A 20-foot cut slope is made in a soft saturated clay with its sides rising at an angle of 55° to the horizontal. Given: cu = 800 lb/ft² and y = 120 lb/ft³ 1. What is the factor of safety against failure of this cut? 2. At what depth will this slope fail? 3. What will be the mode of failure?

1: An embankment is to be constructed by making use of sandy clay having the following properties: c' = 35 kN/m², o' = 25° and y= 19.5 kN/m³. The height of the embankment is 20 m with a slope of 30° with the horizontal as shown in below Figure. Estimate the factor of safety by the method of slices for the trial circle shown in the figure. 20 m 77 ww 25° A 30° 35° c'= 35 kN/m² p' = 25° y = 19.5 kN/m³ 25 m

Using ₁ = 0 analysis and assuming planar failure as shown, the minimum factor of safety against shear failure of a vertical cut of height 4 m in a pure clay having C₁ = 120 kN/m2 and Ysat Y sat = 20 kN/m³ is 77XXX77XXX Potential failure plane 4 m →

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 was excavated in saturated clay for installation of water piping system. The angle Bof cut slope is 45° with the horizontal. During excavation works, slope failure was occurred when the cut reached a depth of 4.0 m. Previous soil explorations showed that a hard layer was located at a depth of 6 m below the ground surface. The unit weight of saturated clay is 17.5 kN/m. (Assuming an undrained condition). 1. a) Determine the undrained cohesion of the clay. b) Determine Factor of Safety (FoS) if the depth of cut slope is 3m. c) What is the maximum depth of saturated clay can be excavated if the Factor of Safety (FoS) 1.4 to be implemented?

Que.5 A long natural slope in overconsolidated fissured clay of saturated unit weight 22kN/m³, is inclined at 14 degrees to the horizontal. The water table is at the surface and seepage is approximately parallel to the slope. A slip surface has developed on a plane parallel to the surface at a depth of 4m. Determine the factor of safety along the slip plane using the residual strenght parameter, o'r=20 degrees.

A 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° = 9m

A cutting of 12 m depth is to be made for a road as shown in below figure. The soil properties are: c'=24 kN/m, o'= 15°, and y= 18 kN/m. Estimate the factor of safety by the method of slices for the trial circle shown in the figure. 7.5 m R=24 m E 75m - 12 m Trial circle -

A cut slope was excavated in saturated clay for installation of water piping system. The angle Bof cut slope is 45° with the horizontal. During excavation works, slope failure was occurred when the cut reached a depth of 4.0 m. Previous soil explorations showed that a hard layer was located at a depth of 6 m below the ground surface. The unit weight of saturated clay is 17.5 kN/m³. (Assuming an undrained condition). 1. а) Determine the undrained cohesion of the clay. b) Determine Factor of Safety (FoS) if the depth of cut slope is 3m. What is the maximum depth of saturated clay can be excavated if the Factor of Safety (FoS) 1.4 to be implemented? c)

A vertical cut is to be made in a purely cohesive clay deposit (c'=30kPa, φ'=0deg, γ=16kN/m3). Find the maximum height of the cut which can be temporarily supported. From the stability chart, the stability number can be used as 0.261.