Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Chapter 15, Problem 15.16P
To determine
Find the maximum depth of cut to ensure a factor of safety of 3.0 against sliding.
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2. A cut is to be made in a soil that has γ = 17 kN/m3, c = 40 kPa, φ = 20°. The side of the cut slope will make an angle of 25° with the horizontal. Using a factor of safety of 2.0.
a. Determine the stability number.
b. Determine the stability factor.
c. Determine the depth of the cut slope.
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 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?
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 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_forward1. A vertical cut is made through a homogeneous soil mass (c=20 kPa,ϕ=20°,γ=16.50 ?? ?3 ⁄ ). Estimate thesafe depth of the cut, taking a factor of safety of 2 using Culmann’s methodarrow_forward
- A cut slope of h= 7m was excavated in a saturated clay with a slope of angle beta = 45 degrees with the horizontal. The Previous explorations showed that a rock layer was located at a depth of 14 m. Given that, c = 60 kN/m² and unit weight = 17 kN / (m ^ 3) . Determine the factor of safety.arrow_forwardA 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_forwardA 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_forward
- 4. 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_forwardAs part of port expansion project, a temporary vertical cut is to be made in a submerged clay deposit. A cross section and soil parameters are shown below. (Note: if needed, you can use a single wedge failure mechanism with an inclination of 45° for the failure surface) Clay c = 450 psf, o = 0° Y = 100 pcf H Note: the slope is submerged a) How deep can you make the cut with a factor of safety of = 1.5 for temporary construction? b) If the project was abandoned after the vertical cut was made, would you still expect the vertical cut to be standing 20 years? Why or why not? Support your answer with calculation.arrow_forwardA vertical cut is made through a homogeneous soil mass (c=20kPa, φ=20deg, γ=16.5kN/m3). Using Culmann's method, determine the safe depth of the cut, taking a factor of safety of 2.0.arrow_forward
- From the figure shown the soil has a unit weight of 28.5 kN/m^3 and the undrained shear strength cu = 20 kN/m^2. The slope makes an angle 62 degrees with the horizontal. Assume a stability number m = 0.16761. Compute the stability factor. a. 6.78b. 6.7c. 5.99d. 5.012. Compute the maximum depth up to which the cut could be made. a. 5.18 mb. 4.97 mc. 4.20 md. 3.84 m3. Compute angle that the failure plane makes with the horizontal if BC is 9 m. a. 22.32 degreesb. 28.25 degreesc. 21.05 degreesd. 20.15 degreesarrow_forwardA trench was cut in a clay slope as shown in the figure. When the trench reached a depth of 3 m, the top portion of the clay suddenly failed. On investigating, the failure was observed to be a slip plane approximately parallel to the original slope. Clay Y = 17.5 kN/m² 3m 36° The undrained shear strength (in kPa) of the clay will be.arrow_forward3. 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_forward
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