Question 9 of 10: Stability of a cantilever retaining wall against overturning and sliding

Question 9: Stability of a cantilever retaining wall against overturning and sliding (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. A cantilever wall 11.0 m high overall on a 9.0 m base. The base slab is 1.0 m thick, the toe projects 1.5 m and the heel 6.0 m, and the stem is 10.0 m tall, 1.5 m thick at its foot and 0.5 m at the top with the back face vertical and the front face battered. Soil 1 is the backfill behind the wall, horizontal at the level of the top of the stem; Soil 2 is the natural ground, which founds the base and covers the toe to a depth of 1.5 m below its own surface. No water table is shown, so the backfill is taken as fully drained.

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

[Figure not reproduced: Figure 4 (redrawn). Cantilever retaining wall, the vertical plane through the heel on which the Rankine active thrust acts, and the principal dimensions. See the official exam paper.]

Approach. Take a vertical plane through the heel, apply the Rankine active thrust of Soil 1 on it, and treat everything to the left of that plane — concrete plus the backfill sitting on the heel — as the resisting body. Overturning is then a moment ratio about the front bottom edge of the toe, and sliding a ratio of base friction and adhesion, plus the passive resistance of the soil covering the toe, to the horizontal thrust.

1. Compute the Rankine active pressure on the vertical plane through the heel. With horizontal backfill, $$\begin{aligned}K_{a1} &= \tan^2\!\left(45^{\circ} - \frac{36^{\circ}}{2}\right) = \tan^2 27^{\circ} = 0.2596 \\ \sqrt{K_{a1}} &= 0.5095\end{aligned}$$Because the backfill has a cohesion intercept, the active pressure is $$\sigma'_a(z) = K_{a1}\gamma_1 z - 2c'_1\sqrt{K_{a1}}$$which is negative near the surface and reaches $\sigma'_a(11) = 0.2596(20)(11) - 2(5)(0.5095) = 52.02$ kPa at the base.
2. Allow for the tension crack and resolve the thrust. Soil cannot sustain the tensile pressure, so the diagram is truncated at the depth where it changes sign: $$z_0 = \frac{2c'_1}{\gamma_1\sqrt{K_{a1}}} = \frac{2(5)}{20(0.5095)} = 0.981\ \text{m}$$Ignoring the tensile block entirely, which is the conservative and usual assumption, leaves a triangle of height $H - z_0 = 10.019$ m: $$\begin{aligned}P_a &= \tfrac{1}{2}(52.02)(10.019) = 260.6\ \text{kN/m} \\ \bar{y} &= \frac{H-z_0}{3} = 3.340\ \text{m above the base}\end{aligned}$$Retaining the negative area would give 258.1 kN/m, only one per cent less, so the choice is not critical here.
3. Assemble the vertical forces and their moments about the toe. The resisting body is the concrete plus the backfill standing on the heel; the 0.5 m of Soil 2 covering the toe is included as a small extra. Moment arms are measured from the front bottom edge of the base.

   Vertical forces and their moments about the toe

   Component	Area (m2)	Unit weight (kN/m3)	W (kN/m)	Arm from toe (m)	Moment (kN·m/m)
   Stem — rectangular part	5.000	24.0	120.00	2.7500	330.00
   Stem — battered (triangular) part	5.000	24.0	120.00	2.1667	260.00
   Base slab	9.000	24.0	216.00	4.5000	972.00
   Backfill (Soil 1) over the heel	60.000	20.0	1200.00	6.0000	7200.00
   Soil 2 cover over the toe	0.750	18.0	13.50	0.7500	10.12
   Sums			1669.50	—	8772.12

4. Factor of safety against overturning. The overturning moment is the active thrust times its lever arm about the same point: $$M_O = P_a \bar{y} = 260.6(3.340) = 870.2\ \text{kN}\cdot\text{m/m}$$so $$FS_{overturning} = \frac{\sum M_R}{M_O} = \frac{8772}{870.2} = \boxed{10.08}$$This is far above the usual minimum of 2.0, and the reason is plain from the table: the 6 m heel carries 1200 kN/m of backfill acting at a lever arm of 6.0 m, which alone supplies 82 per cent of the resisting moment. The wall is heel-dominated.
5. Passive resistance in front of the toe. The toe is buried 1.5 m in Soil 2, so $$K_{p2} = \tan^2\!\left(45^{\circ} + \frac{30^{\circ}}{2}\right) = \tan^2 60^{\circ} = 3.0$$$$P_p = \tfrac{1}{2}K_{p2}\gamma_2 D^2 + 2c'_2\sqrt{K_{p2}}\,D = \tfrac{1}{2}(3.0)(18)(1.5^2) + 2(10)(1.7321)(1.5) = 112.7\ \text{kN/m}$$
6. Factor of safety against sliding. Base friction and adhesion are mobilized against the foundation soil, and the customary practice is to take only two thirds of its strength on a cast-against-soil concrete base: $$FS_{sliding} = \frac{V\tan\!\left(\tfrac{2}{3}\phi'_2\right) + B\left(\tfrac{2}{3}c'_2\right) + P_p}{P_a}$$Substituting $V = 1669.5$ kN/m, $\tan(20^{\circ}) = 0.36397$ and $B = 9.0$ m: $$FS_{sliding} = \frac{607.6 + 60.0 + 112.7}{260.6} = \frac{780.4}{260.6} = \boxed{2.99}$$Neglecting the passive term altogether, as many designers do because the soil in front can be excavated for services, still leaves $FS = 2.56$, comfortably above the usual minimum of 1.5.
7. Check the resultant and the bearing pressure. The line of action of the resultant crosses the base at $$\bar{x} = \frac{\sum M_R - M_O}{V} = \frac{8772 - 870.2}{1669.5} = 4.733\ \text{m from the toe}$$so the eccentricity is $e = \bar{x} - B/2 = 0.233$ m, that is 0.233 m towards the heel. It is well inside the middle third ($B/6 = 1.5$ m), so the whole base stays in compression, with $$\begin{aligned}q_{toe} &= \frac{V}{B}\left(1 - \frac{6e}{B}\right) = 156.7\ \text{kPa} \\ q_{heel} &= \frac{V}{B}\left(1 + \frac{6e}{B}\right) = 214.3\ \text{kPa}\end{aligned}$$Both must be checked against the allowable bearing capacity of Soil 2, which is the third stability check the question does not ask for but a design would require.