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 15.17P
To determine
Find the factor of safety
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Sand is placed on a rock slope, as shown in Figure Q2.
(a) Show that sand will be stable (i.e., no sliding sliding) if
16.9 Use Taylor's method to determine the factor of safety
of the slope shown in Figure P16.9.
1
Very stiff clay
FIGURE P16.9
$ = 20 kPa
18 kN/m²
A 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.
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_forwardFor a railway track 7 m high embankment is required. The clay to be used for the embankment was found to have c = 20 kN/m2and unit weight = 19 kN/m3 Compute the critical maximum side slope angle for the embankment if a hard rocky stratum was found 3.5 m below the ground level. Assume ϕ for the clay equal to zero. The following values are given from Taylor's chart for depth factor D = 1.5 :arrow_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_forward
- Q.3. A cut slope was excavated in a saturated clay with a slope angle β = 57 ͦ with the horizontal. Slope failure occurred when the cut reached a depth of 8 m. Previous soil explorations showed that a rock layer was located at a depth of 2 m below the ground surface. Assuming an undrained condition and that ɣsat = 19 kN/m3 : Determine the undrained cohesion of the clay (Fig 3). What was the nature of the critical circle? With reference to the top of the slope, at what distance did the surface of the sliding intersect the bottom of the excavation?arrow_forward(Note:Depth of rock layer is 2m not 12 m.) Q A cut slope was excavated in a saturated clay with a slope angle β = 57 ͦ with the horizontal. Slope failure occurred when the cut reached a depth of 8 m. Previous soil explorations showed that a rock layer was located at a depth of 2 m below the ground surface. Assuming an undrained condition and that ɣsat = 21 kN/m3 : Determine the undrained cohesion of the clay (Fig 3). What was the nature of the critical circle? With reference to the top of the slope, at what distance did the surface of the sliding intersect the bottom of the excavation?arrow_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
- 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 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_forwardThe infinite sand slope shown in the figure is on theverge of sliding failure. The ground water tablecoincides with the ground surface. Unit weight of water Y = 9.81kN/m² The value of the effective angle of internal friction (indegrees, up to one decimal place) of the sand is 5 m 20° 7sat = 21 kN/m³arrow_forward
- A 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_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 maximum depth in meters up to which the cut could be made?arrow_forwardA 40° slope is excavated to a depth of 8 m in a deep layer of saturated clay having strength parameters, c = 60 kN/m², p = 0, and y- 19 kN/m³. Determine the factor of safety for the trial failure surface shown in below figure using Swedish circle method. 8 m 40° 95⁰ R = 10.2 m 4m- W = 1050 kN THE Warrow_forward
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