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

Given Data:

• Effective cohesion: \[ c' = 15 \, \text{kN/m}^2 \]
• Effective angle of internal friction: \[ \phi' = 15^\circ \]
• Undrained cohesion: \[ c_u = 75 \, \text{kN/m}^2 \]
• Unit weights: \[ \gamma_{\text{above}} = 19 \, \text{kN/m}^3, \quad \gamma_{\text{below}} = 20 \, \text{kN/m}^3, \quad \gamma_w = 9.81 \, \text{kN/m}^3 \]
• Shear stress at point A: \[ \tau = 60 \, \text{kN/m}^2 \]

Calculation:

For long-term stability, effective shear parameters will be used. The factor of safety (FOS) is given by:

\[ \text{FOS} = \frac{C' + \sigma_n \tan\phi'}{\tau} \]

Step 1: Normal Stress Calculation

The normal stress \( \sigma_n \) is calculated as:

\[ \sigma_n = (5 \gamma_{\text{above}} + 6.5 \gamma_{\text{sat}} - 6.5 \gamma_w) \]

Substituting values:

\[ \sigma_n = (5 \times 19 + 6.5 \times 20 - 6.5 \times 9.81) = 161.235 \, \text{kN/m}^2 \]

Step 2: Factor of Safety Calculation

Using the FOS equation:

\[ \text{FOS} = \frac{15 + 161.235 \times \tan 15^\circ}{60} \]

Solving:

\[ \text{FOS} = 0.97 \]

Final Answer:

Factor of Safety: \( \boxed{0.97} \) (rounded to two decimal places).