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 15.5P
To determine
Find the factor of safety against sliding on the rock surface.
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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?
Example 13.1
An infinite slope is shown in Figure 13.4. The shear strength parameters
at the interface of soil and rock are as follows: c = 18 kN/m², ' = 25°.
a. If H-8 m and 3 = 20°, find the factor of safety against sliding on
the rock surface.
b. If 3= 30°, find the height, H, for which F,-1.
FIG. 13.4
#1
Density.p1500 kg
Rock
(iv
Given 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.
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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Similar questions
- 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?arrow_forwardPlease answer 15.14arrow_forwardA infinite soil slope with an inclination of 35° is subjected to seepage parallel to its sur-face. The soil has C'= 100 kN/m² and ¤'= 30°. Using the concept of mobilized cohesion and friction, at a factor of safety of 1.5 with respect to shear strength, the mobilized friction angle isarrow_forward
- Sand is placed on a rock slope, as shown in Figure Q2. (a) Show that sand will be stable (i.e., no sliding sliding) ifarrow_forwardFigure below gives details of an embankment to be made of cohesive soil with c, = 20 kPa. Unit weight of soil is 19 kN/m³. For the trial circle shown, determine the factor of safety against failure. The weight of the sliding sector is 329 kN acting at an eccentricity of 4.8 m from the centre of rotation. What would be the factor of safety if the shaded portion of the embankment were removed? Assume no tension crack develops. 71° R = 9 m e = 4.8 m 3 m 1.5 m 1.1 3 m VINISarrow_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_forwardRefer to Figure 15.50. Given that β = 55°, γ = 19 kN/m3, ϕ′ = 17°, c′ = 57.4 kN/m2, and H = 13.7 m, determine the factor of safety with respect to sliding. Assume that the critical sliding surface is a plane.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_forwardQ.2 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 .arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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