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. All dimensions are read from Figure 3, per metre run of wall.

Find. The factor of safety against overturning about the toe, and the factor of safety against sliding along the base.

[Figure not reproduced: Question 8 (redrawn from Figure 3): the cantilever wall, the sloping surcharged backfill, and the vertical Rankine plane through the back edge of the heel on which the active thrust is computed. See the official exam paper.]

Approach. Take the vertical plane through the back edge of the heel as the pressure surface, so that everything to the left of it — concrete plus the wedge of soil standing on the heel — is the resisting mass. Compute the Rankine active thrust for a sloping backfill on that plane, resolve it, take moments about the toe for overturning, and resolve horizontally for sliding.

1. Establish the height of the pressure surface. The Rankine plane runs from the underside of the base up to the ground surface at the back edge of the heel. The backfill rises at 8° across the 3.0 m heel, so $$H' = 0.90 + 6.10 + 3.00\tan 8^\circ = 7.00 + 0.4216 = 7.4216\ \text{m}$$
2. Compute the Rankine active coefficient for a sloping backfill. With $\alpha = 8^\circ$ and $\phi' = 32^\circ$, $$K_a = \cos\alpha \frac{\cos\alpha - \sqrt{\cos^{2}\alpha - \cos^{2}\phi'}}{\cos\alpha + \sqrt{\cos^{2}\alpha - \cos^{2}\phi'}} = 0.3159$$ The thrust it produces acts parallel to the ground surface, that is at 8° above the horizontal.
3. Evaluate the active thrust, soil and surcharge separately. The surcharge extends well beyond the base, so it acts as a uniform vertical stress over the full height of the plane: $$P_{a,soil} = \tfrac{1}{2}\gamma H'^{2}K_a = \tfrac{1}{2}(18)(7.4216)^{2}(0.3159) = 156.58\ \text{kN/m}$$ $$P_{a,q} = K_a q H' = (0.3159)(20)(7.4216) = 46.88\ \text{kN/m}$$ giving a total $P_a = 203.46\ \text{kN/m}$.
4. Resolve the thrust and locate it. The triangular soil component acts at $H'/3 = 2.474\ \text{m}$ above the base and the rectangular surcharge component at $H'/2 = 3.711\ \text{m}$, so the resultant acts at $$\bar{y} = \frac{156.58(2.474) + 46.88(3.711)}{203.46} = 2.759\ \text{m}$$ Resolving at 8° gives the horizontal and vertical components $$P_h = P_a\cos 8^\circ = 201.48\ \text{kN/m}, \qquad P_v = P_a\sin 8^\circ = 28.32\ \text{kN/m}$$ The vertical component acts down the plane, at the back edge of the heel, and it helps the wall in both checks.
5. Compute the overturning moment about the toe. Only the horizontal component overturns: $$M_o = P_h\,\bar{y} = 201.48(2.759) = 555.9\ \text{kN}\cdot\text{m/m}$$
6. Assemble the resisting weights and their moments about the toe. Distances are measured from the front edge of the base slab.

   Resisting forces and moments about the toe, per metre run

   Component	Calculation	$W$ (kN/m)	$x$ (m)	$M_R$ (kN·m/m)
   Base slab	$4.80 \times 0.90 \times 23.5$	101.52	2.400	243.65
   Stem, rectangular part	$0.40 \times 6.10 \times 23.5$	57.34	1.600	91.74
   Stem, battered taper	$\tfrac{1}{2}(0.36)(6.10) \times 23.5$	25.80	1.280	33.03
   Soil on the heel, rectangle	$3.00 \times 6.10 \times 18$	329.40	3.300	1087.02
   Soil on the heel, sloping wedge	$\tfrac{1}{2}(3.00)(0.4216) \times 18$	11.38	3.800	43.26
   Vertical component of the thrust	$P_a\sin 8^\circ$	28.32	4.800	135.92
   Totals	—	553.76	—	1634.6

   The 0.1 m of soil sitting over the toe is neglected, which is conservative and standard, because it can be removed by future excavation or erosion.
7. Factor of safety against overturning. Dividing the resisting moment by the overturning moment, $$\boxed{FS_{overturning} = \frac{1634.6}{555.9} = 2.94}$$ which comfortably exceeds the usual requirement of 2.0.
8. Factor of safety against sliding along the base. The figure gives the base friction angle directly as $\delta' = 15^\circ$; the base adhesion is taken as two-thirds of the foundation cohesion, $c_a = \tfrac{2}{3}(30) = 20\ \text{kPa}$, which is Das's $k_2$ convention for concrete cast against soil. Passive resistance in front of the toe is neglected, because 1 m of soil over a toe cannot be relied on for the life of the wall. Then $$FS_{sliding} = \frac{\sum V\tan\delta' + Bc_a}{P_h} = \frac{553.76(0.2679) + 4.80(20)}{201.48} = \frac{148.38 + 96.00}{201.48}$$ $$\boxed{FS_{sliding} = 1.21}$$
9. Test the sensitivity of the sliding result. If the 1.0 m of soil in front of the toe is credited, with $K_p = \tan^{2}(45^\circ + \delta'/2) = 1.698$ and the foundation cohesion, $$P_p = \tfrac{1}{2}K_p\gamma_2 D^{2} + 2c'\sqrt{K_p}\,D = 14.27 + 78.19 = 92.46\ \text{kN/m}$$ which lifts the factor of safety to $(244.38 + 92.46)/201.48 = 1.67$. The design conclusion sits between these two numbers, and it is not comfortable.

Interpretation and recommendation. The wall is secure against overturning at 2.94 but marginal against sliding: 1.21 with passive resistance neglected is below the 1.5 that is normally required, and even the optimistic 1.67 leans on a cohesion intercept and on a metre of soil that a future service trench could remove overnight. The eccentricity of the base resultant is $e = B/2 - (M_R - M_o)/\sum V = 2.40 - 1.948 = 0.452\ \text{m}$, well inside the middle third ($B/6 = 0.80\ \text{m}$), so the base pressures are wholly compressive at 180.5 kPa under the toe and 50.2 kPa under the heel, and no tension develops. The economical fix is therefore not a wider base but a shear key beneath the base slab, which mobilises passive resistance in undisturbed soil below the level of any future excavation and would take the sliding factor of safety past 1.5 without changing the wall proportions. Extending the heel by half a metre would also help, by adding roughly 55 kN of soil weight and 15 kN of base friction. Whichever is chosen, the backfill must be a free-draining granular material with a functioning drain behind the stem: the analysis above contains no water pressure at all, and even a partial build-up of hydrostatic pressure behind this wall would overwhelm both checks.