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 15.32P
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
Question No. 16
A long slope is formed in a soil with shear strength
parameters: c' = 0 and d' = 34°. A firm stratum lies
below the slope and it is assumed that the water
table may occasionally rise to the surface, with
seepage taking place parallel to the slope. Use
Ysat = 18 kN/m³ and Yw = 10 kN/m³. The maximum
slope angle (in degrees) to ensure a factor of
safety of 1.5, assuming a potential failure surface
parallel to the slope, would be
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?
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 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_forwardThe 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.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_forward
- 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.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 Cu= 50 kN/m², qu=0° and y = 19 kN/m³ Determine the factor of safety. [Take area of wedge = 39 m²]arrow_forwardQuestion 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_forward
- This 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_forwardThe figure below shows a slope with an inclination of B = 56°. If AC represents a trial failure plane inclined at an angle 0 = 33° with the horizontal, determine the factor of safety against sliding for the wedge ABC. Given H = 8.4 m; y = 20 kN/m³: ' = 21°; and c' = 36 kN/m?. (Round your final answer to two decimal places.) F, =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
- A sloping line 392.40 m long was measured using a 50 m steel tape weighing 0.075 kg/m which was standardized at 15°C and 25 N pull. The difference in the elevation between the initial and end point was 2.25 m. If the measurement was undertaken with a temperature of 20°C and the tape was supported at both ends with an applied pull of 50 N. The cross-sectional area of the tape is 5.25 mm^2 and E=200GPa. Determine the corrected length of the horizontal line.arrow_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_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_forward
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