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MindTap Engineering for Das/Sobhan's Principles of Geotechnical Engineering, SI Edition, 9th Edition, [Instant Access], 2 terms (12 months)
9th Edition
ISBN: 9781305971264
Author: Braja M. Das; Khaled Sobhan
Publisher: Cengage Learning US
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Question
Chapter 15, Problem 15.19P
To determine
Find the maximum height of a slope and the nature of the critical circle.
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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.
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?
Question 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 be
Chapter 15 Solutions
MindTap Engineering for Das/Sobhan's Principles of Geotechnical Engineering, SI Edition, 9th Edition, [Instant Access], 2 terms (12 months)
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 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_forwardA 9m cut slope is shown in the figure. The unit weight of soil is 17kN/m3. Friction angle and cohesion along the rock surface are 20 degrees and 24kPa respectively. The slope makes an angle of 300 from horizontal and the failure plane is at 150. Determine the developed frictional force on the failure plane.arrow_forwardQ.1. Refer to the infinite slope shown in Figure 1. Given: β = 19 ͦ, ɣ = 20 kN/m3 , Ø = 33 ͦ, and c’ = 47 kN/m2 . Find the height, H, such that a factor of safety, Fs = 3.1 is maintained against sliding along the soil-rock interface.arrow_forward
- 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_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 30° slope has a height of 10 m as shown in the figure below. The soil in the slope has the following parameters c = 20 kPa, ϕ = 0°, γ = 18 kN/m . Calculate the factor of safety for the slip surface shown in the figure.arrow_forward
- 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, =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_forwardProblem 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.arrow_forward
- 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° Carrow_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_forwardA 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_forward
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