A long slope is formed in a soil with shear strength parameter C' = 0, ' = 34°. Firm strate lies below the slope and it is assumed that water table may occasionally rise to the surface, with seepage taking place parallel to the slope. Use Yeat 10 kN/m³. maximum slope angle (in degree) to ensure the factor of safety 1.5. Assuming a potential failure surface parallel to the slope would be 18 kN/m³ and Y (a) 45.3 (c) 12.3 1 = (b) 44.7 (d) 11.3

A long slope is formed in a soil with shear strength parameter C' = 0, ' = 34°. Firm strate lies below the slope and it is assumed that water table may occasionally rise to the surface, with seepage taking place parallel to the slope. Use Yeat 10 kN/m³. maximum slope angle (in degree) to ensure the factor of safety 1.5. Assuming a potential failure surface parallel to the slope would be 18 kN/m³ and Y (a) 45.3 (c) 12.3 1 = (b) 44.7 (d) 11.3

Fundamentals of Geotechnical Engineering (MindTap Course List)

5th Edition

ISBN:9781305635180

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter13: Slope Stability

Section: Chapter Questions

Problem 13.27CTP

See similar textbooks

Similar questions

An infinite slope of clay soil has a slope of 29 degrees. The soil has a specific gravity of 2.7 and a void ratio of 0.47. The soil has a depth of 5m over ledge rock measured vertically. Effective friction angle and soil cohesion are 14 degrees and 25kPa respectively. Assuming the ground water table is in the ground surface, determine the limiting depth of this soil, measured vertically over ledge rock. A 1.25m B 5.95m 3.67m D 6.88m
A long natural slope in overconsolidated fissured clay of saturated unit weight 22 kN/m³, is inclined at 14 degrees to the horizontal. The water table is at the surface and seepage is approximately parallel to the slope. A slip surface has developed on a plane parallel to the surface at a depth of 4 m. Determine the factor of safety along the slip plane using the residual strength parameter, o', = 20 degrees.
A vertical cut is to be made in a purely cohesive clay deposit (c'=30kPa, φ'=0deg, γ=16kN/m3). Find the maximum height of the cut which can be temporarily supported. From the stability chart, the stability number can be used as 0.261.
2. A cut slope is to be made as shown. The unit weight of soil is 18KN/cu.m and the angle of internal friction is 22°. The soil has cohesion of 30 Kpa. The cut slope makes an angle of 30° with the horizontal and the height is 12 m. If the failure plane is 15° from horizontal. Determine a. Frictional force along the failure plane b. Cohesive force along the failure plane Factor of safety against sliding B 24 m A trial failure plane 12 m 30 15° C
3. A cut slope consist of soil materials that has a unit weight of 16 KN/cu.m and undrained shear strength = 28 KN/sq.m. The slope makes an angle of 50° with the horizontal. Assume a stability number of 0.17. Determine the stability factor b. Determine the maximum height to which the cut could be made Determine the angle that the failure plane makes with the horizontal if BA=7.6m. a. C. В 7.6 m А Hcr = 12.3 failure plane
A cut slope was excavated in saturated clay for installation of water piping system. The angle Bof cut slope is 45° with the horizontal. During excavation works, slope failure was occurred when the cut reached a depth of 4.0 m. Previous soil explorations showed that a hard layer was located at a depth of 6 m below the ground surface. The unit weight of saturated clay is 17.5 kN/m³. (Assuming an undrained condition). 1. а) Determine the undrained cohesion of the clay. b) Determine Factor of Safety (FoS) if the depth of cut slope is 3m. What is the maximum depth of saturated clay can be excavated if the Factor of Safety (FoS) 1.4 to be implemented? c)
e 120 m GL. 118 m GWL (2) A ground soil profile is shown in the right 7 Y =19 KN/m³ Y-21 KN/m³ figure. Soil layer information and their unit weights clay v 116 m are presented in the figure. The groundwater level Y-20 KN/m³ sand is 2 m below the ground surface. Find the vertical 111 m. effective overburden aressure and plot its clay Y=21 KN/m³ distribution curve. 107 m.
A cut slope was excavated in a saturated clay. The slope made an angle of 39.55 degree with the horizontal. Slope failure occurred when the cut reached a depth of 6 m. Previous soil explorations showed that a rock layer was located at a depth of 10 m below the ground surface. Assuming an undrained condition and γsat = 18 kN/m3, Analyze the following. a. undrained cohesion of the clay.b. nature of the critical circle?c. With reference to the toe of the slope, at what distance did the surface of sliding intersect the bottom of the excavation?
A fully submerged infinite sandy slope has an inclination of 30° with the horizontal. The saturated unit weight and effective angle of internal friction of sand are 18 kN/m² and 38°, respectively. The unit weight of water is 10 kN/m². Assume that the seepage is parallel to the slope. Against shear failure of the slope, the factor of safety (round off to two decimal places) is,
Problem 1Explain whether the following are true or false. 1. A lower safety factor implies that a greater fraction of the shear strength is mobilized along the failure surface. 2. In an undrained slope of 1.0 (horizontal): 1.5 (vertical), the critical circle is always a toe circle. 3. The mid-point circle passes through the toe. 4. Taylor’s stability charts can be used only for homogeneous soils.
A cut slope was excavated in saturated clay for installation of water piping system. The angle Bof cut slope is 45° with the horizontal. During excavation works, slope failure was occurred when the cut reached a depth of 4.0 m. Previous soil explorations showed that a hard layer was located at a depth of 6 m below the ground surface. The unit weight of saturated clay is 17.5 kN/m. (Assuming an undrained condition). 1. a) Determine the undrained cohesion of the clay. b) Determine Factor of Safety (FoS) if the depth of cut slope is 3m. c) What is the maximum depth of saturated clay can be excavated if the Factor of Safety (FoS) 1.4 to be implemented?
3. A cut slope consist of soil materials that has a unit weight of 16 KN/cu.m and undrained shear strength 28 KN/sq.m. The slope makes an angle of 50° with the horizontal. Assume a stability number of 0.17. a. Determine the stability factor b. Determine the maximum height to which the cut could be made c. Determine the angle that the failure plane makes with the horizontal if BA=7.6m.