Question 7 of 9: Cantilever retaining wall: overturning and bearing capacity (24 marks)

Question 7 — Cantilever retaining wall: overturning and bearing capacity (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 scaled from Figure 2 of the paper.

Find. The factor of safety against overturning about the toe, and the factor of safety against a bearing-capacity failure of the base, both per metre run of wall.

[Figure not reproduced: Figure Q7. Cantilever wall of Figure 2, redrawn with the free body used in the analysis. The Rankine active thrust is taken on the vertical plane through the heel and acts parallel to the backfill surface; the soil wedge between that plane and the stem is counted as resisting weight. See the official exam paper.]

Approach. Take the Rankine active thrust on the virtual vertical plane through the heel, compute the resisting weights of the wall and of the soil standing on the heel, take moments about the toe for overturning, then convert the resultant into an eccentricity, a maximum base pressure and, through Meyerhof's effective-width formulation with depth and load-inclination factors, an ultimate bearing capacity.

1. Convert the concrete density to a unit weight. $$\gamma_c = \rho_c\,g = \frac{2358 \times 9.81}{1000} = 23.132\ \text{kN/m}^3 .$$
2. Establish the height of the vertical plane through the heel. The backfill rises at 10° from the top of the stem, so over the 2.00 m heel it gains $$H_1 = 2.00\tan 10^\circ = 0.3527\ \text{m},$$ and the full height on which the Rankine thrust acts is $$H' = H_1 + H_2 + H_3 = 0.3527 + 5.00 + 0.60 = 5.9527\ \text{m}.$$
3. Compute the Rankine active coefficient for the inclined backfill. With $\alpha = 10^\circ$ and $\phi'_1 = 30^\circ$, $$K_a = \cos\alpha\, \frac{\cos\alpha - \sqrt{\cos^{2}\alpha - \cos^{2}\phi'_1}} {\cos\alpha + \sqrt{\cos^{2}\alpha - \cos^{2}\phi'_1}} .$$ Here $\cos^{2}10^\circ = 0.96985$ and $\cos^{2}30^\circ = 0.75000$, so the surd is $\sqrt{0.21985} = 0.46888$ and $$K_a = 0.98481\times\frac{0.98481 - 0.46888}{0.98481 + 0.46888} = 0.34952 .$$
4. Compute the active thrust and resolve it. $$P_a = \tfrac{1}{2}\gamma_1 H'^{2} K_a = \tfrac{1}{2}(20)(5.9527)^{2}(0.34952) = 123.85\ \text{kN/m}.$$ The thrust acts parallel to the backfill surface, so $$P_h = P_a\cos 10^\circ = 121.97\ \text{kN/m},\qquad P_v = P_a\sin 10^\circ = 21.51\ \text{kN/m},$$ with $P_h$ acting at $H'/3 = 1.9842$ m above the underside of the base and $P_v$ acting on the vertical plane, that is at $x = B = 3.50$ m from the toe.
5. Assemble the vertical forces and their moments about the toe. The free body is the wall plus the soil standing on the heel, bounded on the right by the virtual vertical plane. Taking the toe C as origin:

   Component	Area (m2)	Unit weight (kN/m3)	W (kN/m)	Arm from C (m)	Moment (kN·m/m)
   1. Stem, rectangular part (0.50 × 5.00)	2.500	23.132	57.83	1.2500	72.29
   2. Stem, battered wedge (½ × 0.25 × 5.00)	0.625	23.132	14.46	0.9167	13.25
   3. Base slab (3.50 × 0.60)	2.100	23.132	48.58	1.7500	85.01
   4. Backfill over the heel, rectangle (2.00 × 5.00)	10.000	20	200.00	2.5000	500.00
   5. Backfill over the heel, sloping wedge (½ × 2.00 × 0.3527)	0.353	20	7.05	2.8333	19.98
   6. Vertical component of the thrust, Pv	—	—	21.51	3.5000	75.27
   Totals			ΣV = 349.42	—	ΣMR = 765.81

6. Take the overturning moment and form the first factor of safety. Only the horizontal component overturns: $$\sum M_O = P_h\frac{H'}{3} = 121.97 \times 1.9842 = 242.01\ \text{kN}\cdot\text{m/m},$$ $$FS_{\text{overturning}} = \frac{\sum M_R}{\sum M_O} = \frac{765.81}{242.01} = \boxed{3.16}.$$ This comfortably exceeds the usual requirement of 2 (Das) to 2.5, so the wall is safe against overturning.
7. Locate the resultant on the base. The line of action of $\sum V$ meets the base at $$\bar{x} = \frac{\sum M_R - \sum M_O}{\sum V} = \frac{765.81 - 242.01}{349.42} = 1.4990\ \text{m},$$ so the eccentricity measured from the centre of the base is $$e = \frac{B}{2} - \bar{x} = 1.7500 - 1.4990 = 0.2510\ \text{m}.$$ Since $e = 0.251$ m is less than $B/6 = 0.583$ m the resultant stays within the middle third and the whole base remains in compression.
8. Compute the base pressures. $$q_{\max,\min} = \frac{\sum V}{B}\left(1 \pm \frac{6e}{B}\right) = \frac{349.42}{3.50}\left(1 \pm \frac{6\times0.2510}{3.50}\right),$$ giving $q_{\max} = 142.79$ kPa at the toe and $q_{\min} = 56.88$ kPa at the heel.
9. Compute the bearing-capacity factors for the foundation soil. With $\phi'_2 = 20^\circ$, $$N_q = e^{\pi\tan\phi'_2}\tan^{2}\!\left(45^\circ + \frac{\phi'_2}{2}\right) = 6.399, \qquad N_\gamma = 2\left(N_q + 1\right)\tan\phi'_2 = 5.386 ,$$ and $N_c = \left(N_q - 1\right)\cot\phi'_2 = 14.83$, which plays no part because $c'_2 = 0$.
10. Apply Meyerhof's effective width and the depth factors. The eccentrically loaded base is replaced by a centrally loaded strip of width $$B' = B - 2e = 3.50 - 2(0.2510) = 2.998\ \text{m},$$ carrying the surcharge $q = \gamma_2 D = 20 \times 2.00 = 40.0$ kPa. The depth factors, computed on $B'$, are $$F_{qd} = 1 + 2\tan\phi'_2\left(1 - \sin\phi'_2\right)^{2}\frac{D}{B'} = 1 + 2(0.36397)(0.43294)\frac{2.00}{2.998} = 1.2102, \qquad F_{\gamma d} = 1 .$$
11. Apply the load-inclination factors. The resultant on the base is inclined to the vertical by $$\psi = \tan^{-1}\!\left(\frac{P_h}{\sum V}\right) = \tan^{-1}\!\left(\frac{121.97}{349.42}\right) = 19.24^\circ ,$$ so $$F_{qi} = \left(1 - \frac{\psi}{90^\circ}\right)^{2} = 0.6181, \qquad F_{\gamma i} = \left(1 - \frac{\psi}{\phi'_2}\right)^{2} = \left(1 - \frac{19.24}{20}\right)^{2} = 0.00144 .$$ The second factor has all but vanished, because the obliquity of the base resultant is within three quarters of a degree of the friction angle of the foundation soil. That is the governing feature of this wall and is discussed below.
12. Assemble the ultimate bearing capacity and the second factor of safety. With $c'_2 = 0$, $$q_u = qN_qF_{qd}F_{qi} + \tfrac{1}{2}\gamma_2 B' N_\gamma F_{\gamma d}F_{\gamma i} .$$ The surcharge term is $40.0 \times 6.399 \times 1.2102 \times 0.6181 = 191.49$ kPa and the width term is $\tfrac{1}{2}(20)(2.998)(5.386)(1)(0.00144) = 0.23$ kPa, so $q_u = 191.72$ kPa and $$FS_{\text{bearing}} = \frac{q_u}{q_{\max}} = \frac{191.72}{142.79} = \boxed{1.34}.$$
13. Interpret the result. The wall is safe against overturning but deficient in bearing: a factor of safety of 1.34 falls far short of the 3.0 normally required on a retaining-wall base. The cause is not the magnitude of the load but its obliquity — a horizontal thrust of 122 kN/m against a vertical load of 349 kN/m tilts the resultant to 19.2°, and the foundation soil offers only $\phi'_2 = 20^\circ$. Remedies, in order of effectiveness, are to found the wall deeper (the surcharge term is the only one left alive, and it scales with $D$), to widen the base and lengthen the heel so that $\sum V$ grows and $e$ falls, to improve or replace the foundation soil, or to add a base key so that part of the horizontal thrust is carried in passive resistance rather than obliquity on the base. Reporting the deficiency is the answer; it is not an arithmetic error to hunt.