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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 15.29P
To determine
Find the factor of safety
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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 Cu= 50 kN/m², qu=0° and y = 19 kN/m³
Determine the factor of safety. [Take area of wedge = 39 m²]
The rock slope has been mapped and analysed. The data of slope geometry and rock
Q1
parameters tabulated in Table 1. The face slope facing toward south and the upper slope is
flatted. The discontinuity sets are as follow (dip direction / dip angle); set 1: 10/46', set 2:
190%40, and set 3: 140 /70°.
Parameters
Values
Face slope dip angle
65°
Slope height
50 meter
Slope length
150 meter
Unit weight of the rock
25 kN/m3
Depth of tension crack
2 m
Unit weight of water
9.81 kN/m³
The cohesion of all discontinuities
50 kPa
Friction angle for all discontinuities
30°
) Calculate the factor of safety for the planar failure mode when the tension crack is
completely filled with water.
13.20 The slope shown in Figure 13.53 has recently failed. A geotechnical investigation indicates
the failure surface was as shown. Assuming the failure occurred while undrained conditions
prevailed in the slope, back-calculate the value of su. Use the Swedish slip circle method
with the cross-section that existed immediately before it failed. Use y = 119 lb/ft³.
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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- Consider a slope given below with 10 slices, which has a circular failure surface. Each slice has 5 meters wide, and the weight of each slice was estimated on the basis of average height and a unit weight of soil ? = 16 kN/m3. The shear strength of the soil is assumed to be ? = 20 kPa and ? = 20°. The pertinent data for these ten slices is given in the table below. a) Determine the factor of safety using Janbu’s simplified method by assuming the initial ?? = ?. ?. b) Discuss how the factor of safety (stability of the Slope) will change due to the rainfall. You start to consider what environmental conditions will change first.arrow_forwardThe factor of safety of an infinite soil slope shown in the figure having the properties c = 0, 0 = 35°, 16 kN/m3 and Ysat %3D = = 20 kN/m3 is Ydry approximately equal to 30° 8 m 10 marrow_forwardProblem 6: A cut slope is to be made as shown in the figure with the respective shear strength parameters: 0 = 29° B = 50° $ = 10° Unit weight of soil is 16 kN/m3 %3D 3.0 m Failure plane Soil cohesion = 25 kPa Height of slope = 3 m %3D Determine the factor of safety against sliding.arrow_forward
- Briefly explain the meaning of FS = 1.0 for sand followed by the infinite slope analysis.arrow_forwardA slope is shown in the figure below. If AC represents a trial failure plane, determine the factor of safety against sliding for the wedge ABC. Given: β= 570 , ϒ = 17.5 kN/m3 , φ’ = 13.50 , and c’ =27 kN/m2Please use the Culmann’s method B. Briefly Comment on how to analyze a slope and determine the factor of safety against overturning.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_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_forwardThe factor of safety (upto 2 decimal places) of an infinite soil slope shown in the figure having the properties c = 0. ф= 35°, Ydry =15 kN/m³, Ysat =19 kN/m³arrow_forwardThe factor of safety of an infinite soil slope shown in the figure having the properties c = 0, o = 35°, %| Ydry = 16 kN/m3 and Ysat = 20 kN/m3 is approximately equal to 30° 8 m 10 marrow_forward
- Question 2 Calculate the slope at point C in degrees. E- 200 GPa, I= 1240 x 106 mm4 Express your answer in two decimal places. No need to indicate sign convention (just in your solution sheet) 1.5 m 10 kN/m 20 kN/m 10 kN/m 1.5 m C D A 2 m 5 m 4 marrow_forwardGEOTECHNICAL ENGINEERING Determine the height of a 1: 1 slope.Saturated clay Cu = 25 kn / m2Ysat = 18 kn / m3D = 1.2FS = 2.0arrow_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_forward
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