16-Civ-B3 Geotechnical Design · December 2017

Question 6 of 9: Stability of a cantilever retaining wall

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

Notes on this paper

Paper format. National Examinations, December 2017 — 16-Civ-B3 Geotechnical Design; three hours, open book, any non-communicating calculator. Section A holds five discussion questions worth 7 marks each of which four are marked; Section B holds four design questions worth 24 marks each of which three are marked, so 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 marked script.

Reference texts. B. M. Das, Principles of Foundation Engineering, 8th–9th ed. (Cengage); B. M. Das, Principles of Geotechnical Engineering, 9th ed.; R. F. Craig / J. Knappett, Craig's Soil Mechanics, 9th ed.; Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed.; D. P. Coduto, Foundation Design: Principles and Practices; J. E. Bowles, Foundation Analysis and Design; ASTM D1586 (SPT), D5778 (CPTu), D2573 (field vane).

Source of design charts and assumed values (page 1, Note 6). Rankine active and passive coefficients from Das, Principles of Foundation Engineering, Ch. 7 (Eqs. 7.11 and 7.30); cantilever-wall stability procedure from Das Ch. 8 (Eqs. 8.11–8.14). Drilled-shaft adhesion factor alpha* = 0.55 and the 1.5 m surface exclusion from Reese & O'Neill (1989), tabulated in Das Ch. 12; bearing factor Nc* = 9 from Skempton (1951); block-failure check from Das Ch. 11 (Eq. 11.55). Compression index from Skempton's Cc = 0.009(LL − 10). Strain-influence factors and the C1, C2 corrections from Schmertmann, Hartman & Brown (1978) as presented in Das Ch. 5; Es = 500(N60 + 15) kPa from Das Table 5.7 (Bowles). Friction angle from the SPT via Wolff (1989), phi' = 27.1 + 0.3(N1)60 − 0.00054[(N1)60]², with the Liao & Whitman (1986) overburden correction; Vesic bearing-capacity, shape and depth factors from Das Tables 4.2 and 4.3. Every assumed value (unit weights of concrete, specific gravity for the void ratio, pile spacing, factors of safety) is stated in a callout beside the step that uses it.

Question 6: 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. The cantilever wall of Figure 1, read off the drawing as follows.

Item	Symbol	Value
Overall height, top of stem to underside of base	H	6.0 m
Base slab thickness	t	0.5 m → stem height 5.5 m
Base width (toe 1.0 + stem 0.5 + heel 2.0)	B	3.5 m
Depth of soil in front of the wall, to underside of base	D	1.5 m
Backfill unit weight and strength	gamma₁, c', phi'₁	18 kN/m³, 0, 30°
Founding soil unit weight and strength	gamma₂, c', phi'₂	20 kN/m³, 0, 35°
Base friction angle (soil against concrete)	delta	25°
Concrete unit weight	gamma_c	24 kN/m³

The backfill surface is horizontal and level with the top of the stem, there is no surcharge, and no water table is shown, so the backfill is taken as drained.

Find. The factor of safety against sliding along the base and the factor of safety against overturning about the toe, using Rankine earth pressures.

[Figure not reproduced: Figure 1 (redrawn). Cantilever wall geometry, soil zones and the Rankine pressures acting on the virtual vertical plane C–B through the heel. The label gamma = 24 kN/m 3 on the exam figure has a leader to the stem and is the concrete, not a soil stratum — it carries no c' or phi', which . See the official exam paper.]

Approach. Take Rankine active pressure on the virtual vertical plane C–B through the heel, so the soil wedge between the stem and that plane is counted as part of the free body; sum the vertical forces and their moments about the toe A; then compare the resisting sliding force (base friction plus the passive wedge in front of the toe) and the resisting moment against the horizontal thrust and its overturning moment.

Earth-pressure coefficients. Rankine's coefficients for a horizontal ground surface and a vertical plane, with the active value governed by the backfill and the passive value by the founding soil in front of the wall: $$K_{a1}=\frac{1-\sin\phi'_1}{1+\sin\phi'_1}=\frac{1-\sin 30^{\circ}}{1+\sin 30^{\circ}}=0.3333, \qquad K_{p2}=\frac{1+\sin\phi'_2}{1-\sin\phi'_2}=\frac{1+\sin 35^{\circ}}{1-\sin 35^{\circ}}=3.690$$ Using the vertical plane through the heel is what makes Rankine legitimate here: the plane is vertical, the soil above the heel moves with the wall, and the thrust on it is horizontal because the backfill surface is level.
Active thrust on the plane C–B. The active pressure grows linearly from zero at C to $K_{a1}\gamma_1 H = 0.3333(18)(6.0)=36.0$ kPa at B, so $$P_a=\tfrac{1}{2}K_{a1}\gamma_1 H^{2}=\tfrac{1}{2}(0.3333)(18)(6.0)^{2}=\boxed{108.0\ \text{kN/m}}$$ acting horizontally at $H/3 = 2.0$ m above the underside of the base.
Overturning moment about the toe. With the thrust horizontal, the lever arm is simply its height above the base: $$M_o=P_a\left(\frac{H}{3}\right)=108.0(2.0)=216.0\ \text{kN}\!\cdot\!\text{m/m}$$
Vertical forces and their moments about the toe A. The free body is the concrete plus the soil standing on the heel and on the toe. Each area is multiplied by its unit weight and by the horizontal distance of its centroid from A.

Component	Calculation	W (kN/m)	Arm from A (m)	Moment (kN·m/m)
Stem	0.5 × 5.5 × 24	66.0	1.25	82.5
Base slab	3.5 × 0.5 × 24	42.0	1.75	73.5
Backfill on the heel	2.0 × 5.5 × 18	198.0	2.50	495.0
Soil on the toe	1.0 × 1.0 × 20	20.0	0.50	10.0
Totals		326.0		661.0

The soil on the toe is only 1.0 m thick, not 1.5 m: the dimension D is measured from the front ground surface to the underside of the base, and the top 0.5 m of that depth is occupied by the base slab itself.

Factor of safety against overturning. Dividing the restoring moment by the overturning moment, $$FS_{(overturning)}=\frac{\sum M_R}{\sum M_o}=\frac{661.0}{216.0}=\boxed{3.06}$$ which is comfortably above the usual requirement of 1.5 to 2.0. Omitting the soil on the toe, as many designers do, changes this only to 3.01, so the conclusion does not depend on that judgement.
Sliding resistance from base friction. With $c'=0$ in the founding soil, the only interface resistance is friction, and the drawing gives the wall friction angle directly rather than requiring the usual $\delta = \tfrac{2}{3}\phi'_2$ assumption: $$F_{R(friction)}=\left(\sum V\right)\tan\delta = 326.0\tan 25^{\circ}=152.0\ \text{kN/m}$$
Passive resistance in front of the toe. The founding soil stands 1.5 m deep against the front face of the base and the buried part of the stem, giving a passive pressure of $K_{p2}\gamma_2 D = 3.690(20)(1.5)=110.7$ kPa at the base and $$P_p=\tfrac{1}{2}K_{p2}\gamma_2 D^{2}=\tfrac{1}{2}(3.690)(20)(1.5)^{2}=83.0\ \text{kN/m}$$ acting at $D/3 = 0.5$ m above the underside of the base.
Factor of safety against sliding. Collecting the two resistances against the active thrust, $$FS_{(sliding)}=\frac{F_{R(friction)}+P_p}{P_a}=\frac{152.0+83.0}{108.0}=\boxed{2.18}$$ If the passive wedge is discounted entirely — the prudent assumption whenever the soil in front may be excavated for services, eroded or loosened — the same calculation gives $FS = 152.0/108.0 = 1.41$, which is below the usual requirement of 1.5.
Check the base pressure distribution, since the answer to (a) depends on it. The resultant of the vertical forces acts at $\bar{x}=(\sum M_R-\sum M_o)/\sum V=(661.0-216.0)/326.0=1.365$ m from the toe, so the eccentricity is $e=B/2-\bar{x}=1.75-1.365=0.385$ m, comfortably inside the middle third ($B/6=0.583$ m). The whole base therefore stays in compression, $$q_{max,min}=\frac{\sum V}{B}\left(1\pm\frac{6e}{B}\right)=\frac{326.0}{3.5}\left(1\pm\frac{6(0.385)}{3.5}\right)=155\ \text{and}\ 32\ \text{kPa}$$ which confirms that the full base width is available to mobilise friction, as step 6 assumed.

Quantity	Value	Requirement	Verdict
Active thrust P_a	108.0 kN/m at 2.0 m above the base	—	—
Total vertical force ΣV	326.0 kN/m	—	—
Restoring / overturning moment	661.0 / 216.0 kN·m/m	—	—
(b) FS against overturning	3.06	≥ 1.5 to 2.0	satisfactory
Base friction / passive resistance	152.0 / 83.0 kN/m	—	—
(a) FS against sliding, with P_p	2.18	≥ 1.5	satisfactory
(a) FS against sliding, ignoring P_p	1.41	≥ 1.5	marginal — governs the design
Eccentricity e / q_max	0.385 m (< B/6) / 155 kPa	base in full compression	satisfactory

Check — assumptions and the value that actually governs. (1) The label gamma = 24 kN/m³ in Figure 1 has a leader line into the stem and is quoted without c' or phi', so it is the reinforced concrete, not a soil stratum; every soil label on the drawing carries strength parameters beside it. Reading it as a buried soil would change both factors of safety. (2) The soil in front of the wall is taken as the founding stratum (20 kN/m³, phi' = 35°), which is what the figure's lower annotation describes; if the front soil were instead the 18 kN/m³, phi' = 30° material, K_p would fall to 3.00 and P_p to 60.8 kN/m, giving FS_(sliding) = 1.97 — still above 1.5, so the conclusion is unchanged. (3) No water table is shown; if the backfill were to become saturated to the top of the stem the thrust would rise to 0.3333(18 − 9.81)(6)²/2 + 9.81(6)²/2 = 49.1 + 176.6 = 226 kN/m; the factor of safety against overturning would drop to 1.46 and against sliding to 0.67 (1.04 even counting the passive wedge), and once the corresponding base uplift (a triangle from 58.9 kPa at the heel to zero at the toe) is added both fall below unity, to 0.96 and 0.83. A drainage blanket, weepholes and a filter behind the stem are not optional details but part of the design. (4) The design value is the 1.41 obtained without the passive wedge. The remedies, in order of preference, are a 0.4 to 0.6 m deep shear key beneath the base (which moves the failure surface into the soil and lets the undiscounted founding strength phi' = 35° act), widening the heel, or extending the base to found deeper.

Topic. External stability of a cantilever retaining wall by Rankine's theory on the virtual back plane.

Key relations. K_a = (1 − sin phi')/(1 + sin phi') and K_p = 1/K_a for a level surface; P_a = ½K_agamma H² at H/3; FS_{(overturning)} = ΣM_R/ΣM_o; FS_(sliding) = (ΣV tan delta + B c'_a + P_p)/P_a; e = B/2 − (ΣM_R − ΣM_o)/ΣV; q_max,min = (ΣV/B)(1 ± 6e/B).

Why this works. Rankine's solution assumes a semi-infinite mass in a state of plastic equilibrium bounded by a plane on which no shear stress acts. A vertical plane rising from the heel of a cantilever wall satisfies that condition when the backfill surface is horizontal, because the soil wedge above the heel moves down with the wall and no relative shear develops across the plane. Everything to the left of that plane — stem, base and the soil sitting on the heel — is then one free body, which is why the backfill weight over the heel is such a large part of the restoring moment (495 of 661 kN·m/m here). The battered face or stepped back of the stem never enters the calculation.

Common pitfall. Applying Coulomb's coefficient with wall friction to the concrete back face while also counting the soil over the heel as resisting weight — the two treatments belong to different free bodies and mixing them double-counts. A second is measuring the depth D to the top of the base instead of its underside, which inflates the passive wedge; a third is taking the full 1.5 m as soil thickness over the toe.

Source. Das, Principles of Foundation Engineering, Ch. 8, Sections 8.2–8.5 (Eqs. 8.11 to 8.14) with the Rankine coefficients of Ch. 7; Canadian Foundation Engineering Manual, 4th ed., Ch. 27.

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