An infinite slope, with a slope angle of 14° made up of a cohesionless soil having o = 30° and y, = 20 kN/m³. It experiences seepage with the water table at the surface. (Assume, unit weight of water as Y„ = 10 kN/m³) %3D FOS against failure without seepage FOS against failure with seepage

An infinite slope, with a slope angle of 14° made up of a cohesionless soil having o = 30° and y, = 20 kN/m³. It experiences seepage with the water table at the surface. (Assume, unit weight of water as Y„ = 10 kN/m³) %3D FOS against failure without seepage FOS against failure with seepage

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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4. Apparent cohesion does not arise from a. Negative pore water pressures b. Negative excess pore water pressures due to dilation C. Apparent mechanical forces d. None 5. In a drained condition test a. Pore water pressure is greater than hydrostatic pressure b. Pore water pressure is less than hydrostatic pressure Pore water pressure is equal to hydrostatic pressure d. All may apply C. 6. In an Undrained condition test a. Excess pore water pressure will not develop b. Excess pore water pressure will be negative during dilation C. Excess pore water pressure will be negative during contraction d. Excess pore water pressure developed will not depend on volume change
Two footing "A" and “B" rest in a layer of sand 2.7 m thick. The bottom of the footings are 0.90m below the ground surface. Beneath the sand layer is 1.8m clay layer. Beneath the clay layer is a hard pan. The water table is at a depth of 1.8m below the ground surface. The layer of granular soil is composed of material having a saturated unit weight of 20.46 kN/m³ and a unit weight of 18.45 kN/m³. The clay saturated unit weight is 16.52 kN/m³. A laboratory compression test indicates that the clay has a pre-consolidation pressure of 67 kPa. Compression index is 0.30 and the value of swell index is 0.05. Void ratio of clay is 1.50. Determine: 1. The stress increase at the center of clay layer assuming that footing "A", with a dimension 2m x 1.8m, distributes a pressure at an angle of 2 vertical to 1 horizontal. The load of the footing is 350OKN. 2. The size of the square footing “B" so that the settlement of the clay layer is the same beneath footing "A" and "B". 3. Determine the…
A soil profile consists of 12 m of gravel overlying a clay layer of 3.5 m thickness,which in turn rests on a 4 m sand layer. The dry and saturated unit weights of thegravel are 17.5 kN/m3 and 20.5 kN/m3. Unit weights of the clay and the sand are20 kN/m3 and 19.5 kN/m3 respectively. Below the sand layer lies impermeablebedrock. If a wide excavation of 3 m depth is to be made in a relatively short time,draw the effective, total and pore pressure diagrams before and a long time afterthe excavation is made. Ground water table is 5 m below the ground surface.
A cut is made into a homogeneous soil as depicted in the figure below, at an angle of 60° with the horizontal. Calculate the resisting shear force in kN along the failure plane given that the shear force applied by the overlying soil wedge is 120 kN. The failure plane is at an angle of 40° from the horizontal direction and the ground water level is below this plane. Assume drained conditions for this question.
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Q.2 having k=1 cm /sec, dia. -0.75 m, n = 0.33 for soill and k= 2 cm /sec, dia. -0.6m, n=0.28 for soil2 in the setup shown, Compute: 1. Total head, elevation head, pressure head for points A, B, C, D and E. 2. Seepage velocity along soil 1 and soil 2 3. Force in soil 1 and soil 2 B so il 2 soil 1 E D 15 CH 25 cn T 7 cn 1 20 cn datum
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A 650-kN column load is supported on a 1.5 m wide by 2.0 m long by 0.5 m deep spread foot-ing. The soil below is a well-graded, normally consolidated sand with g = 18.0 kN/m3 and the following SPT N60 values:Depth (m) 1.0 2.0 3.0 4.0 5.0N60 12 13 13 8 22 The groundwater table is at a depth of 25 m. Compute settlement of the footing using the gen-eralized elastic method with Christian and Carrier’s (1978) influence factors.