For a given soil, the following are given in Table: Specific Gravity Gs, bulk unit weight γ), and moisture content (w). Use data from the table according to your registration number. (You can solve answer on paper and paste neat clear image in the answer). Student Registration Number Gs γ (kN/m3) moisture content (w) % 102 2.72 17.2 20 Determine: Dry unit weight Void ratio Porosity Degree of saturation

For a given soil, the following are given in Table: Specific Gravity Gs, bulk unit weight γ), and moisture content (w). Use data from the table according to your registration number. (You can solve answer on paper and paste neat clear image in the answer). Student Registration Number Gs γ (kN/m3) moisture content (w) % 102 2.72 17.2 20 Determine: Dry unit weight Void ratio Porosity Degree of saturation

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

See similar textbooks

Similar questions

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
5.5 The modified Proctor test (ASTM D 698, method C) was performed for a soil with G, = 2.70, and the results are as follows: Moist Weight in Mold, gf Water Content, % 6.5 3250 9.3 3826 12.6 4293 14.9 4362 17.2 4035 18.6 3685 (a) Plot the Ya versus w relation. (b) Determine Ya,max and wopt- (c) Calculate S and e at the maximum dry unit weight point. (d) What is Y, at Wopt? What is the range of water content if the relative compaction (R.C.) is required to be 95% of the modified Proctor Ya.max?
5.5 The modified Proctor test (ASTM D 698, method C) was performed for a soil with G, = 2.70, and the results are as follows: Moist Weight Content, % in Mold, gf Water 6.5 3250 9.3 3826 12.6 4293 14.9 4362 17.2 4035 18.6 3685 (a) Plot the Ya versus w relation. (b) Determine Ya,max and wopt- (c) Calculate S and e at the maximum dry unit weight point.
Q5) A: Laboratory test results on the soil (L.L=35% and P.L-21%) is given in the following Table below. Classify the soil by using the unified classification system (10 mark) (showing all steps). Sieve Size(mm) Weight retained (gm) 4.75 2 0.425 0.25 0.075 Pan 135 70 60 30 142.5 462.5 B: A fine-grained soil has a liquid limit of 200% and a plastic limit of 45%, The natural water content of the soil in the field is 60% and the clay content is 63%, Calculate the plasticity index, the liquidity index and the activity. (10 mark)
Answer ALL Questions Time: 2 Hours Q3: Classify the following soil by using the Unified soil classification system. Give the group symbols and the group names. Table 1. Sample 1 Grain Size Results (ASTM D422). Table 2. Atterberg Limits Results (ASTM D4318). Sleve No. Dia. (mm) % Passing Sample PL LL % in 12.7 100 G0 40 14 4.75 97 #10 2 94 #20 0.85 84 # 40 0.425 57 #60 0.25 32 140 0.106 15 #200 0.075 100 90 Atterberg Limit Results: LL = 60%, PL = 40% E 80 * 70- 60 - 50 40 30- a 20 10 1 30 a40 100 0.1200 10 80 0.01 Particle Diameter (mm) Answer ALL Questions Time: 2 Hours 70 CH or OH 60 A-line PI = 0.73(LL - 20) Or 50 40 30 CL MH OL 20 or CL-ML OH ML 10 - or OL 10 16 20 30 40 50 60 70 80 90 100 Liquid limit Figure 5.3 Plasticity chart Percent (%) Finer by Weight Plasticity index U-line PI = 0.(LL - 8) O Cengage Learning 2014
The following are the results from the liquid limit and plastic limit test for a sail Liquid limit = 39.1% Plastic limit=18.7% a. b. C. No. of Blows 15 20 28 LI Moisture Content 42 40.8 39.1 What is the plasticity index of the soil? What is the liquidity index if the water content is 24%? What is the consistency index?
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
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
In a partially saturated soil, the moisture content W% is 66.67%, the relative specific gravity of solids Ss is 2.5, and the saturated specific gravity of the sample is 1.45 g/cm3. If the volume of air is equal to 8 cm3. Determine the other quantities of the scheme, and calculate: e, n, Gw%, Ym, Yd, Ysat, Y'. Note: Kindly answer all with clear calculations.
PROBLEM 7: The laboratory results of a soil test are as follows: Sieve No. # 4 #10 #40 #200 LL PL Percent passing 100 93.20 81 55 45 17 Find the group index according to AASHTO classification system. Hint: GI = (F-35)[0.20 + 0.005(LL – 40)] + 0.01(F – 15)(PI – 10)
16. The results for sieve analysis is shown in the following table: Sieve No. 4 10 20 40 60 100 200 % Passing 68.5 36.1 Liquid Limit 34.1 Plastic Limit 16.5 21.9 a. Determine the classification of soil according to AASHTO. b. Determine the classification of soil according to USCS.
5.5 The modified Proctor test (ASTM D 698, method C) was performed for a soil with G, = 2.70, and the results are as follows: Moist Weight in Mold, gf Water Content, % 6.5 3250 9.3 3826 12.6 4293 14.9 4362 17.2 4035 18.6 3685 (d) What is y at wopi? (e) What is the range of water content if the relative compaction (R.C.) is required to be 95% of the modified Proctor Ya.max?