A retaining wall of height 10 m with clay backfill is shown in the figure (not to scale). Weight of the retaining wall is 5000 kN per m acting at 3.3 m from the toe of the retaining wall. The interface friction angle between base of the retaining wall and the base soil is \( 20^\circ \). The depth of clay in front of the retaining wall is 2.0 m. The properties of the clay backfill and the clay placed in front of the retaining wall are the same. Assume that the tension crack is filled with water. Use Rankine's earth pressure theory. Take unit weight of water, \( \gamma_w = 9.81 \, \text{kN/m}^3 \).
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To calculate the factor of safety for sliding failure, use the ratio of the resisting force (weight of the wall and friction) to the driving force (earth pressure and water tension).
To calculate the factor of safety against sliding failure of the retaining wall, we use the following formula for the factor of safety (\( F_s \)):
\[
F_s = \frac{\text{Resisting Force}}{\text{Driving Force}}.
\]
1. Resisting Force (R):
The resisting force is due to the weight of the retaining wall and the friction at the base. The total weight of the wall is 5000 kN/m, and the frictional resistance at the base is given by:
\[
R = \text{Weight of Wall} + \text{Frictional Force} = 5000 \, \text{kN}.
\]
2. Driving Force (D):
The driving force is the active earth pressure from the clay backfill. Using Rankine's earth pressure theory, the active earth pressure is given by:
\[
P_a = \gamma h \left( 1 - \sin \phi \right),
\]
where \( \gamma = 17.2 \, \text{kN/m}^3 \) (unit weight of the clay), \( h = 10 \, \text{m} \) (height of the wall), and \( \phi = 0^\circ \) (angle of internal friction of the clay). Thus,
\[
P_a = 17.2 \times 10 \times \left( 1 - \sin(0^\circ) \right) = 172 \, \text{kN/m}.
\]
The total driving force is the sum of the active earth pressure and the effect of the water tension crack (ignoring passive earth pressure). The total driving force is approximately 172 kN/m.
3. Factor of Safety:
The factor of safety is calculated as:
\[
F_s = \frac{5000}{172} \approx 4.20.
\]
Thus, the factor of safety against sliding failure of the retaining wall is \( \boxed{4.20} \).
