An unsupported slope of height 15 m is shown in the figure (not to scale), in which the slope face makes an angle of 50° with the horizontal. The slope material comprises purely cohesive soil having undrained cohesion 75 kPa. A trial slip circle KLM, with a radius 25 m, passes through the crest and toe of the slope and it subtends an angle 60° at its center O. The weight of the active soil mass (W, bounded by KLMN) is 2500 kN/m, which is acting at a horizontal distance of 10 m from the toe of the slope. Consider the water table to be present at a very large depth from the ground surface.
Considering the trial slip circle KLM, the factor of safety against the failure of the slope under undrained condition (rounded off to two decimal places) is \(\underline{\hspace{1cm}}\).
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The factor of safety (FoS) for a slope under undrained conditions can be calculated using the ratio of resisting forces (due to cohesion) to driving forces (due to weight of the soil).
To determine the factor of safety (FOS) against the failure of the slope, we use the following relationships in the context of the slip circle method:
1. Inputs: - Radius of the slip circle, \( R = 25 \, \text{m} \). - Angle subtended by the slip circle at the center, \( \theta = 60^\circ \). - Weight of the soil mass, \( W = 2500 \, \text{kN/m} \). - Cohesive strength of soil, \( c = 75 \, \text{kPa} = 75 \, \text{kN/m}^2 \).
3. Calculate Resisting Force: The resisting force due to cohesion along the slip surface is given by: \( \text{Resisting Force} = c L = 75 \times \frac{25 \pi}{3}\).
4. Calculate Driving Force: The driving force is the component of the weight \( W \) of the soil mass that causes sliding. \(\text{Driving Force} = W \times \sin(50^\circ) = 2500 \times \sin(50^\circ)\).
