Question 8 of 9: Overturning and sliding stability of a cantilever retaining wall

Question 8: Overturning and sliding stability of a cantilever retaining wall (24 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The factor of safety of the wall against overturning about the toe and against sliding along the base, both per metre run, together with a judgement on whether the section as drawn is acceptable.

Approach. Take a vertical virtual plane through the back of the heel and treat the wall, the base and the wedge of soil standing on the heel as one rigid body; apply Rankine active pressure to that plane using effective stresses, add the hydrostatic water pressure and the base uplift separately, and take moments about the front bottom edge of the toe.

1. Assume the density values the question asks for. Only the bulk unit weight of the backfill is given, but a saturated value is needed below the water table. Treating $\gamma = 18$ kN/m$^3$ as the dry unit weight and assuming a specific gravity of solids $G_s = 2.65$ — typical for a quartz sand — the void ratio and the saturated unit weight follow:$$\begin{aligned}e &= \frac{G_s\gamma_w}{\gamma_d} - 1 = \frac{2.65(9.81)}{18} - 1 = 0.444 \\ \gamma_{sat} &= \frac{G_s + e}{1+e}\gamma_w = \frac{2.65+0.444}{1.444}(9.81) = 21.02\ \text{kN/m}^3\end{aligned}$$$$\gamma^{\prime} = 21.02 - 9.81 = 11.21\ \text{kN/m}^3 .$$
2. Geometry of the virtual back plane. The plane rises from the back bottom corner of the base, 4.8 m from the toe edge, to the backfill surface. Because the surface climbs at $\beta = 8^{\circ}$ across the 3.0 m heel, the plane is$$H^{\prime} = 7.0 + 3.0\tan 8^{\circ} = 7.0 + 0.422 = 7.422\ \text{m}$$high, and the water table stands 3.1 m above the underside of the base.
3. Rankine active coefficient for a sloping backfill. With $c^{\prime} = 0$, $\phi^{\prime} = 30^{\circ}$ and $\beta = 8^{\circ}$,$$K_a = \cos\beta\,\frac{\cos\beta - \sqrt{\cos^2\beta - \cos^2\phi^{\prime}}}{\cos\beta + \sqrt{\cos^2\beta - \cos^2\phi^{\prime}}} = 0.99027\,\frac{0.99027 - 0.48024}{0.99027 + 0.48024} = 0.3435 .$$The resultant thrust on the plane acts parallel to the ground surface, i.e. inclined at $8^{\circ}$ below the horizontal.
4. Effective active pressure at the three key levels. Working down the plane with $\sigma^{\prime}_a = K_a\sigma^{\prime}_v$:$$\begin{aligned}\text{surface } (z=0): & \quad \sigma^{\prime}_v = q = 20.0, & \sigma^{\prime}_a &= 6.87\ \text{kPa}\\\text{water table } (z=4.322): & \quad \sigma^{\prime}_v = 20.0 + 18(4.322) = 97.79, & \sigma^{\prime}_a &= 33.59\ \text{kPa}\\\text{base } (z=7.422): & \quad \sigma^{\prime}_v = 97.79 + 11.21(3.1) = 132.53, & \sigma^{\prime}_a &= 45.52\ \text{kPa}\end{aligned}$$
5. Resultant earth thrust and its moment. Splitting the diagram into a rectangle and a triangle above the water table and again below it gives four components whose sum and first moment about the underside of the base are$$P_a = 29.69 + 57.73 + 104.12 + 18.50 = 210.03\ \text{kN/m},$$$$\textstyle\sum P_i h_i = 29.69(5.261) + 57.73(4.541) + 104.12(1.550) + 18.50(1.033) = 598.8\ \text{kN}\cdot\text{m/m}.$$Resolving at $8^{\circ}$, the horizontal component is $P_h = 210.03\cos 8^{\circ} = 207.99$ kN/m with an overturning moment of $598.8\cos 8^{\circ} = 593.0$ kN·m/m, and the vertical component $P_v = 210.03\sin 8^{\circ} = 29.23$ kN/m acts downwards on the plane at $x = 4.8$ m from the toe.
6. Water pressure on the back plane. Below the water table the pore pressure is hydrostatic and acts in addition to the effective earth pressure:$$P_w = \tfrac12\gamma_w h^2 = \tfrac12(9.81)(3.1)^2 = 47.14\ \text{kN/m} \quad \text{at } h/3 = 1.033\ \text{m above the base},$$giving an overturning moment of 48.7 kN·m/m.
7. Uplift on the base. With water standing 3.1 m above the underside of the base at the heel and the toe side drained, the pore pressure on the base is taken to fall linearly from $\gamma_w(3.1) = 30.41$ kPa at the heel to zero at the toe:$$U = \tfrac12(30.41)(4.8) = 72.99\ \text{kN/m} \quad\text{acting at } x = \tfrac23(4.8) = 3.20\ \text{m from the toe},$$an overturning moment of 233.6 kN·m/m. Uplift is included because the foundation is a silty clay of finite permeability and no cut-off is shown; the effect of omitting it is quantified at the end.
8. Total overturning moment about the toe.$$M_o = 593.0 + 48.7 + 233.6 = 875.2\ \text{kN}\cdot\text{m/m}.$$
9. Stabilising weights and their moments. The wall, the base and everything standing on the heel act about the same point:

10. Factor of safety against overturning.$$FS_{overturning} = \frac{\sum M_R}{\sum M_O} = \frac{1924.1}{875.2}$$$$\boxed{FS_{overturning} = 2.20}$$which comfortably exceeds the usual requirement of 1.5 to 2.0.
11. Effective normal force on the base. The uplift is subtracted from the total weight before any friction is claimed:$$N^{\prime} = \sum V - U = 654.66 - 72.99 = 581.7\ \text{kN/m}.$$
12. Base shear resistance. Adopting the customary reductions $\delta = \tfrac23\phi^{\prime}_2 = 10^{\circ}$ and $c_a = \tfrac23 c^{\prime}_2 = 10$ kPa for a concrete base cast against the silty clay,$$F_R = N^{\prime}\tan\delta + c_a B = 581.7(0.1763) + 10.0(4.8) = 102.6 + 48.0 = 150.6\ \text{kN/m}.$$
13. Passive resistance in front of the toe. Over the 1.0 m of embedment, with $K_p = \tan^2(45^{\circ} + 7.5^{\circ}) = 1.698$,$$P_p = \tfrac12 K_p\gamma_2 D^2 + 2c^{\prime}_2\sqrt{K_p}\,D = \tfrac12(1.698)(19)(1.0)^2 + 2(15)(1.303)(1.0) = 16.1 + 39.1 = 55.2\ \text{kN/m}.$$
14. Factor of safety against sliding. The driving force is the horizontal earth thrust plus the water thrust, $F_D = 207.99 + 47.14 = 255.1$ kN/m, so$$FS_{sliding} = \frac{F_R + P_p}{F_D} = \frac{150.6 + 55.2}{255.1}$$$$\boxed{FS_{sliding} = 0.81 \quad (0.59 \text{ if } P_p \text{ is neglected})}$$
15. Position of the resultant and base pressures. The net moment about the toe places the resultant at$$\begin{aligned}\bar{x} &= \frac{\sum M_R - \sum M_O}{N^{\prime}} = \frac{1924.1-875.2}{581.7} = 1.803\ \text{m} \\ e &= \frac{B}{2} - \bar{x} = 2.40 - 1.80 = 0.597\ \text{m}\end{aligned}$$which is inside the middle third ($B/6 = 0.80$ m), so the whole base stays in compression with $q_{max} = 211.6$ kPa at the toe and $q_{min} = 30.8$ kPa at the heel. Those are high for a $\phi^{\prime} = 15^{\circ}$, $c^{\prime} = 15$ kPa foundation and the bearing capacity would also have to be checked.
16. Assessment and remedy. The wall is safe against overturning but fails in sliding by a wide margin, and the reason is visible in the numbers: the water standing in the backfill contributes 47.1 kN/m of thrust and 73.0 kN/m of uplift, while the foundation offers only $\tan 10^{\circ}$ of friction. Even with a fully drained backfill the factor of safety only reaches 0.98, so drainage alone is not enough. Two measures are needed. First, install a full-height granular or geocomposite drain against the stem discharging through weep holes, which removes $P_w$ and most of the uplift. Second, cast a shear key under the base so that sliding is forced through the soil rather than along the concrete interface (raising the available strength to $N^{\prime}\tan\phi^{\prime}_2 + c^{\prime}_2 B = 227.9$ kN/m) and so that the passive block is deepened; a key taking the effective embedment to about 2.1 m, i.e. roughly 1.1 m deep, restores $FS_{sliding} \ge 1.5$. Widening the heel would also help, since it recruits more backfill weight without increasing the thrust.