For the cut slope described in Problem 15.11, how deep should the cut be made to ensure a factor of safety of 2.5 against sliding?
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A: find the factor of safety with respect to sliding
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Please answer 15.12
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- 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?A long slope is made up of a cohesionless soil with the internal friction angle of 34°. If the water table rises to the surface of the slope, then determine the maximum possible value of the slope angle to get a factor of safety of 1.5. Take saturated unit weight as 18 kN/m3 and the unit weight of water as 10 kN/m3.Figure 2 shows a slope with an inclination of : β = 58 ͦ. If AC represents a trial failure plane inclined at an angle θ = 32 ͦ with the horizontal, determine the factor of safety against sliding for the wedge ABC. Given: H = 6 m; ɣ = 19 kN/m3, Ø =21 ͦ, and c’= 38 kN/m2
- Figure 15.47 shows an infinite slope with H = 9.14 m and the groundwater table coinciding with the ground surface. If there is seepage through the soil, determine the factor of safety against sliding along the plane AB. The soil properties are Gs = 2.7, e = 0.8, β = 16°, ϕ′ = 21°, and c′ = 59.8 kN/m2.Q.1. Refer to the infinite slope shown in Figure 1. Given: β = 19 ͦ, ɣ = 20 kN/m3 , Ø = 33 ͦ, and c’ = 47 kN/m2 . Find the height, H, such that a factor of safety, Fs = 3.1 is maintained against sliding along the soil-rock interface.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 maximum depth in meters up to which the cut could be made?
- An infinite slope consists of bed rock inclined at 15° to the horizontal that is covered by 4 m (measured vertically) soil with ?′=10?N/?2, ?′=24° and ?=19.0?N/?3.a. What is the factor of safety of this slope?b. If there is steady state seepage through the soil, with the water table coinciding with the ground level, what would be the factor of safety? Assume ?sat = 19.0?N/?3.A cut slope is planned in a soil having properties as follows: γ = 19.3 kN/m3, ϕ′ = 19°, and c′ = 55 kN/m2. The cut slope makes an angle 60° with the horizontal. Assuming a planar failure surface, what would be the maximum permissible slope height, H, if the desired factor of safety is 2.5?An infinite slope is inclined at 22 degree. A plane of failure has developed at depth of 8 m from the surface. Given dry unit weight = 18kN/m3, c’ = 12kN/m2, angle of friction = 23.5 degree and saturated unit weight = 23kN/m3. Calculate the slope factor of safety if: (i) water table exists far from the slip line
- 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 cement lined rectangular channel 6m wide carries water at the rate of 30m3/s. Find the value of Manning’s constant, if the slope required to maintain a depth of 1.5m is 0.00117A cut slope is to be made in a saturated clay at an angle β = 60° with the horizontal. Assuming that the critical sliding surface is circular, determine the maximum depth up to which the cut could be made. Given: undrained shear strength, cu = 34 kN/m2 (ϕ = 0 condition), and γ = 17 kN/m3 . What is the nature of the critical circle (toe, slope, or midpoint)?