EBK PRINCIPLES OF GEOTECHNICAL ENGINEER
9th Edition
ISBN: 8220103611718
Author: SOBHAN
Publisher: CENGAGE L
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Chapter 15, Problem 15.15P
To determine
Find maximum depth
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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 maximum depth in meters up to which the cut could be made?
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.
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/m³
Chapter 15 Solutions
EBK PRINCIPLES OF GEOTECHNICAL ENGINEER
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 deep cut of 7 m has to be made in a clay with unit weight 16 kN/m' and a cohesion of 25 kN/m² . What will be the factor of safety if one has to have a slope angle of 30°? Stability number is given to be 0.178 (from Taylor's chart) for a depth factor of 3.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_forwardA cut is to be made in pure clay having cohesion of 40 kN/m². The unit weight of soil is 20 kN/m³. Side slope of cut is 1.5H: 1V. Taylor's stability number for the given side slope is 0.17. To provide a safety factor of 1.5, the maximum depth (in meters) for the cut isarrow_forward
- 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 planearrow_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_forwardA cut slope is to be made in a saturated clay at an angle β = 60° with the horizontal. Assuming that the critical sliding surface is circular, determine the maximum depth up to which the cut could be made. Given: undrained shear strength, cu = 34 kN/m2 (ϕ = 0 condition), and γ = 17 kN/m3 . What is the nature of the critical circle (toe, slope, or midpoint)?arrow_forward
- Provide solution with explanation with detail arrow_forwardPlease answer 15.11arrow_forward2. 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_forward
- A cut slope is to made. The unit weight of soil is 17 kN/m^3 and the angle of internal friction is 20°. The soil has a cohesion of 24 kPa. The cut slope makes an angle of 30° with the horizontal and the height is 9 m. if the trial failure plane is 15°from the horizontal. Determine the following: 1. Frictional force along the failure plane. 2. Cohesive force along the failure plane. 3. Factor of safety against sliding.arrow_forward4. A cut is to be made in a soil having γ = 18 kN/m3, c = 40 kPa, and φ = 15°. The cut slope makes an angle of 30°. a. If H = 6 m and θ = 15°, find the force that tend to cause sliding.b. Find the frictional force.c. Find the cohesive force.d. Find the factor of safety.e. For a factor of safety of 2.0, find the value of the critical angle along which the maximum developed cohesion occurs. f. Determine the depth of cut for a factor of safety of 2.0g. Determine the developed shear stress along the failure plane h. Determine the developed normal stress alongthe failure plane.arrow_forwardI need the answer as soon as possiblearrow_forward
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