Question 8 of 9: Factors of Safety of a Cantilever Retaining Wall against Overturning and Sliding

Question 8: Factors of Safety 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.

Find. (a) the factor of safety against overturning about the toe, (b) the factor of safety against sliding along the base, a comment on both, and the measures needed if they fall short of the usual criteria ($FS_{ot} \ge 2.0$, $FS_{sl} \ge 1.5$).

[Figure not reproduced: Figure 3 (redrawn from the printed dimensions; the original is not to scale). The dashed red line is the Rankine vertical plane through the heel, which the exam figure itself draws — the pressure system is computed on that plane, not on the stem face. See the official exam paper.]

Approach. Because the backfill surface is horizontal and the figure draws a vertical plane through the heel, use Rankine active pressure on that plane over the full 5.0 m height, with water pressure added below the table; take moments about the toe for overturning; and compare base friction $\sum V \tan\delta'$ with the total horizontal thrust for sliding.

1. Establish the two unit weights of the backfill. The question gives $\gamma_d$, $w$ and $G_s$, which is exactly enough to build both. Above the water table the soil is moist, $$\gamma_{moist} = \gamma_d(1+w) = 18.0(1.10) = 19.8\ \text{kN/m}^3 .$$ The void ratio follows from the dry unit weight, $$e = \frac{G_s\gamma_w}{\gamma_d} - 1 = \frac{2.70(9.81)}{18.0} - 1 = 0.4715 ,$$ and hence below the table $$\gamma_{sat} = \frac{(G_s+e)\gamma_w}{1+e} = \frac{(2.70+0.4715)(9.81)}{1.4715} = 21.14\ \text{kN/m}^3, \qquad \gamma' = 21.14 - 9.81 = 11.33\ \text{kN/m}^3 .$$ As a consistency check the degree of saturation above the table is $S = wG_s/e = 0.10(2.70)/0.4715 = 57$ per cent, a plausible field value for a drained granular backfill; had it exceeded 100 per cent the data would have been contradictory.
2. Compute the Rankine active coefficient. With a horizontal backfill surface and a vertical plane of reference, $$K_a = \frac{1-\sin\phi'}{1+\sin\phi'} = \frac{1-\sin 36^{\circ}}{1+\sin 36^{\circ}} = \frac{0.41221}{1.58779} = 0.2596 ,$$ and the companion passive coefficient is $K_p = 1/K_a = 3.852$. Because $c' = 0$ there is no $2c'\sqrt{K_a}$ term and no tension zone.
3. Build the lateral pressure diagram on the vertical plane. Effective vertical stress is carried down through the surcharge, the moist soil and then the submerged soil, and multiplied by $K_a$ at each break: $$\sigma'_{h}(0) = K_a q = 0.2596(20) = 5.19\ \text{kPa},$$ $$\sigma'_{h}(2.5) = K_a(q + \gamma_{moist}z_w) = 0.2596(20+49.5) = 18.04\ \text{kPa},$$ $$\sigma'_{h}(5.0) = K_a(q + \gamma_{moist}z_w + \gamma'(H-z_w)) = 0.2596(69.5+28.34) = 25.40\ \text{kPa}.$$ The water pressure is not multiplied by $K_a$ — water has no shear strength, so it pushes equally in all directions: $u(5.0) = \gamma_w(H-z_w) = 9.81(2.5) = 24.53$ kPa. Notice that the water alone contributes almost as much pressure at the base as the entire submerged soil skeleton, which is why drainage is the first thing to consider later.

4. Resolve the pressure diagram into five thrusts. Each is taken with its own lever arm above the base underside:

   Component	Force (kN/m)	Arm (m)	Moment (kN·m/m)
   Surcharge rectangle, $5.19 \times 5.0$	25.96	2.500	64.90
   Soil triangle, upper 2.5 m	16.06	3.333	53.55
   Soil rectangle carried down, lower 2.5 m	32.13	1.250	40.16
   Buoyant soil triangle, lower 2.5 m	9.20	0.833	7.66
   Pore water triangle, lower 2.5 m	30.66	0.833	25.55
   Totals	114.00		191.82

   Adding the five moments independently gives $64.90+53.55+40.16+7.66+25.55 = 191.82$ kN·m/m, confirming the sum.
5. Assemble the resisting vertical forces. The free body is the concrete plus everything standing on the heel, out to the vertical plane already used for the pressures. Distances are measured from the toe:

   Component	Force (kN/m)	Arm from toe (m)	Moment (kN·m/m)
   Base slab, $4.0(0.35)(23.5)$	32.90	2.000	65.80
   Stem, $0.30(4.65)(23.5)$	32.78	1.850	60.65
   Moist backfill on heel, $2.0(2.5)(19.8)$	99.00	3.000	297.00
   Saturated backfill on heel, $2.0(2.15)(21.14)$	90.92	3.000	272.75
   Surcharge on heel, $20(2.0)$	40.00	3.000	120.00
   Totals	295.60		816.20

   The soil standing on the toe (only 0.15 m deep) and any passive resistance in front of the wall are conservatively omitted; both are quantified in the callout. Note that total unit weights have been used below the water table, so the base uplift that accompanies them must be treated explicitly — also in the callout — and the two must never be mixed.
6. Part (a) — factor of safety against overturning. Taking moments about the toe, $$FS_{overturning} = \frac{\sum M_R}{\sum M_O} = \frac{816.2}{191.8} \qquad \Rightarrow \qquad \boxed{\,FS_{ot} = 4.26\,}$$ which comfortably exceeds the customary requirement of 2.0 (CFEM; Das §8.4).
7. Part (b) — factor of safety against sliding. With a cohesionless foundation soil the only resistance on the base is friction, $$FS_{sliding} = \frac{\sum V\tan\delta'}{P_h} = \frac{295.60\tan 27^{\circ}}{114.00} = \frac{295.60(0.5095)}{114.00} = \frac{150.62}{114.00} \qquad \Rightarrow \qquad \boxed{\,FS_{sl} = 1.32\,}$$ against a requirement of 1.5. The wall is inadequate in sliding.
8. Locate the resultant and check the base pressures. The line of action of the resultant crosses the base at $$\bar{x} = \frac{\sum M_R - \sum M_O}{\sum V} = \frac{816.2-191.8}{295.6} = 2.112\ \text{m from the toe},$$ so the eccentricity is $e = 2.112 - 2.000 = 0.112$ m toward the heel, well inside the middle third ($B/6 = 0.667$ m). The contact pressures are therefore $$q = \frac{\sum V}{B}\left(1 \pm \frac{6e}{B}\right) = 73.9(1 \pm 0.168) \ \Rightarrow\ q_{heel} = 86.3\ \text{kPa},\quad q_{toe} = 61.5\ \text{kPa}.$$ The whole base stays in compression and the pressures are modest, so bearing is not the problem here. It is worth noting that the maximum falls under the heel rather than the toe: with a 1.7 m toe and only a 2.0 m heel the wall is carrying an unusually large amount of concrete out in front of the stem.

Comment on the factors of safety, and the measures recommended

Comment. The wall is stable against overturning by a wide margin and marginal against sliding, which is the normal signature of a cantilever wall with a generous base width and a water table in the backfill. The reason is visible in the pressure diagram: the pore-water triangle alone supplies 30.7 kN/m, 27 per cent of the total horizontal thrust, while adding nothing whatever to the vertical force that generates friction. Every kilopascal of water pressure is pure demand with no matching supply. The eccentricity check confirms that the base is fully in compression at modest pressure, so neither bearing capacity nor tension at the heel is critical; the deficiency is specifically in horizontal equilibrium.

Measure 1 — drain the backfill (the primary recommendation). If the granular fill behind the wall is drained by a vertical or inclined drainage blanket, a geocomposite drain against the stem, a perforated collector at the base of the heel and weep holes at 1.5 to 3 m centres, the water table is prevented from rising and both the pore-water triangle and the buoyancy of the fill disappear. Recomputing with a moist backfill throughout gives $P_h = 90.2$ kN/m, $\sum V = 289.8$ kN/m, and $$FS_{ot} = 4.64, \qquad FS_{sl} = \frac{289.8(0.5095)}{90.2} = 1.64 ,$$ which satisfies both criteria. Drainage is by far the cheapest of the available remedies and should be provided in any case; the drain must be filter-compatible with the retained fill and its outlet protected against freezing, which in a British Columbia interior climate means daylighting the collector below frost depth or providing a heated outfall.

Measure 2 — add a shear key. If the design must tolerate a rise of the water table — a prudent assumption if the drain could ever clog — a key cast monolithically beneath the base moves the sliding surface off the concrete interface and into the foundation soil, where the full $\phi'$ rather than $0.75\phi'$ is mobilised, and mobilises passive resistance in front of the key. Requiring $FS_{sl} = 1.5$ with the water thrust retained needs an extra $1.5(114.0) - 150.6 = 20.4$ kN/m of resistance; solving $\tfrac12 K_p\gamma_{moist}D^2 = 20.4$ gives $D = 0.73$ m measured from the front ground surface, that is a key extending only 0.23 m below the base underside. If the more severe case with full base uplift is adopted, the requirement rises to 45.4 kN/m and $D = 1.09$ m, a key 0.59 m deep. A key of 0.6 m depth by 0.4 m width placed beneath the stem would cover both cases with a wide margin.

Measure 3 — other options, in order of cost. Roughening or castellating the underside of the base, or casting against undisturbed soil rather than a blinding layer, raises the mobilised $\delta'$ toward $\phi'$; taking $\delta' = \phi' = 36^{\circ}$ instead of $27^{\circ}$ would give $FS_{sl} = 1.88$ on its own. Extending the heel would add both weight and friction, though it also lengthens the pressure plane. Battering the base a few degrees into the slope resolves part of the thrust into the base. Relieving the surcharge, or moving it back from the wall, removes 25.96 kN/m of thrust directly. A tie-back or a series of ground anchors is available but is disproportionate for a 5 m wall.

Recommendation. Provide a properly filtered drainage system behind the wall and a 0.6 m deep shear key. Drainage alone brings $FS_{sl}$ to 1.64 in the design condition; the key ensures the wall remains safe if the drain ever ceases to function, which is the failure mode that actually brings retaining walls down.