. Number of Blows 6 12 20 28 32 Water content (%) 52.5 47.1 42.3 38.6 37.5 What is the liquidity index of this clay? (Use Interpolation) Group of answer choices 0.895 0.96 1.08 0.90 Please answer this asap. For upvote. Thanks
6.Table 2 shows the results of the Standard Proctor Test TABLE 2 RESULTS OF ORI TEST Water Content (%) Weight (Kg) 14 1.45 16.5 1.71 18.5 1.87 20 1.88 21 1.76 22.5 1.67 If this test uses a sample mold with a size of 943 cm^3 , answer a few questions this: a. interpret the results of this proctor test, and determine the optimum moisture content and maximum dry density. (Gs = 2.7) b. Determine the range of moisture content for 90% compaction in the field c. Between soil A and soil B in Figure 1, which is easier to compact? Explain your answer.
The following results were obtained from a liquid limit test on a clay using the Casagrande cup device. The natural water content of this clay is 38% and the plastic limit is 21%. No. of blows 6 12 20 28 32 Water 52.5 47.1 42.3 38.6 37.5 content (%) What is the liquid limit of this clay? b. What is the plasticity index of this clay? What is the liquidity index of this clay?
11. The range of water content between the liquid and plastic limits is termed as: A. liquid limit B. plasticity index C. plastic limit D. shrinkage limit 12. A container of volume 3.25 x 10-3 m³ weighs 10.8 N. Dry sand was poured to fill the container. The container and the sand weigh 62.5 N. Calculate the porosity of the sand. Assume Gs = 2.72. A. 67.74% C. 43.56% B. 40.38% 0 mod D. 54.25%
Q2: A 0.01 m' of saturated soil sample has a unit weight of 1.6g/cm3. After it is dried in the oven, the weight of the soil sample reduced to 13.5kg. Use a phase diagram, determine the following: Weight of water for the saturated soil, Ww Water content, wc Ws WATER Void ratio, e Porosity, n SOLID befare trng W. Va SOLID efter ening
b. Compute the plastic limit. C. Compute the natural water content. d. Compute the plasticity index. e. Compute the liquidity index. f. Compute the consistency index. PROBLEM NO. 4 From the table shows a summary of a liquid limit and plastic limit test. LIQUID LIMIT TEST Soil Wt. of wet Wt. of dry Water No. of Sample Soil (kg) Soil (kg) A. Content Blows 9.5 6.7 28 B 7.95 5.54 20 8.21 5.69 18 D. 12.85 8.83 13 PLASTIC LIMIT TEST Wt. of Dry Soil(kg) Soil Wt, of Wet Sample Soil(kg) 9.05 7.35 2 8.66 7.01 Natural Water Content Scil Wt. of Wt. of Sample Wet Soil Dry Soi! 1 9.69 7.0 9.47 6.86 a. Compute the liquid limit. b. Compute the plastic limit. C. Compute the natural water content. d. Compute the plasticity index. e. Compute the liquidity index GEOTECHNICAL ENGINEERING 1
2. Given the laboratory results of the Atterberg Limits Test in table below. You may use a chart to determine the liquid limit by plotting the Casagrande graph or you may also use ratio and proportion or calculator in determining the LL. a. Determine the nearest value to the Liquid Limit of the soil. b. Determine the nearest value to the Plastic Limit of the soil. c. Determine the nearest value to the Liquidity Index of the soil. A. Liquid Limit Test Number Cone penetration, mm Weight of Wet Soil +Container, g Weight of Dry Soil +Container, g Weight of Container, g Weight of Water, g Weight of Dry Soil, g Water Content, % B. Plastice Limit and Natural Water Content Test Number Weight of Wet Soil +Container, g Weight of Dry Soil +Container, g Weight of Container, g Weight of Water, g Weight of Dry Soil, g Water Content, % Average, % 1 3 38 20 22.47 21.29 21.27 19.44 18.78 18.75 12.74 13.24 13.2 2 29 Plastic Limit 1 2 23.2 22.8 20.42 20.19 12.9 12.95 4 14 26.12 22.1 13.27 Natural Water…
The laboratory tests on a soil sample gave the following results: 1). Natural water content, W₁ = 25% 2). Liquid limit, WL = 60% 3). Plastic limit, Wp = 35% 4). Percentage of particle less than 2μ = 20% The value of activity number is