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 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?

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 4.3P

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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.
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 w. (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? pt? opr?
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?
liquid limit and plastic limit test. Compute the plasticity index. Liquid Limit Test Wt. of Soil Wt. of Wet No. of Dry Soil (kg) Sample Soil (kg) Blows 9.5 6.7 28 7.95 5.54 20 C 8.21 5.69 18 12.85 8.83 13 Plastic Limit Test Wt. of Dry Soil (kg) Soil Wt. of Wet Sample Soil (kg) 1 9.05 7.35 2 8.66 7.01 Natural Water Content Soil Wt. of Wet Wt. of Dry Sample Soil (kg) Soil (kg) 1 9.69 7.0 9.47 6.86
. Problem 3- From the following data of the grain size analysis Sieve size system #4 #8 # 10 #25 #60 # 100 #200 Pan Sieve opening (mm) 4.76 2.36 2.00 0.71 0.25 0.149 0.074 Determine; a) Cu and Cc % Finer (by weight) 80 56 51 35 18 13 7 I b) classify the soil using the unified soil classification
Q3. A 1200 g sample of Gravel in the SSD condition in air and weighed 770 g when immersed in water. Also, a 500 g sample of Sand in the SSD condition in air and weighed 300 g when immersed in water Calculate the BSG(SSD) of the sand and gravel 1) the BSG (SSD) of the Gravel * 1.2.791 2.4.71 3.3.703 4.2.50 2) the BSG (SSD) of the Sand * 1.2.221 2.3.590 3.2.500 4.3.232 Choose one answer and write the answer number for me
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
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
You are very disappointed that a compressibility experiment is no longer part of the CIE 334 laboratory curriculum. Taking matters into your own hands, you break into the soils lab over the weekend and obtain the following data. Assume the specific gravity of solids is 2.70. The initial height of the sample at zero stress was 22.5mm. The water content of the soil at the start of the test is 78%. Plot the e-log o' and e-log o' curves and calculate Cc. Effective Vertical Stress, o' (kPa) 20 50 100 200 400 800 Total Change in Height (mm) 0.23 0.87 1.90 3.62 5.55 7.25
|Laboratory testing was performed on two soil samples (A and B). (a) Determine the USCS classification symbol for Sample A. (b) Determine the AASHTO classification for Sample B. Sieve No. Sieve Opening (mm) A- Percent Passing B - Percent Passing 3 inch 76.2 100 1.5 inch 38.1 98 0.75 inch 19.1 96 4 4.75 77 100 10 2.00 96 20 0.85 55 94 40 0.425 73 100 0.15 30 200 0.075 18 55 Liquid Limit 32 52 Plastic Limit 25 32 A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate the wet density of the CL sample. (b) A small piece of the CL sample had a wet weight of 140.9 grams and a weight of 85.2 grams after drying. Compute the water content. (c) Compute the dry density of the CL sample A Standard Proctor test was performed on a soil with a specific gravity of 2.71. For the water content and wet soil unit weight in the following table: a) Plot the moisture-dry unit weight curve b) Find the maximum dry density and optimum…
AASHTO SOIL CLASSIFICATION (by AASHTO) AASHTO = American Association of State Highway and Transport Officials Situation 2 Classification of Highway Subgrade Materials % FINER SOIL B SIEVE General classification Granular materials (35% or less of total sample passing No. 200) SOIL A SOIL C 10 83 100 90 A-1 А-2 Procedure: 40 48 92 76 Group classification A-1-a А-1-b А-3 A-2-4 A-2-5 A-2-6 A-2-7 O Group Classification with required test data in mind, proceed from left to right, in chart. The correct group will be found by process of elimination. The first group from the left consistent with the test data is the corect classification. 86 34 Sieve analysis (percentage passing) No. 10 200 20 LL 20 70 37 50 max. 30 max. No. 40 No. 200 Characteristics of fraction PI 32 18 50 max. 51 min. 15 max. 25 max. 10 max. 35 max. 35 max. 35 max. 35 max. Using AASHT0 method: 1. Determine the group symbol and index for soil A. 2. Determine the group symbol and index for soil B. 3. Determine the group symbol…
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% Classify the soil by the AASHTO system. O A-2-7 (15) O A-7-6 (18) O A-7-6 (32) O A-2-5 (6)