EBK PRINCIPLES OF GEOTECHNICAL ENGINEER
9th Edition
ISBN: 9781337517218
Author: SOBHAN
Publisher: VST
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Chapter 15, Problem 15.30P
To determine
Find the factor of safety
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Find slope at D in degrees using double integration method. Consider 250 mm x 400 mm (b x h) section and E = 120 GPa. Use a = 3.2 m, b = 19.1 kN/m, and c = 25.5 kN
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]
A 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.
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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- The 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_forwardA 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.arrow_forwardConsider 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_forward
- Briefly explain the meaning of FS = 1.0 for sand followed by the infinite slope analysis.arrow_forwardThis is a question from slope stability bishop method Please Note i am not asking for the entire problem to be completed, i am simply seeking guidance on how to calculate slice 1 entire row that is highlighted in red. example how did we derive the values of (W) in the table. formula ....W=ybz but im not getting the values see diagram attached Question A compacted earth fill is constructed on a soft, saturated clay (figure below). The fill was compacted to an average dry unit weight of 19 kN m3 ⁄ and water content of 15%. The shearing strength of the fill was determined by CU tests on samples compacted to representative field conditions. The shear strength parameters are ?? = 45 kPa, ??′ = 34°, and ???′ = 28°. The variation of undrained shear strength of the soft clay with depth as determined by simple shear tests is shown in the figure below, and the friction angle at the critical state is ???′ = 30°. The average water content of the soft clay is 40%. Compute the factor of safety…arrow_forwardCalculate the safety factor with the following conditions in a rock slope, likewise this slope requires a long-term static SF project equivalent to 2 based on the RCCDMX. b Zw B H a 02 A Rock mass properties: -Cohesion =3.3 T/m2 ; Internal friction angle: = -ymat = 3.8 T/m3; b=3m; B=6m; a=60° - 01=45° ; 0=72°; Zw=4m; Ax=12 m2 35° 01arrow_forward
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