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?
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.
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
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 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 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_forwardSand is placed on a rock slope, as shown in Figure Q2. (a) Show that sand will be stable (i.e., no sliding sliding) ifarrow_forward1.) A concrete dam retaining water is shown. If the specific weight of the concrete is 23.5 kN/m3. Assume there is a hydrostatic uplift that varies uniformly from full hydrostatic head at the heel of the dam to zero at the toe and that the coefficient of friction between dam and foundation soil is 0.45. What is the magnitude of the uplift force at the bottom of the dam? CHOICES: a.) 1530.10 b.) 3010.50 c.) 1,030.05 d.) 5030.10arrow_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_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_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_forwardE A 40° slope is excavated to a depth of 10 m is a deep layer of saturated clay of unit weight 20 kN/m3; the relevant shear strength parameters are c, = 72 kN/m2 and o, = 0. The rock ledge is at %3D a great depth. The Taylor's stability coefficient for O, = 0 and 40° slope angle is 0.18. The factor of safety of the load is %3Darrow_forward3. An infinite slope has shear strength parameters at the interface: c = 30 kPa, φ = 30°, γ = 17 kN/m3 (Assume 1 m strip perpendicular to the paper and 1 m strip along the inclination of the infinite slope). a. If H = 5 m, and β = 20°, find the sliding force. b. If H = 5 m, and β = 20°, find the force that resist sliding on the rock surface. c. If HCR = 8 m, find the minimum angle β so that the slope is stable.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 % = 9.81kN/m² The value of the effective angle of internal friction (indegrees, up to one decimal place) of the sand is. 111 7sat = 21 kN/m²³arrow_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_forward7.17 A soil profile consists of a clay layer underlain by a sand layer, as shown in Figure P7.17. If a tube is inserted into the bottom sand layer and the water level rises to 1 m above the ground surface, determine the vertical effec- tive stresses and porewater pressures at A, B, and C. If K, is 0.5, determine the lateral effective and lateral total stresses at A, B, and C. What is the value of the pore- water pressure at A to cause the vertical effective stress there to be zero? GWL 11 m Y=18.5 kN/m? Clay 2 m Y= 19.0 kN/m³ 2: 1,5m Y =17.0 kN/m Sand 2 m FIGURE P7.17arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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