Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Question
Chapter 13, Problem 13.6P
(a)
To determine
Find the factor of safety of the slip plane.
(b)
To determine
Find the factor of safety of the slip plane using the ultimate friction angle.
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Students have asked these similar questions
The excavation shown below will be made on the clay for the construction of a highway. Potential Slope stability analysis is required on the slip plane. The ground is silty clay and c '= 18 kPa,' '= 20o and water = 100 kPa. At point a If the shear stress is 60 kPa, calculate the safety coefficient for this point.
A 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 W
A cut slope is to made. The unit weight of soil is 17 kN/m^3 and the angle of internal friction is 20°. The soil has a cohesion of 24 kPa. The cut slope makes an angle of 30° with the horizontal and the height is 9 m. if the trial failure plane is 15°from the horizontal. Determine the following:
1. Frictional force along the failure plane.
2. Cohesive force along the failure plane.
3. Factor of safety against sliding.
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 cut slope is to be excavated in saturated clay with sides rising at an angle a = 55° to the horizontal, as shown below B Saturated Clay Ysat = 115 Ib/ft Unconfined Compressive Strength q,= 1000 Ib/ft? a = 55°, B = 35° H. B A a. The depth H of the excavation for a factor of safety of 1 when considering the trial failure plane AC is ft. b. The depth H of the excavation for a factor of safety of 1 when considering the most critical failure plane is ft.arrow_forwardDetermine the factor of safety of the slope against shear failure if water percolates in a direction parallel to surface of slope.arrow_forward2. A cut slope is to be made as shown. The unit weight of soil is 18KN/cu.m and the angle of internal friction is 22°. The soil has cohesion of 30 Kpa. The cut slope makes an angle of 30° with the horizontal and the height is 12 m. If the failure plane is 15° from horizontal. Determine a. Frictional force along the failure plane b. Cohesive force along the failure plane Factor of safety against sliding B 24 m A trial failure plane 12 m 30 15° Carrow_forward
- Please answer 15.15arrow_forwardAn embankment consists of clay fill for which c=25 kPa and angle of internal friction is 260 9from consolidated undrained test with pore pressure measurement) The weight of fill per unit volume is 18.64 kN/m3. Estimate the shear strength of the material on a horizontal plane at a point 20 m below the surface of the embankment. Include fbdarrow_forward(Note:Depth of rock layer is 2m not 12 m.) Q A cut slope was excavated in a saturated clay with a slope angle β = 57 ͦ with the horizontal. Slope failure occurred when the cut reached a depth of 8 m. Previous soil explorations showed that a rock layer was located at a depth of 2 m below the ground surface. Assuming an undrained condition and that ɣsat = 21 kN/m3 : Determine the undrained cohesion of the clay (Fig 3). What was the nature of the critical circle? With reference to the top of the slope, at what distance did the surface of the sliding intersect the bottom of the excavation?arrow_forward
- 1. Determine the gradient of each point from point 1. ( 2540 2540 2290 650-m from point 1-2 850-m from point 1-3 1550-m from point 1-4 Contour map units: meters 2. Given a density of a cubical quart rock to be 2650 kg/m³, determine the lithostatic stress on 800,000 cm³ volume if the acceleration is 980 cm/sec² with a depth of burial to be 6 miles./arrow_forwardGeotechnical engineeringarrow_forwardProblem 1Explain whether the following are true or false. 1. A lower safety factor implies that a greater fraction of the shear strength is mobilized along the failure surface. 2. In an undrained slope of 1.0 (horizontal): 1.5 (vertical), the critical circle is always a toe circle. 3. The mid-point circle passes through the toe. 4. Taylor’s stability charts can be used only for homogeneous soils.arrow_forward
- Thanks for the solution.arrow_forwardAnswer Q13 to Q 15 based on following description A cutting in a saturated clay is inclined at a slope 1 vert:1.2 horizontal and has a vertical height of 10.0 m. The bulk unit weight of the soil is 18 kN/m³ and its undrained cohesion is 35 kN/m³. Undrained friction angle, -0. Area of slip mass = 51.2 m², d= 7 m r=12m 2m 1 W 10m FoS of a slope along a circular slip surface is given by; cular F = Wd Calculate the arc length along the slip surface, La a-964.8 m O b. 16.8 m O c. 114.0 m O d. 8.4 m d=5m 1.2arrow_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_forward
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