Question

A cut slope is made in a silty clay soil for a new road project, as shown in the figure. The locations of the ground water table (GWT) and potential failure surface are shown in the figure. After the cut is made, the excess pore water pressure is fully dissipated, and the shear stress at the point A is 60 kN/m². The factor of safety at the point A for long-term stability is .......... (rounded off to 2 decimal places).

Hint:

When calculating the factor of safety, always ensure you use the effective shear strength parameters and consider the depth of the water table for accurate normal stress calculations.

Answer 1

Correct Answer: 0.96

\[ c' = 15 \, {kN/m}^2, \, \phi' = 15^\circ, \, c_u = 75 \, {kN/m}^2 \] \[ \gamma_{{above}} = 19 \, {kN/m}^3, \, \gamma_{{below}} = 20 \, {kN/m}^3, \, \gamma_{{w}} = 9.81 \, {kN/m}^3 \] \[ {Shear stress at point A} = 60 \, {kN/m}^2 \] For long-term, effective shear parameters will be used. The factor of safety (FOS) is given by: \[ {FOS} = \frac{C' + \sigma_n \tan\phi'}{\tau} \] Where \( \sigma_n \) is the normal stress. To calculate \( \sigma_n \), we use the following formula: \[ \sigma_n = (5y_B + 6.5y_{{sat}} - 6.5y_w) \] Substituting the values: \[ \sigma_n = (5 \times 19 + 6.5 \times 20 - 6.5 \times 9.81) = 161.235 \, {kN/m}^2 \] Now, calculate the factor of safety: \[ {FOS} = \frac{15 + 161.235 \times \tan 15^\circ}{60} = 0.97 \] Thus, the factor of safety is \( \boxed{0.97} \) (rounded to two decimal places).