Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 13, Problem 13.18P
To determine
Find the critical height of the slope.
Find the nature of the critical circle.
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Chapter 13 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
Ch. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Prob. 13.4PCh. 13 - Prob. 13.5PCh. 13 - Prob. 13.6PCh. 13 - Prob. 13.7PCh. 13 - Prob. 13.8PCh. 13 - Prob. 13.9PCh. 13 - Prob. 13.10P
Ch. 13 - Prob. 13.11PCh. 13 - Prob. 13.12PCh. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - Prob. 13.15PCh. 13 - Prob. 13.16PCh. 13 - Prob. 13.17PCh. 13 - Prob. 13.18PCh. 13 - Prob. 13.19PCh. 13 - Prob. 13.20PCh. 13 - Prob. 13.21PCh. 13 - Prob. 13.22PCh. 13 - Prob. 13.23PCh. 13 - Prob. 13.25PCh. 13 - Prob. 13.26PCh. 13 - Prob. 13.27CTPCh. 13 - Prob. 13.28CTPCh. 13 - Prob. 13.29CTP
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- 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_forwardPlease solve with stepsarrow_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_forward
- Please answer 15.22arrow_forwardA 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_forwardGiven an infinite slope with the following properties: γ = 17 kN/m3, H = 5 m, β = 34°, φ = 25° and c = 20 kPa. a. Determine the normal stress. b. Determine the shear stress. c. Determine the factor of safety.arrow_forward
- Determine the factor of safety of the slope shown in the figure below using an ESA and a TSA. The point of rotation is indicated by O and the line representing the top of the stiff soil is a tangent to the failure plane. 7 m 2 3 5m T 3 m 2 m Ysat = 18 kN/m³ 3 m Soft soil S₁ = 25 kPa = 28° Ysat = 18.5 kN/m³ Stiff soil S₁ = 70 kPa = 22°arrow_forwardFigure 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/m2arrow_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 stability factor?arrow_forward
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