GATE 2025 CE (CE2) – Question 55

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

The bank of a canal has the profile shown in the figure. The material is a homogeneous clay with a bulk unit weight of 20 kN/m$^3$, undrained cohesion of 30 kPa and it is fully saturated ($\phi_u = 0$). For the trial slip circle shown, the area ABCDEA is 150 m$^2$ and the centroid is at P. A tension crack (DE) of 2.5 m deep was also observed. Assume unit weight of water is 9.81 kN/m$^3$ and consider 1 m run of the bank for the analysis.
Considering the canal is empty and tension crack is completely filled with water, the factor of safety against slope failure of the bank is ______ (round off to two decimal places).

a canal bank ABCD with a slip circle of centre O and radius R = 14 m that cuts the ground at A and E, with the angle 80° at O. The centre O is 5.5 m above the level of CD, the tension crack DE is 2.5 m deep, and the centroid P of the sliding mass is 2.5 m horizontally from the vertical through O.

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Correct answer: 1.05 to 1.07

Explanation

Use the total stress method with $\phi_u = 0$. The weight of the sliding mass is $W = 20 \times 150 = 3000$ kN per metre, acting 2.5 m from the vertical through the centre O, so its driving moment is $3000 \times 2.5 = 7500$ kN m. The water in the tension crack gives the force $\frac{1}{2}\gamma_wh^2 = \frac{1}{2} \times 9.81 \times 2.5^2 = 30.7$ kN acting at $\frac{2}{3}$ of the crack depth below the ground, which is $5.5 + 1.67 = 7.17$ m above the centre, with the moment $30.7 \times 7.17 = 220$ kN m. The driving moment is $7500 + 220 = 7720$ kN m. The slip arc subtends $80°$, so its length is $14 \times 80° \times \frac{\pi}{180} = 19.55$ m. The resisting moment is $cLR = 30 \times 19.55 \times 14 = 8210$ kN m. The factor of safety is $\frac{8210}{7720} = 1.06$.