Question 8 of 9: Overturning stability of a gravity retaining wall

Question 8: Overturning stability of a gravity 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. Figure 4 dimensions the base of the wall as a chain of five segments — 0.6, 1.0, 1.2, 1.2 and 2.0 m — with two dashed verticals dropped from the top corners of the stem. Those dashed lines are the key to reading the figure: the segments between a dashed line and the adjacent stem foot are the batter offsets of the two faces, the segment between the dashed lines is the stemʹs top width, and only the outer two segments are the toe and the heel. The base is therefore 6.0 m wide, not the 3.4 m that a reading of “toe + stem + heel” would give.

Find. The factor of safety against overturning about the toe using Coulombʹs theory, followed by the four commentaries (i) to (iv).

[Figure not reproduced: The wall of Figure 4, redrawn to the printed dimension chain. The two dashed verticals dropped from the top corners of the stem identify the middle three segments as batter offset, stem top width and batter offset, so only the outer 0.6 m and 2.0 m are toe and heel. See the official exam paper.]

Approach. Establish the inclination of the back face, compute Coulombʹs active coefficient for that inclination, resolve the thrust into components, and take moments about the toe against the weight of the wall and of the soil standing on the heel.

1. Fix the inclination of the back face. Walking up the back face from its foot, the face rises 10.0 m while moving 1.2 m towards the toe. Measuring from the horizontal ray that runs into the backfill, $$\beta = 90^{\circ} + \tan^{-1}\!\left(\frac{1.2}{10.0}\right) = 90^{\circ} + 6.84^{\circ} = 96.84^{\circ}$$ Because β exceeds 90° the face leans away from the fill, and Coulombʹs coefficient will fall below its vertical-face value — the opposite of the common intuition that any batter increases the thrust.
2. Evaluate Coulombʹs active coefficient. With a horizontal backfill surface (α = 0), $$K_a = \frac{\sin^{2}(\beta + \varphi^{\prime})}{\sin^{2}\beta\,\sin(\beta - \delta)\left[1 + \sqrt{\dfrac{\sin(\varphi^{\prime} + \delta)\sin(\varphi^{\prime} - \alpha)}{\sin(\beta - \delta)\sin(\beta + \alpha)}}\,\right]^{2}}$$ Substituting β = 96.84°, φʹ = 30° and δ = 20.1° gives $$\boxed{K_a = 0.2512}$$ against 0.2973 for an otherwise identical vertical face — a reduction of 16 per cent. The check that the quadrant has been read correctly is that K_a(β) must pass through the vertical-face value at β = 90°.
3. Compute the active thrust and its direction. Taking the pressure surface over the full height of the wall, $$P_a = \tfrac{1}{2}\gamma H^{2}K_a = \tfrac{1}{2}(19)(11.25)^{2}(0.2512) = 302.0\ \text{kN/m}$$ The resultant acts on the face at δ to its normal, which for an inclined face means it is inclined at (β − 90°) + δ below the horizontal — the (β − 90°) term is easy to drop and shifts the vertical component by about 9 per cent: $$\theta = (96.84^{\circ} - 90^{\circ}) + 20.1^{\circ} = 26.94^{\circ}$$ so that P_ah = 302.0 cos26.94° = 269.3 kN/m and P_av = 136.9 kN/m.
4. Assemble the resisting weights. The stem is a trapezoid 3.4 m wide at its foot and 1.2 m at the crest, so its area is ½(3.4 + 1.2)(10.0) = 23.0 m²; the base is 6.0 × 1.25 = 7.50 m²; and the backfill standing over the heel, bounded by the battered back face, is 26.0 m². Using γ_c = 24 kN/m³ for the concrete: $$\begin{aligned} W_{\text{base}} &= 180 \\ W_{\text{stem}} &= 552 \\ W_{\text{soil}} &= 494\ \text{kN/m} \end{aligned}$$ summing to ∑W = 1226 kN/m, acting at 3.000, 2.258 and 4.677 m from the toe respectively.
5. Take moments about the toe. The overturning moment is the horizontal thrust acting at H/3 = 3.75 m above the base underside, and the resisting moment is the sum of the weight moments plus the vertical component of the thrust at its lever arm of 3.70 m: $$\begin{aligned}M_o &= P_{ah}\,\frac{H}{3} = 269.3(3.75) = 1010\ \text{kN}\cdot\text{m/m} \\ M_r &= \sum W_i x_i + P_{av}x_{av} = 4097 + 136.9(3.70) = 4603\ \text{kN}\cdot\text{m/m}\end{aligned}$$ giving $$\boxed{FS_{\text{overturning}} = \frac{M_r}{M_o} = \frac{4603}{1010} = 4.56}$$

(i) Comment on the factor of safety obtained

Part (i) — the value is comfortably adequate, and overturning is not the mode that governs this wall. A factor of safety of 4.56 is well above the 1.5 that CFEM treats as the minimum and above the 2.0 that Das recommends when the active pressure is computed by Coulombʹs method. Three supporting checks make the point concrete. The resultant of all vertical forces is ∑V = 1363 kN/m acting at 2.637 m from the toe, so the eccentricity is e = 0.363 m, comfortably inside B/6 = 1.000 m — the whole base remains in compression and no tension develops at the heel. The corresponding base pressures are q_max = 310 kPa at the toe and q_min = 145 kPa at the heel. Sliding, checked with two-thirds of the founding soilʹs strength on the concrete interface, gives a factor of safety of about 2.0, and bearing, checked through Meyerhofʹs equation with the load-inclination factors that a 11.2° resultant obliquity demands, gives about 4.6. So the section passes every mode, with sliding the least comfortable — which is the usual ranking for a gravity wall and the reason that overturning alone is never a sufficient check.

The honest qualification is that 4.56 is a generous number rather than a tight one, for the reason set out in the callout below: the conventional calculation counts the whole soil column over the heel as resisting weight while simultaneously applying Coulomb pressure to the back face, and those two steps are not consistent with one another. A self-consistent free body gives 2.73. Either way the wall is safe against overturning, and the section could in fact be slimmed — but the design value to quote is the lower one.

(ii) The factor of safety Rankineʹs theory would give

Part (ii) — Rankine would give a noticeably lower factor of safety. No calculation is needed to see why; three effects all push the same way. First, Rankineʹs coefficient for a horizontal backfill, K_a = (1 − sinφʹ)/(1 + sinφʹ) = 1/3, is larger than the 0.2512 obtained above, because Rankine assumes a smooth wall and so discards the wall friction that helps carry the fill. Second, and for the same reason, Rankine gives a thrust that is horizontal, so there is no vertical component to add to the resisting weight — the 137 kN/m of P_av and its contribution to M_r simply disappear. Third, Rankine takes no account of the fact that this back face leans away from the fill, which is precisely the geometric feature that reduced Coulombʹs coefficient here. A larger thrust, applied horizontally, with no vertical relief, must give a smaller ratio of resisting to overturning moment. The reduction is substantial rather than marginal — of the order of 40 per cent — but the wall would still pass, so the conclusion is unchanged and only the margin narrows.

(iii) Which method is more appropriate here

Part (iii) — Coulombʹs method is the appropriate choice for this wall. The reason is that Coulombʹs theory is formulated for exactly the two features this wall possesses and Rankineʹs is not: an inclined back face, and a rough soil–wall interface. Rankineʹs solution is derived for a smooth, vertical boundary and a stress state that is uniform along it; applying it to a face battered at 6.8° with δ = 0.67φʹ violates both assumptions. Coulombʹs wedge analysis, by contrast, admits β, δ and a sloping backfill directly, which is why it is the standard tool for gravity and semi-gravity walls with masonry or mass-concrete back faces. The question itself directs the candidate to Coulomb, and the geometry vindicates that instruction.

Rankine is not therefore useless here — it is the right tool for a different idealisation. For a cantilever wall the usual practice is to apply Rankine on a virtual vertical plane through the heel and to include all the soil inboard of that plane in the free body, and that construction is both simpler and self-consistent. For this wall it is worth carrying out as a check (it is the 2.73 quoted above) precisely because it avoids the double-counting that the conventional Coulomb treatment involves. Two further points of judgement: Coulombʹs solution assumes a planar failure surface and therefore overestimates passive resistance badly, so it should not be used for the passive side without correction; and neither classical theory accounts for the backfillʹs cohesion, which brings us to the next remark.

(iv) Effect of water rising 4 m in the backfill

Part (iv) — the factor of safety decreases. The pressure distribution diagram below shows why, and the mechanism has two parts that pull in opposite directions but do not cancel.

Below the new water table the soil is buoyant, so the effective vertical stress grows with the submerged unit weight γʹ = 19 − 9.81 = 9.19 kN/m³ rather than with 19 kN/m³. The active earth-pressure diagram therefore kinks at the water table and becomes much flatter below it, so the earth component of the thrust falls. Acting against that, the pore water itself now presses on the wall with a full hydrostatic triangle, and this is the term that decides the answer: water pressure is not reduced by any earth-pressure coefficient. Where the soil skeleton transmits only K_aσʹ_v — about a third of the vertical effective stress — the water transmits γ_wh_w in full. Losing roughly a third of 9.81h_w from the earth term while gaining the whole of it as water pressure is a net increase, and a substantial one.

The geometry of the change makes it worse than the resultant alone suggests. The hydrostatic triangle is concentrated in the lowest 4 m of the wall, but its centroid still sits at h_w/3 = 1.33 m above the base, and because the earth pressure that it displaces was distributed over the full height, the combined resultant moves down. A larger thrust acting lower would reduce the overturning moment for a fixed resultant, but here the thrust grows enough that the moment grows too. Meanwhile the resisting side is either unchanged or slightly worse: the submerged soil over the heel weighs less, so M_r falls, and if water reaches beneath the base it generates uplift, which reduces the effective vertical force still further and attacks the sliding and bearing checks at the same time.

So the answer is unambiguous: the factor of safety decreases. The design lesson is the one that follows from it — the drainage provisions behind the wall are not a detail but a structural assumption. The question states that the groundwater table is very deep, and the wall passes on that basis; a blocked weep hole converts the same section into a marginal one. Any wall of this height should carry a granular drainage layer or geocomposite against the stem, weep holes at 1.5 to 3 m centres, a perforated collector discharging to a positive outlet, and a sealed surface behind the crest.