Question 6: 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.

[Figure not reproduced: Figure 1 redrawn to scale. The dashed red line C-B is the vertical virtual plane on which the Rankine active thrust is taken; the pressure triangle reaches 36.0 kPa at the underside of the base. See the official exam paper.]

Given. The cantilever wall of Figure 1, dimensions scaled from the figure and soil properties as printed on it.

Find. The factors of safety of the wall against (a) sliding along its base and (b) overturning about the toe, using Rankine earth-pressure theory.

Approach. Take the Rankine active thrust on the vertical virtual plane C–B through the heel; sum the vertical forces and their moments about the toe A for the concrete and for the soil resting on the base; then form the two factors of safety, resisting over disturbing.

1. Establish the Rankine active pressure coefficient for the backfill. The backfill surface is horizontal and the virtual plane is vertical, so the Rankine condition applies directly and the thrust is horizontal: $$K_a = \frac{1-\sin\phi^{\prime}}{1+\sin\phi^{\prime}} = \frac{1-\sin 30^{\circ}}{1+\sin 30^{\circ}} = \frac{0.5}{1.5} = 0.333 .$$
2. Compute the active thrust on the virtual plane C–B. The plane is the full height of the wall, H = 6.0 m, and the backfill is dry, cohesionless and carries no surcharge, so the pressure varies linearly from zero at C to $K_a\gamma H = 0.333(18)(6.0) = 36.0$ kPa at B: $$P_a = \tfrac{1}{2}K_a\gamma H^{2} = \tfrac{1}{2}(0.333)(18)(6.0)^{2} = \boxed{108.0\ \text{kN/m}}$$ acting horizontally at $H/3 = 2.0$ m above the underside of the base.
3. Form the overturning moment about the toe A. With a horizontal thrust the lever arm is simply the height of its point of application: $$M_o = P_a\left(\frac{H}{3}\right) = 108.0(2.0) = 216.0\ \text{kN}\cdot\text{m/m}.$$
4. Sum the vertical forces and their moments about A. Divide the wall and the soil it carries into four rectangles. Moment arms are measured from A, the front edge of the base.

   Vertical forces and moments about the toe A

   Component	Calculation	W (kN/m)	Arm x (m)	MR (kN·m/m)
   1. Base slab	3.5 × 0.5 × 24	42.0	1.75	73.5
   2. Stem	0.5 × 5.5 × 24	66.0	1.25	82.5
   3. Backfill on the heel	2.0 × 5.5 × 18	198.0	2.50	495.0
   4. Soil on the toe	1.0 × 1.0 × 20	20.0	0.50	10.0
   Totals		326.0		661.0

   The stem height is H minus the base thickness, 6.0 − 0.5 = 5.5 m, and the soil on the toe is 1.0 m wide by (1.5 − 0.5) = 1.0 m deep.
5. Factor of safety against overturning about the toe. Dividing the restoring moment by the overturning moment, $$FS_{overturning} = \frac{\sum M_R}{M_o} = \frac{661.0}{216.0} = \boxed{3.06} \;\; \ge 2.0 \ \text{, satisfactory.}$$
6. Locate the resultant and confirm there is no tension under the heel. The line of action of the resultant meets the base at $$\bar{x} = \frac{\sum M_R - M_o}{\sum V} = \frac{661.0-216.0}{326.0} = 1.365\ \text{m from A},$$ so the eccentricity is $e = B/2 - \bar{x} = 1.75 - 1.365 = 0.385$ m, comfortably inside the middle third, $B/6 = 0.583$ m. The whole base therefore stays in compression, which is the condition under which the sliding calculation below is valid.
7. Base friction available against sliding. The foundation soil is cohesionless, so adhesion contributes nothing and the entire resistance on the base is frictional: $$F_R = \left(\sum V\right)\tan\delta_f = 326.0\tan 25^{\circ} = 326.0(0.4663) = 152.0\ \text{kN/m}.$$
8. Factor of safety against sliding, ignoring passive resistance. Taking the conservative view that the soil in front of the wall may be removed for a service trench at some time in the wall's life, $$FS_{sliding} = \frac{F_R}{P_a} = \frac{152.0}{108.0} = \boxed{1.41} \;\; \lt 1.5 \ \text{, inadequate.}$$
9. Recompute with the passive resistance in front of the wall. The front soil stands 1.5 m above the underside of the base with $\phi^{\prime} = 35^{\circ}$, so $$K_p = \frac{1+\sin 35^{\circ}}{1-\sin 35^{\circ}} = 3.690, \qquad P_p = \tfrac{1}{2}K_p\gamma D_f^{2} = \tfrac{1}{2}(3.690)(20)(1.5)^{2} = 83.0\ \text{kN/m}.$$ Full passive pressure needs several tens of millimetres of wall movement, so normal practice is to count only half of it. Reporting both bounds, $$FS_{sliding} = \frac{152.0 + 0.5(83.0)}{108.0} = \boxed{1.79} \qquad\text{and}\qquad \frac{152.0+83.0}{108.0} = 2.18 .$$

The wall is therefore secure against overturning by a wide margin but only marginally secure against sliding, and it relies on the soil in front of the toe to reach the usual 1.5 threshold. If that reliance is unacceptable — and next to a services corridor it usually is — the standard remedies are a shear key 0.5 to 0.7 m deep beneath the base (which forces the failure surface into the soil, replacing tan 25 degrees by tan 35 degrees and raising the frictional resistance to 228 kN/m and the factor of safety to 2.11 with no passive contribution at all), or widening the heel, which adds backfill weight and therefore friction.