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, ${\gamma }_w=9.81$ kN/m${}^3$.


The factor of safety (round off to two decimal places) against sliding failure of the retaining wall after ignoring the passive earth pressure will be ________

In the range: 4.2 - 4.35

To find the factor of safety against sliding failure for the retaining wall, we need to determine the total driving forces (active earth pressure and water pressure in the tension crack) and the resisting forces (base friction), while ignoring the passive earth pressure.

1. Given Data:

- Height of the wall ($H$): $10\text{ m}$
- Backfill properties (Clay): $\gamma =17.2{\text{ kN/m}}^3$, $\varphi =0^{\circ }$, $c=30{\text{ kN/m}}^2$
- Weight of the wall ($W$): $5000\text{ kN/m}$
- Interface friction angle ($\delta$): $20^{\circ }$
- Unit weight of water (${\gamma }_w$): $9.81{\text{ kN/m}}^3$

2. Active Earth Pressure Calculations (Rankine’s Theory):

For a purely cohesive soil ($\varphi =0^{\circ }$), the coefficient of active earth pressure is:

$$K_a={\tan }^2\left(45^{\circ }-\frac{\varphi }{2}\right)={\tan }^2(45^{\circ })=1$$

The depth of the tension crack ($z_c$) is calculated where the active pressure is zero:

$$z_c=\frac{2c}{\gamma \sqrt{K_a}}=\frac{2\times 30}{17.2\times 1}\approx 3.488\text{ m}$$

The active earth pressure at the base of the wall ($z=10\text{ m}$) is:

$${\sigma }_a=K_a\gamma H-2c\sqrt{K_a}=(1\times 17.2\times 10)-(2\times 30)=172-60=112{\text{ kN/m}}^2$$

**3. Driving Forces ($F_d$):

Since the tension crack is filled with water, the total horizontal driving force is the sum of the soil pressure (acting below the crack) and the water pressure (acting within the crack depth):

- Active thrust from soil ($P_a$):** $$P_a=\frac{1}{2}\times {\sigma }_a\times (H-z_c)=\frac{1}{2}\times 112\times (10-3.488)=56\times 6.512=364.672\text{ kN/m}$$
- **Hydrostatic force from water in the crack ($P_w$):** $$P_w=\frac{1}{2}\times {\gamma }_w\times z_c^2=\frac{1}{2}\times 9.81\times (3.488{)}^2\approx 59.69\text{ kN/m}$$

Total Driving Force:

$$F_d=P_a+P_w=364.672+59.69=424.362\text{ kN/m}$$

**4. Resisting Force ($F_r$):**

The resisting force is provided by the friction between the base of the wall and the soil:

$$F_r=W\tan (\delta )=5000\times \tan (20^{\circ })\approx 5000\times 0.36397=1819.85\text{ kN/m}$$

**5. Factor of Safety ($FoS$):**

The factor of safety against sliding is the ratio of resisting forces to driving forces:

$$FoS=\frac{F_r}{F_d}=\frac{1819.85}{424.362}\approx 4.288$$

Rounding to two decimal places, we get:

Factor of Safety = 4.29