Please answer situation 3 TNTN 351 9:30K/sSITUATION 1A 100 cc mold was used to collect a wet soil sample. Itsweight was found to be 200- grams. Its oven dried weightwas recorded as 189- grams. Use Gs= 2.70.(a) Determine the wet unit weight of the soil in g/ cc.(b) Determine the moisture content of the soil.(c) Determine the degree of saturation of the soil inpercentage.SITUATION 2A specimen of moist soil weighing 122 g has an apparentspecific gravity of 1.82. The specific gravity of the soilsolids is 2.53. After the specimen is oven dried, the weightbecomes 104 g.(a) Calculate the void ratio of the original mass of the soil.(b) Determine the water content of the original mass of thesoil.(c) What is the dry density of the soil mass in g/cc?SITUATION 3A soil has a bulk unit weight of 18.7 kN/ cu.m and amoisture content of 12.5%. Assume the specific gravity ofsoil solids to be 2.67.(a) Calculate the submerge unit weight of the soil samplein kN/cu.m.?(b) Calculate the buoyant unit weight of the soil sample inkN/cu.m.?(c) Determine the mass of water to be added in a 2.5 cubicmeters of soil to make the soil 90% saturated.SITUATION 4A soil sample has the following properties:@ S= 50% its mass density (moist)= 1690 kg/ cu.m.@ S= 75% its mass density (moist)= 1808 kg/ cu.m.Calculate the following:(a) specific gravity(b) void ratioSITUATION 5For loose uncompacted sand:Dr= 40%, e(min)= 0.46, e(max)= 0.90, Gs= 2.65

Answered: Image /qna-images/answer/d8c7e83d-c796-4db1-bcfa-4e36b125e707.jpg

Answered: SITUATION 3 A soil has a bulk unit…

Fundamentals of Geotechnical Engineering (MindTap Course List)

5th Edition

ISBN:9781305635180

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter5: Soil Compaction

Section: Chapter Questions

Problem 5.17CTP

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A cylindrical soil sample 50 mm in diameter and 200 mm in height is used as thepermeameter in a falling head permeability test setup. In one minute of observation, thehydraulic head in the 10 mm diameter standpipe through which test water went reducedfrom 900mm to 500 mm. During that time, 1.5 liters of water were collected in the graduate. Calculate the soil sample's permeability coefficient in cm/s.
The data from a standard Proctor test are shown in the table below.a. Determine the maximum dry unit weight and the optimum water content.b. Plot the zero air voids line.c. Determine the degree of saturation at the maximum dry unit weight.Diameter of mold 101.4 mmHeight of mold 116.7 mmMass of mold 4196.50 gramsSpecific gravity, Gs 2.722
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
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.
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
The mass of the wet sample of the soil and its container is 14.5 kg. The drymass of the soil and its container is 12.5 kg. The mass of the container is 0.5 kg, and its volume is 7.5 x 10-3 m3 . The specific gravity of the soil is 2.67. Calculate the: a) water content, wb) bulk unit weight, ?c) dry unit weight, ??d) void ratio, ee) saturated unit weight, ????f) degree of saturation, Sg) porosity, nh ) How much air void is in the soil?i ) Determine the weight of water required to saturate 1 m3 of this soil.
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
A permeameter of 8.2 cm diameter contains a sample of soil of length 35 cm. It can be used either for constant head or falling head tests. The standpipe used for the latter has a diameter of 2.5 cm. In the constant head test the loss of head was 116 cm measured on a length of 25 cm when the rate of flow was 2.73 ml/s. Find the coefficient of permeability of the soil in mm/s.
A150-cm3 sample of wet soil has a mass of 250 g when 100 % saturated. When oven-dried, the mass is 162 g. Calculate the dry density, dry unit weight, water content, void ratio, and Gs.
A permeameter of 8.2 cm diameter contains a sample of soil of length 35 cm. It can be used either for constant head or falling head tests. The standpipe used for the latter has a diameter of 2.5 cm. In the constant head test the loss of head was 116 cm measured on a length of 25 cm when the rate of flow was 2.73 ml/s. Find the coefficient of permeability of the soil in mm/s. a. 4.8 x 10^-1 b. 5.9 x 10^-2 c. 2.15 x 10^-2 d. 1.1 x 10^-1
Calculate the coefficient of permeability of a soil sample of 6 cm is height and 50cm2 in cross sectional area, if the quantity of water equal to 450ml passed down in 10minutes under an effective constant head of 40cm. On oven drying, the test specimen weigh 495g. Take Gs as 2.65, calculate the seepage velocity of water during test.
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. a.41 b.43 c.50 d.47

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