2. The resulting of the Standard Proctor Test is as follows: Vol. of Proctor Mold Wt. of wet soil in mold Moisture (cu.m) (kg) (%) 943.3 1.76 12 943.3 1.86 14 943.3 1.92 16 943.3 1.95 18 943.3 1.93 20 943.3 1.90 22 a. Determine the maximum dry unit weight of compaction b. Determine the optimum moisture content

2. The resulting of the Standard Proctor Test is as follows: Vol. of Proctor Mold Wt. of wet soil in mold Moisture (cu.m) (kg) (%) 943.3 1.76 12 943.3 1.86 14 943.3 1.92 16 943.3 1.95 18 943.3 1.93 20 943.3 1.90 22 a. Determine the maximum dry unit weight of compaction b. Determine the optimum moisture content

Principles of Foundation Engineering (MindTap Course List)

8th Edition

ISBN:9781305081550

Author:Braja M. Das

Publisher:Braja M. Das

Chapter16: Soil Improvement And Ground Modification

Section: Chapter Questions

Problem 16.3P

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The resulting of the Standard Proctor Test is as follows: Vol. of Proctor Mold (cu.m) Wt. of wet soil in mold (kg) Moisture (%) 943.3 1.76 12 943.3 1.86 14 943.3 1.92 16 943.3 1.95 18 943.3 1.93 20 943.3 1.90 22 a. Determine the maximum dry unit weight of compaction b. Determine the optimum moisture content
A fine-grained soil has 60% clay with LL 220%, PL 545%, and a natural water content of 6%. Astandard Proctor test was carried out in the laboratory and the following data were recordedDiameter of mold = 101.40 mm Height of mold = 116.70 mm Mass of mold = 4196.50 gram Specific gravity, Gs = 2.69 a) What field equipment would you specify to compact the soil in the field, and why?b) How would you check that the specified dry unit weight and water content are achievedin the field?
The following are the data of the compaction test: Volume of mold = 0.000945 m^3 Determine the degree of saturation at the optimum moisture content if Gs= 2.60
A modified Proctor compaction test was carried out using a cylindrical moldwith a 1000 ml volume. Table 2 summarizes the results. a. Plot the compaction and ZAV curves (RD=2.75 and ρwater=1000 kg/m3)b. Determine the maximum dry density and the optimum moisture content Table 2 Moisture Content (%)Mass of WetSample (g)10.5 2033.211.8 2146.613.0 2203.514.9 2183.116.4 2147.617.6 2116.8
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 25 and the in situ moisture content is 45%. Determine the liquidity index. a. 0.56 b. 0.91 c. 1.25 d. 0.59
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 20 and the in situ moisture content is 35%. Determine the liquidity index. a.0.56 b.0.91 c.1.25 d.0.59
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 20 and the in situ moisture content is 45%. Determine the plasticity index. a.22 b.17 c.27 d.12
A 72-cm3sample of moist soil weighs 141.5 g. When it is dried out in an oven,it weighs 122.7 g. The specific gravity of solids is found to be 2.66. Find the watercontent, void ratio, porosity, degree of saturation, and wet and dry unit weights
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 30 and the in situ moisture content is 35%. Determine the plasticity index. A. 27 b. 12 C. 17 d. 22
A block of soil in a permeability test has a length of 120 mm and a cross-sectional area of 600 mm2. The water level at one end is 200 mm above a fixed datum and at the other end is 30 mm above the same plane. The flow rate was 2 ml in 90 seconds. Compute the coefficient of permeability of the soil in m/s. a.2.61 x 10-5 m/s b.0.98 x 10-5 m/s c.1.78 x 10-5 m/s d.3.97 x 10-5 m/s e.1.89 x 10-5 m/s
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL=25 and the in situ moisture content is 45%. Determine the liquid limit. O a. 41 O b. 47 O c. 43 O d. 50
A WET OF SOIL SAMPLE HAS A VOLUME OF 4.85 x 10^-4 m3 AND WEIGHS 8.5N. AFTER OVEN DRYING, THE WEIGHT REDUCES TO 7.5N. IF sg = 2.7DETERMINE THE UNIT WEIGHT. DETERMINE THE MOISTURE CONTENT. DETERMINE THE DEGREE OF SATURATIONAnswer in 4 decimal places with unit