Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Question
Chapter 15, Problem 15.7P
To determine
Find the factor of safety
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The figure below shows a slope with an inclination of B = 56°. If AC represents a trial failure plane inclined at an angle 0 = 33° with the horizontal, determine the factor of safety against sliding
for the wedge ABC. Given H = 8.4 m; y = 20 kN/m³: ' = 21°; and c' = 36 kN/m?.
(Round your final answer to two decimal places.)
F, =
Figure 2 shows a slope with an inclination of : β = 58 ͦ. If AC represents a trial failure plane inclined at an angle θ = 32 ͦ with the horizontal, determine the factor of safety against sliding for the wedge ABC. Given: H = 6 m; ɣ = 19 kN/m3, Ø =21 ͦ, and c’= 38 kN/m2
A cut slope was excavated in saturated clays as shown in the figure. The soil has a unit weight of 17 kN/m³ and an undrained shear strength cᵤ = 20 kPa. The slope make an angle of 60° with the horizontal. Assume stability number m = 0.185. Slope failure occurs along the plane AC with BC = 8 m. Which of the following most nearly gives the stability factor?
Chapter 15 Solutions
Principles of Geotechnical Engineering (MindTap Course List)
Ch. 15 - Prob. 15.1PCh. 15 - Prob. 15.2PCh. 15 - Prob. 15.3PCh. 15 - Prob. 15.4PCh. 15 - Prob. 15.5PCh. 15 - Prob. 15.6PCh. 15 - Prob. 15.7PCh. 15 - Prob. 15.8PCh. 15 - Prob. 15.9PCh. 15 - Prob. 15.10P
Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32P
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- 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?arrow_forwardA 45 ° slope as shown in figure has been excavated to a depth of 6 m in a saturated clay having the following properties: C₁ = 50 kN/m² ₂ 0 = 0 and y = 19 kN/m²³ Determine the factor of safety. [Take area of wedge 39 m²] ok 3.4 m f 2.65 m 6m 45° 90° = 9marrow_forward3. For the slope shown in Figure below, determine the factor of safety against sliding along the circular slip surface AC. Use the ordinary method of slices. 27 m 7.5 m 3 21 m y= 19 kN/m3 c' = 31 kN/m2 6. 5 d'= 24° A 30°arrow_forward
- A 10 m high slope of dry clay soil (unit weight 20 N/m3), with a slope angle of 45° and the circular slip surface, is shown in the figure (not drawn to the scale). The weight of the slip wedge is denoted by W. The undrained unit cohesion %3D (c,) is 60 kPa. 10m Unit weight = 20 kN/m3 %3D 10m C, = 60 kPa %3D 4.48m 45° W The factor of safety of.the slope against slip failure, isarrow_forwardQuestion No. 16 A long slope is formed in a soil with shear strength parameters: c' = 0 and d' = 34°. A firm stratum lies below the slope and it is assumed that the water table may occasionally rise to the surface, with seepage taking place parallel to the slope. Use Ysat = 18 kN/m³ and Yw = 10 kN/m³. The maximum slope angle (in degrees) to ensure a factor of safety of 1.5, assuming a potential failure surface parallel to the slope, would bearrow_forward3. For the planar wedge shown below: 22 m Slip surface 39° 28° Calculate the factor of safety against plane failure assuming: p (degree) > (kN/m³) c (kPa) 55 2800 1850 c: cohesion of rock mass; y: unit weight of rock mass; p: friction angle of rock mass جیهان کاظیarrow_forward
- A slope is shown in the figure below. If AC represents a trial failure plane, determine the factor of safety against sliding for the wedge ABC. Given: β= 570 , ϒ = 17.5 kN/m3 , φ’ = 13.50 , and c’ =27 kN/m2Please use the Culmann’s method B. Briefly Comment on how to analyze a slope and determine the factor of safety against overturning.arrow_forwardA cut slope is to be made as shown in the figure. The unit weight of soil is 15.74 kN/m3 and an angle of internal friction of 10°. The soil has as cohesion of 28.8 kN/m2. The trial failure plane makes an angle of 30° with the horizontal while the cut slope makes an angle of 50° with the horizontal. If the height of the slope is 3m, Compute the force that causes sliding. Compute the resistance sliding force. Compute the factor of safety against sliding.arrow_forwardA cut slope was excavated in saturated clays as shown in the figure. The soil has a unit weight of 17 kN/m³ and an undrained shear strength cᵤ = 20 kPa. The slope make an angle of 60° with the horizontal. Assume stability number m = 0.185. Slope failure occurs along the plane AC with BC = 8 m. Which of the following most nearly gives the maximum depth in meters up to which the cut could be made?arrow_forward
- A deep cut of 10 m depth is made in sandy clay for a road. The sides of the cut make an angle of 60° with the horizontal. The shear strength parameters of the soil are c' = 20 kN/m², Ø' = 25°, and y= 18.5 kN/m³. If AC is the failure plane (SEE FIGURE BELOW), estimate the factor of safety of the slope using Culmann's method. 10 m 60⁰ 40⁰ 10° c²= 20 kN/m² p' = 25° y = 18.5 kN/m³arrow_forwardThe infinite sand slope shown in the figure is on theverge of sliding failure. The ground water tablecoincides with the ground surface. Unit weight of water % = 9.81kN/m² The value of the effective angle of internal friction (indegrees, up to one decimal place) of the sand is. 111 7sat = 21 kN/m²³arrow_forward3. For the planar wedge shown below: 22 m Slip surface 39° 28° Calculate the factor of safety against plane failure assuming: p (degree) y (kN/m³) c (kPa) 55 2800 1850 c: cohesion of rock mass; y: unit weight of rock mass; q: friction angle of rock massarrow_forward
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