Question 4 of 6: Retaining Structures

Question 4 — Retaining Structures (30 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 retaining wall of total height 6.4 m from the underside of the base to the backfill surface, founded 1.4 m below the ground in front of it on medium dense sand.

Find. The active pressure distribution, and the factors of safety against overturning, sliding and bearing failure, then the bearing factor of safety after the water table rises to base level.

[Figure not reproduced: Figure 4.1 — Cantilever wall section (6.4 m overall: 5.0 m above the front ground plus the 1.4 m founding depth) with the Rankine active pressure distribution on the vertical plane through the back of the heel. Dimensions as read from the examination figure. See the official exam paper.]

Approach. Take a vertical plane through the back of the heel and use Rankine active pressure on it, so that the soil wedge standing on the heel counts as part of the resisting weight. Sum vertical forces and moments about the toe for overturning; compare base friction plus passive resistance with the horizontal thrust for sliding; then convert the resultant into an eccentric, inclined load on an effective base width and apply Vesic's bearing-capacity equation with inclination factors.

1. Part (a) — Rankine active coefficient and pressure distribution. The backfill is horizontal and cohesionless, so $$K_a = \tan^{2}\left(45 - \frac{\phi^{\prime}}{2}\right) = \tan^{2}(28^{\circ}) = 0.2827$$ The pressure on the vertical plane through the heel, which runs the full 6.4 m from the backfill surface to the underside of the base, is the sum of a rectangle from the surcharge and a triangle from self-weight: $$\sigma_a(0) = K_aq = 0.2827 \times 10 = 2.83\ \text{kPa}$$ $$\sigma_a(H) = K_a(q + \gamma H) = 0.2827(10 + 18 \times 6.4) = \boxed{35.40\ \text{kPa at the base}}$$ The water table lies 2.0 m below the base, so no water pressure acts on the wall.
2. Part (a) — resultant thrust and its line of action. Integrating the two components separately, $$P_{a1} = K_aqH = 2.827 \times 6.4 = 18.09\ \text{kN/m at } \bar{y} = H/2 = 3.20\ \text{m}$$ $$P_{a2} = \tfrac{1}{2}K_a\gamma H^{2} = 0.5 \times 0.2827 \times 18 \times 6.4^{2} = 104.22\ \text{kN/m at } \bar{y} = H/3 = 2.133\ \text{m}$$ $$P_a = 18.09 + 104.22 = 122.31\ \text{kN/m (horizontal)}$$
3. Passive resistance available in front of the toe. The 1.4 m of sand in front of the wall is above the water table, so with the adopted moist unit weight of 20.0 kN/m³, $$K_p = \tan^{2}\left(45 + \frac{36^{\circ}}{2}\right) = 3.852,\qquad P_p = \tfrac{1}{2}K_p\gamma D_f^{2} = 0.5 \times 3.852 \times 20 \times 1.4^{2} = 75.50\ \text{kN/m}$$ acting 0.467 m above the base. It is excluded from the overturning check (standard conservative practice, since the soil in front can be excavated for services) and reported both ways in the sliding check.
4. Part (b) — vertical forces and their moments about the toe. Taking moments about the front bottom edge of the base slab, with the stem 5.8 m tall above the base:

   Vertical forces per metre run and moments about the toe

   Component	W (kN/m)	Arm x (m)	MR (kN·m/m)
   Stem, rectangular part (0.30 m by 5.8 m)	41.76	1.150	48.02
   Stem, tapered part (0.30 m by 5.8 m triangle)	20.88	1.400	29.23
   Base slab (3.00 m by 0.60 m)	43.20	1.500	64.80
   Backfill on the heel (1.4 m by 5.8 m)	146.16	2.300	336.17
   Backfill over the stem taper	15.66	1.500	23.49
   Surcharge over the heel (1.7 m width)	17.00	2.150	36.55
   Totals	284.66	—	538.26

5. Part (b) — factor of safety against overturning. The overturning moment is the thrust acting at its own lever arms: $$M_O = 18.09 \times 3.20 + 104.22 \times 2.133 = 57.90 + 222.34 = 280.24\ \text{kN}\cdot\text{m/m}$$ $$\boxed{F_{ot} = \frac{M_R}{M_O} = \frac{538.26}{280.24} = 1.92 \ < 2.0}$$ The wall falls short of the customary minimum of 2.0. Counting the 0.8 m of soil standing over the toe (16 kN/m at 0.5 m) lifts it only to 1.95, so the shortfall is real rather than an artefact of conservative bookkeeping.
6. Part (c) — factor of safety against sliding. The base is cast against sand, so base adhesion is zero and the friction angle on the base is taken as $\delta = \tfrac{2}{3}\phi^{\prime}_2 = 24^{\circ}$: $$F_R = \sum V \tan\delta = 284.66 \times \tan 24^{\circ} = 126.74\ \text{kN/m}$$ $$F_{sl} = \frac{F_R + P_p}{P_a} = \frac{126.74 + 75.50}{122.31} = \boxed{1.65}$$ Neglecting the passive resistance entirely gives $F_{sl} = 126.74/122.31 = 1.04$, and the usual compromise of taking half the passive wedge gives 1.34. The wall therefore meets 1.5 only if the full passive wedge in front of the toe can be relied upon; without it the wall is on the point of sliding. A shear key below the base is needed.
7. Part (d) — eccentricity and base pressure distribution. The resultant acts at $$\bar{x} = \frac{M_R - M_O}{\sum V} = \frac{538.26 - 280.24}{284.66} = 0.906\ \text{m from the toe}$$ $$e = \frac{B}{2} - \bar{x} = 1.500 - 0.906 = 0.594\ \text{m} \ > \ \frac{B}{6} = 0.500\ \text{m}$$ so the resultant falls outside the middle third. The linear formula would give $q_{min} = (\sum V/B)(1-6e/B) = -17.75$ kPa, a tension the soil cannot supply; the heel lifts off, and the pressure is triangular over a contact length $$L_c = 3\left(\frac{B}{2} - e\right) = 3(0.906) = 2.72\ \text{m},\qquad q_{max} = \frac{2\sum V}{L_c} = \frac{2 \times 284.66}{2.72} = 209.36\ \text{kPa}$$
8. Part (d) — effective width and load inclination. Meyerhof's effective-area rule replaces the eccentric load on width B by a centric load on $$B^{\prime} = B - 2e = 3.000 - 1.187 = 1.813\ \text{m}$$ and the resultant is inclined to the vertical by $$\psi = \arctan\frac{P_a}{\sum V} = \arctan\frac{122.31}{284.66} = 23.25^{\circ}$$ giving Vesic's inclination factors $$F_{qi} = \left(1-\frac{\psi}{90}\right)^{2} = 0.5500,\qquad F_{\gamma i} = \left(1-\frac{\psi}{\phi^{\prime}}\right)^{2} = \left(1-\frac{23.25}{36}\right)^{2} = 0.1254$$
9. Part (d) — bearing capacity of the sand. With $\phi^{\prime} = 36^{\circ}$: $N_q = 37.75$ and $N_{\gamma} = 2(N_q+1)\tan\phi^{\prime} = 56.31$; the footing is a strip so all shape factors are unity, and with $D_f/B^{\prime} = 0.772$ the depth factor is $F_{qd} = 1 + 2\tan\phi^{\prime}(1-\sin\phi^{\prime})^{2}(D_f/B^{\prime}) = 1.191$. The water table sits 2.0 m below the base, deeper than $B^{\prime} = 1.813$ m, so it lies below the failure zone and the moist unit weight of 20.0 kN/m³ applies in the $N_{\gamma}$ term without correction. Substituting into Vesic's equation with $q = \gamma D_f = 20 \times 1.4 = 28$ kPa and $c^{\prime} = 0$: $$q_u = qN_qF_{qd}F_{qi} + \tfrac{1}{2}\gamma B^{\prime}N_{\gamma}F_{\gamma i} = 692.3 + 128.0 = 820.3\ \text{kPa}$$ $$\boxed{F_{bc} = \frac{q_u}{q_{max}} = \frac{820.3}{209.36} = 3.92\ \ge 3\ \checkmark}$$ Measured against the uniform pressure on the effective width, $\sum V/B^{\prime} = 157.0$ kPa, the factor is 5.22; the value on $q_{max}$ is the conservative one and is the one reported.
10. Part (d), supplementary — check the soft clay 3.0 m below the base. A strong stratum over a weak one is not covered by the equation above, and there is 30 kPa clay only $1.65B^{\prime}$ beneath the footing, so the check is worth making. Spreading the load at 2:1 from an effective width of 1.813 m to the top of the clay gives a stress increase of $284.66/(1.813+3.0) = 59.1$ kPa on a total overburden of 88.0 kPa, against a gross capacity of $30 \times 5.14 \times 1.366 + 88.0 = 298.6$ kPa. The net factor of safety is 3.56 and the gross factor 2.03, so the soft clay is not critical at working load but is close to being the least safe element of the foundation.
11. Part (e) — water table risen to the base of the wall. With the water table at founding level, the soil below the base is fully submerged so $\gamma = \gamma^{\prime} = 10.2$ kN/m³ in the self-weight term, while the soil above the base and therefore the surcharge term are unchanged, as is the backfill thrust (the backfill sits above base level and stays dry). Only the $N_{\gamma}$ term changes: $$q_u = 692.3 + \tfrac{1}{2}(10.2)(1.813)(56.31)(0.1254) = 692.3 + 65.3 = 757.6\ \text{kPa}$$ $$\boxed{F_{bc} = \frac{757.6}{209.36} = 3.62}$$ a reduction of about 8 per cent. The effect is modest because the surcharge term dominates: with a strongly inclined load, $N_q$ contributes 84 per cent of the capacity, and the surcharge itself is unaffected by a water table at base level. Had the water risen through the backfill instead, the active thrust would rise sharply and a wall that is already short in overturning and sliding would fail in those modes long before bearing became an issue, which is why a properly drained backfill and weep holes are essential on a wall like this.
12. Verdict and remedy. Bearing is adequate in both water-table cases (3.92 and 3.62), but the wall as drawn fails the overturning check (1.92 < 2.0), places its resultant outside the middle third, and relies entirely on the passive wedge for sliding. Lengthening the heel from 1.4 m to 1.6 m (base 3.2 m) is the smallest 0.1 m step that restores both overturning (F = 2.21) and the middle-third rule; 1.5 m is not enough. A shear key beneath the base should be added for sliding.