GATE 2025 CE (CE1) – Question 55

Geotechnical Engineering · Stability of slopes · 2 marks · Numerical answer

A cut slope is made in a silty clay soil for a new road project, as shown in the figure. The locations of 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/m$^2$. The factor of safety at the point A for long-term stability is ___________________ (*rounded off to 2 decimal places*).

Note:

Shear strength properties of silty clay: $c' = 15$ kN/m$^2$, $\phi' = 15°$, and $c_u = 75$ kN/m$^2$

Unit weight of soil above the GWT ($\gamma$) = 19 kN/m$^3$

Unit weight of soil below the GWT ($\gamma_{sat}$) = 20 kN/m$^3$

Unit weight of water ($\gamma_w$) = 9.81 kN/m$^3$

A cut slope above a proposed road. A vertical line through the point A on the potential failure surface is 6.5 m long up to the ground water table and a further 5 m long up to the sloping ground surface. The GWT is drawn as a dashed line.

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: 0.96 to 0.98

Explanation

After the excess pore pressure has dissipated the soil is in the long term, so the effective stress parameters $c'$ and $\phi'$ apply (the undrained $c_u$ is not used). The total vertical stress at A is $19 \times 5 + 20 \times 6.5 = 225$ kN/m$^2$ and the pore water pressure is $9.81 \times 6.5 = 63.77$ kN/m$^2$, so $\sigma' = 161.2$ kN/m$^2$. The shear strength is $\tau_f = c' + \sigma'\tan\phi' = 15 + 161.2 \times 0.2679 = 58.2$ kN/m$^2$. The factor of safety is $\frac{58.2}{60} = 0.97$.