SITUATION 1:A deposit of clay underlain a 4.5 mthick sand . Degree of saturation ofsand is 37% and void ratio is 0.52.Moisture content of clay is 42%.Natural ground water table is located at2m below the Natural ground surface.Assume that specific gravity of sandand clay are both equal to 2.65.3. Compute the unit weight of sandabove water table.a. 17.345 kN/m³ b. 19.345 kN/m³c. 18.345 kN/m³ d. 20.345 kN/m34. Compute the total pressure at point10m below the ground surface.a. 163.92 kPa b. 173.92kPac. 193.92 kPa d. 183.92kPa5. Compute the effective pressure atpoint 10 m below the ground surface.a. 105.445 kPa b. 106.445 kPac. 107.445 kPa d. 108.445 kPaS= 0.372 me = 0.52Gs = 2.65Sand4.5 m2.5 mClay5.5 mPA O= 42%Gs = 2.65

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Answered: A deposit of clay underlain a 4.5 m…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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17. Refer to the soil profile shown. Given H1 = 9.29 m., and H2 = 5.67 m. If the ground water table rises by 3.49 meters, determine the change in effective stress (numerical value only, in kPa) at the bottom of the clay layer. Properties of dry sand: Gs = 2.53, e = 0.60. Properties of clay: Gs = 2.79, e = 0.84. Round off to two decimal places.
The wet weight of 500cm³ of soil is 7.2 N and its dry weight is 6.19N. Determine the: a. the water content b. the degree of saturation c. the void ratio d. the bulk-unit weight (kN/m3) Gs=2.7, Yw = 9.81 kN/m³
or the soil profile shown in the Figure: a. Calculate the effective stress at point C. b. How high should the water table rise so that the effective stress at C is 200 kN/m²? Dry sand Yory-165 AN 6m Groundwater table Saturated sand 13m 7-19.25 AN Dry sand Satted sand Clay
18 Refer to the soil profile shown. Given H1 = 9.80 m., and H2 = 5.17 m. If the ground water table rises by 2.39 meters, determine the change in effective stress (numerical value only, in kPa) at the bottom of the clay layer. Properties of dry sand: Gs = 2.58, e = 0.68. Properties of clay: Gs = 2.72, e = 0.81. Round off to two decimal places.
3.5 A layer of sand extends from ground level to a depth of 9 m and overlies a layer of clay, of very low permeability, 6m thick. The water table is 6 m below the surface of the sand. The saturated unit weight of the sand is 19 kN/m³ and that of the clay 20 kN/m³; the unit weight of the sand above the water table is 16 kN/m³. Over a short period of time the water table rises by 3 m and is expected to remain permanently at this new level. Determine the effective vertical stress at depths of 8 and 12 m below ground level (a) immediately after the rise of the water table and (b) several years after the rise of the water table.
Dry Sand H, Groundwater Table Clay Refer to the soil profile shown. Given H1 = 8.85 m., and H2 4.01 m If the ground water table rises by 2.19 meters, determine the change in effective stress (numerical value only, in kPa) at the bottom of the clay layer Properties of dry sand Gs = 2.51, e = 0.65 Properties of clay. Gs = 2.72, e = 0.84. Round off to two decimal places. Activate Windows Go to Settings to activate Windows. < Question 18 of 20 2 A Moving to another question will save this response. ASUS C8O CD 80 5n E R- T Y H]
Q4 A 0.7m long soil specimen is subject to a steady-state flow, as shown in Figure Q4. The saturated unit weight of the soil is 20 kN/m³. Determine the following: Water 0.3 m 0.8 m А 0.3 m Datum 0.4 m Y Figure Q4 a) Total head (m) at point A, B and C. b) Elevation head (m) at point A, B and C. c) Porewater pressure head (m) at point A, B and C.
Problem 2 Plot the stress profile for total stress, effective stress, and pore water pressure. Show all calculation. Diy sand 0.61 Saturated sand 0.48. G-267 Clay
The soil profile at a site consists of a 2.0 m thick sand layerat the top, underlain by a 3.0 m thick clay layer. The watertable lies at a depth of 1.0 m below the ground level. Thebulk and saturated unit weights of the sand are 16.0 kN/m3and 19.0 kN/m3, respectively. The properties of the clay are:water content = 45.0%, specific gravity of the soil grains =2.70, compression index = 0.65, swelling index = 0.08, andoverconsolidation ratio = 1.5.a. The ground level is raised by placing a 1.5 m high compactedfill with unit weight of 20.0 kN/m3. What is theconsolidation settlement?b. When the consolidation due to the fill is completed,it is proposed to construct a warehouse imposing a uniform pressure of 40.0 kN/m2. What would be theconsolidation settlement due to the construction of thewarehouse?
A ground profile consists of 2.0 meters silty sand underlain by 3.0 meters of clay. The ground water table is 3.0 meters below the ground surface. The sand has a unit weight of 14 kN/cu.meter. The clay has a unit weight of 16 kN/cu.meter above the water table and 20 kN/cu.meter below the water table. Determine the total soil stress at the bottom of the clay layer.
Question 20 A soil profile consists of sand (5-m thick) which overlies a layer of clay (9-m thick). Ground water table is located at the interface of the sand and clay. The effective stress at the bottom of the clay layer was determined. If the water table rises to the top of the ground surface, what is the change in the effective stress (in kPa) at the bottom of the clay layer? For sand layer: Void Ratio = 0.82, Specific gravity = 2.68, Degree of Saturation = 0.6 For clay layer: Void Ratio = 0.92, Specific gravity = 2.86 Round off to two decimal places.
1. A soil profile is shown in Figure below. Calculate the total stress, pore water pressure, and effective stress at points A and D. Dry sand G.-2.68 *»0.6 2 m 2m Groundwater table Clay G₁-2.72 e-09

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