Question

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 is fully saturated (\( \phi = 0 \)). 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 the tension crack is completely filled with water, the factor of safety against slope failure of the bank is ......... (round off to two decimal places).
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Hint:

To calculate the factor of safety for slope failure, use the moment equilibrium approach, where the resisting moment is compared with the overturning moment. Carefully substitute the given values of the area, depth, and unit weights into the respective formulas for accurate calculations.

Answer 1

Correct Answer: 1

The factor of safety (FOS) is given by the ratio of resisting moment to overturning moment: \[ FOS = \frac{{Resisting moment}}{{Overturning moment}} \] Resisting moment is calculated as: \[ R = \left( \frac{80}{180} \times \pi \times 14 \right) \times 30 \] The overturning moment is: \[ {Overturning moment} = \left( (150 \times 1) \times 20 \times 2.5 \right) + \left( \frac{1}{2} \times 9.81 \times 2.5^2 \times \left( 5.5 + \frac{2.5 \times 2}{3} \right) \right) \] Substituting the values into the formula: \[ FOS = \frac{30 \times 14 \times \left( \frac{80}{180} \times \pi \times 14 \right)}{(150 \times 1) \times 20 \times 2.5 + \frac{1}{2} \times 9.81 \times 2.5^2 \times \left( 5.5 + \frac{2.5 \times 2}{3} \right)} \] After calculating: \[ FOS = \frac{8210.028}{7500 + 219.703} = 1.06 \] Thus, the factor of safety against slope failure is 1.06.