Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 13, Problem 13.25P
To determine
Find the minimum factor of safety for the steady-state seepage condition.
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Please calculate the critical factor of safety for the following slope configuration using the following m
ethods: Using slope stability charts
Layer 1
Clay
c=35 kN/sq.m
Friction Angle=0
Unit Weight-18 kN/cu.m
Layer 2
Silty Sand
c=5 kN/sq.m
Friction Angle=35 deg
Unit Weight=18 kN/cu.m
Slope Height=10 m
Slope Angle=30 Degrees
Layer 1
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?
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
Chapter 13 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
Ch. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Prob. 13.4PCh. 13 - Prob. 13.5PCh. 13 - Prob. 13.6PCh. 13 - Prob. 13.7PCh. 13 - Prob. 13.8PCh. 13 - Prob. 13.9PCh. 13 - Prob. 13.10P
Ch. 13 - Prob. 13.11PCh. 13 - Prob. 13.12PCh. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - Prob. 13.15PCh. 13 - Prob. 13.16PCh. 13 - Prob. 13.17PCh. 13 - Prob. 13.18PCh. 13 - Prob. 13.19PCh. 13 - Prob. 13.20PCh. 13 - Prob. 13.21PCh. 13 - Prob. 13.22PCh. 13 - Prob. 13.23PCh. 13 - Prob. 13.25PCh. 13 - Prob. 13.26PCh. 13 - Prob. 13.27CTPCh. 13 - Prob. 13.28CTPCh. 13 - Prob. 13.29CTP
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- 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.arrow_forwardAn infinite slope of granular soil has a slope of 29 degrees. The soil has a dry unit weight of 18.02kN/m^3 and a saturated unit weight of 21.15kN/m^3. The soil has a depth of 5 meters over ledge rock measure vertically. Effective friction angle and soil cohesion are 14 degrees and 25 kPa, respectively. a) Compute the factor of safety considering no seepage. b) Compute the factor of safety considering full seepagearrow_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
- Calculate 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_forwardA 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 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
- 3. 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_forward4. A slope is 5 m high and has an angle of 45°. The soil parameters of the soil are: c'= 18 kN/m?, O'=15° and y = 17.1 kN/m³. Determine the factor of safety, Fs, using Michalowski's (2002) kinematic approach for c'-o' soils.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_forward
- In the problem the slope appeared to be stable with a factor of safety = 2.7. What happens to that factor of safety if the water table rises to the level shown below? Use a unit weight of 20.1 kN/m³, and bedding strength parameters are reduced by the effective parameters ofc' = 15 kPa and p'= 20°. 20.0 m Seepage Original ground surface 1.5 16° 85.0 m 11.3 m 3.2 m Rock's bedding planes Proposed new ground surfacearrow_forward2. A cut is to be made in a soil that has γ = 17 kN/m3, c = 40 kPa, φ = 20°. The side of the cut slope will make an angle of 25° with the horizontal. Using a factor of safety of 2.0. a. Determine the stability number. b. Determine the stability factor. c. Determine the depth of the cut slope.arrow_forwardA long slope is formed in a soil with shear strength parameter C' = 0, o' = 34°. Firm strate lies below the slope and it is assumed that water table may occasionally rise to the surface, with seepage taking place parallel to the slope. Use Yeat = 18 kN/m³ and Y = 10 kN/m³. maximum slope angle (in degree) to ensure the factor of safety 1.5. Assuming a potential failure surface parallel to the slope would be (a) 45.3 (c) 12.3 (b) 44.7 (d) 11.3arrow_forward
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