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 Y versus w relation. (b) Determine Ya,max and wopt- (c) Calculate S and e at the maximum dry unit weight point.
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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 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 (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?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?
- 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 6.86Q3. 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|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…
- A series of identical samples have been tested with drained triaxial test and some of the values are given in the tabulated table 3(Below). Calculate the remaining values as in the blank box. PARAMETER |cell pressure, o3 (kPa) total axial stress, 01 (kPa) pore pressure at failure, u¡ (kPa) |change in volume, Av effective cell pressure, o3' (kPa) effective total axial stress, o,' (kPa) |difference in stress, q (kPa) CD test 150 284 i 1.8 ii i iv O i = 134; ii = 16; iii = 150; iv = 0 O i = 134; ii = 284; iii = 150; iv = 0 O i = 0; ii = 16; iii = 150; iv = 134 O i = 0; ii = 150; iii = 284; iv = 134For 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)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…
- 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. 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 classificationYou 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