16-Civ-B3 Geotechnical Design · December 2016

Question 9 of 9: Overturning and sliding stability of a cantilever retaining wall

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada National Examinations, December 2016 — 98-Civ-B3 Geotechnical Design. Three hours, OPEN BOOK, any non-communicating calculator. Section A carries five discussion questions of 7 marks each (answer any four); Section B carries four design questions of 24 marks each (answer any three); the examinable total is 4 × 7 + 3 × 24 = 100 marks. All nine questions are worked below, because the set is a study resource rather than a timed attempt.

Reference texts (98-Civ-B3 / 16-Civ-B3 Geotechnical Design).

B. M. Das, Principles of Foundation Engineering, 9th ed. — subsurface exploration (Ch. 2), bearing capacity of shallow foundations (Ch. 3), settlement of shallow foundations including Schmertmann's strain-influence method (Ch. 5), retaining walls (Ch. 8), pile foundations (Ch. 11), drilled shafts (Ch. 12).
B. M. Das, Principles of Geotechnical Engineering, 9th ed. — shear strength (Ch. 12), lateral earth pressure (Ch. 13), slope stability (Ch. 15).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for site investigation, in-situ testing, bearing resistance, deep foundations and earth-retaining structures.
R. F. Craig, Craig's Soil Mechanics, 9th ed. — effective stress, undrained versus drained behaviour, earth pressure, and the total-stress circular-arc slope analysis.
M. J. Tomlinson & J. Woodward, Pile Design and Construction Practice, 6th ed. — shaft adhesion in clay and under-reamed (belled) bored piles.
P. K. Robertson & K. L. Cabal, Guide to Cone Penetration Testing, 6th ed. — CPT interpretation and CPT-versus-SPT selection.

Sources of design charts and assumed values (page-1 Note 6). Note 6 requires the candidate to identify the source of every design chart and of every value assumed where the paper supplies none. They are named at the point of use and collected here:

Adhesion factor α = 0.55 for a drilled shaft in clay (Q6) — O'Neill and Reese (1999), reproduced in Das, Principles of Foundation Engineering, 9th ed., Section 12.9; the exclusion of the top 1.5 m and of one shaft diameter above the bell comes from the same source.
Bearing-capacity factor N_c^* = 9 (Q6) — Skempton (1951), as tabulated in Das, Section 11.11.
Overburden correction C_N = √(p_a/σ'_o) (Q7) — Liao and Whitman (1986), Das Principles of Geotechnical Engineering, Section 17.6.
SPT-to-friction-angle correlation (Q7) — Peck, Hanson and Thornburn (1974) as fitted by Wolff (1989); cross-checked against Kulhawy and Mayne (1990). Both are tabulated in Das, Principles of Foundation Engineering, 9th ed., Section 2.9.
Bearing-capacity factors and shape/depth factors (Q7) — Vesic (1973) and De Beer (1970), Das Sections 3.6 and 3.7.
Strain-influence diagram and the C₁, C₂ factors (Q7) — Schmertmann, Hartman and Brown (1978), Das Section 5.6; the modulus correlation E_s = 500(N₆₀ + 15) kPa is Bowles (1996), reproduced in the same section.
Rankine active coefficient for an inclined backfill (Q9) — Das, Principles of Geotechnical Engineering, 9th ed., Eq. (13.35); the base friction and adhesion reductions k₁ = k₂ = 2/3 are Das Section 8.4.
Unit weight of the submerged backfill (Q9) — assumed equal to the printed moist unit weight, 18 kN/m³, in the absence of a saturated value; the consequence of that assumption is bounded in the Q9 callout.

Section A

Question 9: Overturning and sliding stability of a cantilever 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.

Quantity	Symbol	Value
Base width / thickness	B / t	4.8 m / 0.9 m
Toe / stem at base / heel	—	0.76 m / 1.04 m / 3.0 m
Stem height / top thickness	—	6.1 m / 0.4 m
Overall height, base underside to top of stem	H	7.0 m
Backfill slope / surcharge	α / q	8° / 20 kPa
Backfill sand	γ, φ', c'	18 kN/m³, 30°, 0
Water table above top of base	—	2.2 m (3.1 m above base underside)
Concrete	γ_c	23.5 kN/m³
Founding soil	γ₂, φ'₂, c'₂	19 kN/m³, 15°, 15 kPa
Depth of ground in front of the toe	D	1.0 m

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

[Figure not reproduced: Figure 9.1 — The wall of Figure 3, redrawn to scale. The red dashed line at the back of the heel is the Rankine virtual back plane on which the active thrust is computed; the free body taken for stability is everything to the left of it, including the wedge of backfill standing on the heel. See the official exam paper.]

Approach. Use the Rankine method on a virtual vertical plane rising from the back of the heel. The free body is the concrete plus all the soil and surcharge standing on the heel; the earth thrust acts on that vertical plane, parallel to the sloping ground surface. Water enters three times — through the submerged unit weight below the water table, through a hydrostatic thrust on the plane, and through uplift on the base — and all three must be carried consistently. Moments are taken about the front bottom edge of the base (the toe).

Establish the height of the virtual back plane. The ground surface starts at the top of the stem, 7.0 m above the base underside, and rises at 8° across the 3.0 m heel, so at the back of the heel $$H' = 7.0 + 3.0\tan 8^\circ = 7.0 + 0.4216 = 7.4216\ \text{m}$$
Compute the Rankine active coefficient for the sloping backfill. With α = 8° and φ' = 30°, $$K_a = \cos\alpha\, \frac{\cos\alpha - \sqrt{\cos^2\alpha - \cos^2\phi'}} {\cos\alpha + \sqrt{\cos^2\alpha - \cos^2\phi'}} = 0.99027\,\frac{0.99027 - 0.48024}{0.99027 + 0.48024} = 0.3435$$ The resultant thrust on the vertical plane acts parallel to the ground surface, that is inclined at 8° to the horizontal.
Build the effective lateral pressure diagram. The water table lies 3.1 m above the base underside, so it is 7.4216 − 3.1 = 4.3216 m below the surface. Above it the sand has γ = 18 kN/m³; below it the submerged value γ' = 18 − 9.81 = 8.19 kN/m³ applies. Effective vertical stresses on the plane are therefore $$\sigma'_v(0) = q = 20\ \text{kPa},\quad \sigma'_v(\text{WT}) = 20 + 18(4.3216) = 97.79\ \text{kPa},$$ $$\sigma'_v(\text{base}) = 97.79 + 8.19(3.1) = 123.18\ \text{kPa}$$ and multiplying by K_a gives lateral pressures of 6.87, 33.59 and 42.31 kPa at those three levels.
Integrate the earth thrust and locate it. Splitting the diagram into the rectangle and triangle above the water table and the rectangle and triangle below it, $$P_1 = 6.87(4.3216) = 29.69,\quad P_2 = \tfrac12(33.59-6.87)(4.3216) = 57.73,$$ $$P_3 = 33.59(3.1) = 104.12,\quad P_4 = \tfrac12(42.31-33.59)(3.1) = 13.52\ \text{kN/m}$$ $$P_a = 205.05\ \text{kN/m}\quad\text{acting at}\quad \bar{y} = 2.895\ \text{m above the base underside}$$ Resolving at 8° to the horizontal, $$P_{ah} = 205.05\cos 8^\circ = 203.06\ \text{kN/m},\qquad P_{av} = 205.05\sin 8^\circ = 28.54\ \text{kN/m}$$ The vertical component acts downwards on the virtual plane at the back of the heel, so it is a stabilising force with the largest lever arm of all.
Add the water thrust. The pore pressure on the virtual plane is hydrostatic below the water table, reaching u = 9.81(3.1) = 30.41 kPa at the base underside, so $$P_w = \tfrac12 \gamma_w h_w^2 = \tfrac12 (9.81)(3.1)^2 = 47.14\ \text{kN/m} \quad\text{at } 1.033\ \text{m above the base}$$ This acts horizontally and is purely destabilising. Its presence is the single most important feature of this wall.

Figure 9.2 — Lateral pressure on the virtual back plane. Blue to the right of the plane: effective active earth pressure, kinked where the submerged unit weight takes over. Shaded to the left: the hydrostatic water pressure, which adds 47.1 kN/m of horizontal thrust that the drained wall would not carry.

Assemble the vertical forces and their moments about the toe. Total (saturated) unit weights are used for the soil, so uplift must be added explicitly later. This step and the next two constitute the overturning check.

Component	Force (kN/m)	Arm from toe (m)	Moment (kN·m/m)
Base slab, 4.8 × 0.9 × 23.5	101.52	2.400	243.65
Stem (trapezoid 0.4/1.04 × 6.1) × 23.5	103.21	1.416	146.18
Backfill on heel, below the water table	118.80	3.300	392.04
Backfill on heel, above the water table	210.60	3.300	694.98
Sloping wedge above the level of the stem top	11.38	3.800	43.26
Surcharge over the heel, 20 × 3.0	60.00	3.300	198.00
Vertical component of the active thrust	28.54	4.800	136.99
Totals	634.05	—	1855.1

The soil standing on the toe (about 1.4 kN/m) is neglected, which is conservative for overturning.

Compute the overturning moment and the uplift. The two horizontal thrusts give $$M_O = P_{ah}\bar{y} + P_w(1.033) = 203.06(2.895) + 47.14(1.033) = 587.9 + 48.7 = 636.6\ \text{kN}\cdot\text{m/m}$$ Because the backfill is ponded to 3.1 m and the ground in front is only 1.0 m deep and dry, water seeps beneath the base. Taking the standard linear approximation — full pressure 30.41 kPa at the heel falling to zero at the toe — $$U = \tfrac12 (30.41)(4.8) = 72.99\ \text{kN/m}\ \text{acting at }\tfrac23(4.8) = 3.20\ \text{m from the toe}$$ $$M_U = 72.99 \times 3.20 = 233.6\ \text{kN}\cdot\text{m/m}$$ which is a destabilising moment because it lifts the heel end of the base.
Factor of safety against overturning. $$FS_{OT} = \frac{M_R}{M_O + M_U} = \frac{1855.1}{636.6 + 233.6} = \frac{1855.1}{870.2}$$ $$\boxed{FS_{OT} = 2.13}$$ If the base is assumed to be pressure-relieved so that no uplift acts, the same numbers give FS_OT = 1855.1 / 636.6 = 2.91. Either way the wall is adequate against overturning, comfortably above the usual requirement of 1.5 to 2.0.
Assemble the sliding resistance. The sliding check starts here. The net vertical force on the base is reduced by the uplift: $$\sum V_{\text{net}} = 634.05 - 72.99 = 561.07\ \text{kN/m}$$ Following Das Section 8.4, the base friction and adhesion are taken as fractions of the founding soil's strength, $k_1 = k_2 = \tfrac23$, giving $\delta' = \tfrac23(15^\circ) = 10^\circ$ and $c_a = \tfrac23(15) = 10$ kPa: $$F_R = \sum V_{\text{net}}\tan\delta' + B c_a = 561.07(0.17633) + 4.8(10) = 98.9 + 48.0 = 146.9\ \text{kN/m}$$
Evaluate the passive resistance in front of the toe. Over the 1.0 m of founding soil standing in front of the wall, with $K_p = \tan^2(45^\circ + 15^\circ/2) = 1.698$, $$P_p = \tfrac12 K_p \gamma_2 D^2 + 2c'_2\sqrt{K_p}\,D = \tfrac12(1.698)(19)(1.0)^2 + 2(15)(1.303)(1.0) = 16.1 + 39.1 = 55.2\ \text{kN/m}$$ It is standard practice to neglect this term, because the 1.0 m of soil in front of a wall is the material most likely to be removed by a service trench, by scour or by future regrading, and because mobilising it needs far more movement than the active side does (Question 3).
Factor of safety against sliding. The driving force is the sum of the two horizontal thrusts, $P_{ah} + P_w = 203.06 + 47.14 = 250.20$ kN/m, so $$FS_{SL} = \frac{F_R}{P_{ah}+P_w} = \frac{146.9}{250.2}$$ $$\boxed{FS_{SL} = 0.59\quad(\text{0.81 if the passive wedge is counted})}$$ Even with no uplift and the undiscounted base strength (k = 1), with the passive wedge still neglected, the ratio only reaches 0.97; adding the full passive wedge as well takes it to 1.19, still well short of 1.5. The wall as drawn does not satisfy sliding; it is at or beyond the point of failure.
Check the base pressure, and state the remedy. The line of action of the net vertical force is at $\bar{x} = (1855.1 - 636.6 - 233.6)/561.07 = 1.755$ m from the toe, so the eccentricity is $e = 4.8/2 - 1.755 = 0.645$ m, less than B/6 = 0.8 m; the base stays in compression, with $q_{\max} = 211$ kPa and $q_{\min} = 22.7$ kPa. Bearing is therefore not the problem — sliding is, and the diagnosis is unambiguous. Draining the backfill removes both the water thrust and the uplift and lifts the sliding ratio only to 0.73, so drainage alone is not enough; the founding soil, at φ' = 15°, is simply too weak in friction. The effective remedy is a shear key beneath the base, which forces the failure surface into undisturbed soil so that the full φ'₂ and c'₂ may be used, and mobilises passive resistance over the full key depth. Reaching FS_SL = 1.5 requires 375 kN/m of resistance against the 222 kN/m available on the full-strength base, so the key must generate a further 153 kN/m, which corresponds to a key extending to about 2.1 m below the front ground surface, that is roughly 1.1 m below the underside of the base. A weep-holed granular drainage blanket behind the stem should be provided in any case.

Quantity	Value
Height of the virtual back plane, H'	7.4216 m
Rankine active coefficient (α = 8°, φ' = 30°)	K_a = 0.3435
Active thrust P_a (at 2.895 m, inclined 8°)	205.05 kN/m
Horizontal / vertical components	203.06 / 28.54 kN/m
Hydrostatic thrust P_w	47.14 kN/m at 1.033 m
Total vertical force before uplift, ΣV	634.05 kN/m
Resisting moment about the toe, M_R	1855.1 kN·m/m
Overturning moment (thrusts / uplift)	636.6 / 233.6 kN·m/m
Factor of safety, overturning (with uplift)	2.13
Factor of safety, overturning (no uplift)	2.91
Base sliding resistance F_R	146.9 kN/m
Passive resistance available at the toe, P_p	55.2 kN/m (neglected)
Factor of safety, sliding (uplift, P_p neglected)	0.59 — UNSAFE
Factor of safety, sliding (P_p included)	0.81
Eccentricity / maximum base pressure	0.645 m (< B/6) / 211 kPa
Shear key required for FS_SL = 1.5	about 1.1 m below the base

Check: the three modelling choices that move these numbers, and by how much.

Uplift beneath the base. Figure 3 shows ponded water 3.1 m up the back of the wall and dry ground 1.0 m deep in front, so seepage beneath the base is real and a linear uplift from 30.41 kPa at the heel to zero at the toe is the standard simple treatment. Omitting it entirely gives FS_OT = 2.91 and FS_SL = 0.64. The conclusion is unchanged either way: safe in overturning, unsafe in sliding. Both are reported so the reader can see the span.
Unit weight of the submerged backfill. The figure prints one value, γ = 18 kN/m³, and it has been used both above and below the water table. If instead the sand were taken as dry at 18 and saturated at about 21.0 kN/m³ (the value implied by G_s = 2.65 with that dry weight), γ' would rise from 8.19 to 11.2 kN/m³: the thrust would grow by about 5 kN/m and the resisting weight of the heel backfill by about 20 kN/m, moving FS_SL from 0.587 to 0.591. The assumption is immaterial.
Reading of the toe dimension. Figure 3 is not drawn to scale. Its drawn proportions give a toe of about 0.76 m and a stem 1.04 m thick at its base (the reading used above, with toe + stem = 1.8 m), but the arrowed dimension carrying the “1.04 m” label spans the toe projection, which read literally gives a 1.04 m toe and a 0.76 m stem. On that reading the stem weighs 83.1 kN/m instead of 103.2, and FS_OT = 2.11 (2.88 without uplift) and FS_SL = 0.57 (0.79 with the passive wedge). The conclusion is unchanged: adequate in overturning, failing in sliding.
Rankine on the virtual plane versus Coulomb on the stem face. The virtual vertical plane through the heel is the correct free body for a cantilever wall with a projecting heel, and it is the treatment used here. A Coulomb analysis applied to the battered back face of the stem would not be self-consistent with counting the whole soil column over the heel as resisting weight, because that soil lies on the far side of the pressure plane.
Full-strength base for the shear-key sizing. Sizing the key uses φ'₂ = 15° and c'₂ = 15 kPa undiscounted, which is the point of a key: the failure surface no longer lies at a concrete-to-soil interface. If the two-thirds reduction were retained the required key would be substantially deeper.
What was not checked. The question asks only for overturning and sliding. Bearing capacity of the founding soil, overall (deep-seated) slope stability through the weak φ' = 15° stratum, and structural design of the stem and base are all outside the scope but would all be required before construction. Given the sliding result, a deep-seated stability check would be the next thing this designer would run.

Topic. External stability of a cantilever retaining wall with an inclined, surcharged and partly submerged granular backfill.

Key relations. $K_a = \cos\alpha\dfrac{\cos\alpha-\sqrt{\cos^2\alpha-\cos^2\phi'}} {\cos\alpha+\sqrt{\cos^2\alpha-\cos^2\phi'}}$; $FS_{OT} = \dfrac{\sum M_R}{\sum M_O}$; $FS_{SL} = \dfrac{\sum V\tan(k_1\phi'_2) + Bk_2c'_2 + P_p}{P_{ah}+P_w}$; $\gamma' = \gamma_{sat}-\gamma_w$; $e = \dfrac{B}{2}-\dfrac{\sum M_R - \sum M_O}{\sum V}$; $q_{\max,\min} = \dfrac{\sum V}{B}\left(1 \pm \dfrac{6e}{B}\right)$.

Why this works. A cantilever wall is not a gravity wall: most of the weight holding it down is soil sitting on the heel, not concrete. Taking the free body at a vertical plane through the back of the heel is what allows that soil to be counted as part of the wall while the Rankine solution — which is exact for a vertical plane in a semi-infinite mass with a plane surface — supplies the thrust on that plane. Water then has to be handled once and only once: use total weights and add an explicit uplift force, or use submerged weights and omit uplift, but never mix the two.

Common pitfall. Forgetting the hydrostatic thrust altogether. On this wall the water adds 47 kN/m of horizontal force, which is 23 per cent of the earth thrust, and the associated uplift removes 73 kN/m of vertical force; between them they turn a sliding ratio of 0.73 into 0.59. A second pitfall is to include the passive wedge without saying so, which here changes a clearly failing wall (0.59) into one that merely looks marginal (0.81). A third is to take moments about the heel or the base centre instead of the toe.

Source. Das, Principles of Foundation Engineering, 9th ed., Sections 8.3 to 8.6 (proportioning, and the checks for overturning, sliding and bearing); Das, Principles of Geotechnical Engineering, 9th ed., Eq. (13.35) for the inclined backfill; CFEM, 4th ed., Ch. 26 and Ch. 27.

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