07-Str-B1 · May 2013

Question 9 of 10: Overturning and Sliding Stability of a Cantilever Retaining Wall

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

Notes on this paper

National Examinations — May 2013 — 07-Str-B1 Geotechnical Design. Three-hour, OPEN-BOOK exam; any non-communicating calculator permitted (the candidate must record its make and model). Format: Section A carries five short-answer questions of 7 marks each, of which any FOUR are to be answered; Section B carries the long design questions at 24 marks each, of which any THREE are to be answered. The paper instructs candidates to state any interpretive assumptions and to identify the source of every design chart or assumed value used. Every question in both sections is worked below, because the set is intended as a study resource.

Reference texts: Das, B.M., Principles of Foundation Engineering (9th ed., Cengage) — shallow foundations, consolidation settlement, sheet-pile walls, retaining walls and drilled shafts; Das, B.M., Principles of Geotechnical Engineering (9th ed., Cengage) — method of slices, lateral earth pressure, consolidation theory; Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM, 4th ed., 2006) — Canadian practice for SPT interpretation, pile design, limit-states design and tolerable settlement; Craig, R.F. / Knappett, J.A., Craig's Soil Mechanics (8th ed., CRC Press) — effective stress, shear strength and slope stability; Duncan, J.M., Wright, S.G. & Brandon, T.L., Soil Strength and Slope Stability (2nd ed., Wiley) — choice of strength parameters and factors of safety for short- and long-term analyses.

NOTE 1 — question numbering in the source. The printed paper labels the retaining-wall problem (Figure 4) and the drilled-pier problem (Figure 5) both as "Question 9", while the Section B heading reads "answer any THREE of the following FOUR questions". Section B therefore contains five printed problems under four numbers. They are set out below as Question 9 (retaining wall) and Question 10 (drilled pier) in printed order, so that each can be referred to unambiguously; the marks shown are those printed against each problem.

NOTE 2 — dimensions scaled from Figure 1. Figure 1 is a hand-drawn slope on a 1 m × 1 m grid with no written dimensions other than $R=10$ m. The geometry used in Question 6 was scaled from that grid: slope height 7 m over a 9 m horizontal run, a 3 m thick lower layer, and the centre of the trial circle 1.4 m horizontally beyond the toe and 8.0 m above it. Every one of these values reproduces the drawing to within about 0.2 m (one fifth of a grid square). Check against the original if the paper is used for marking rather than study.

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. A cantilever retaining wall with a level backfill of Soil 1 and founded in Soil 2.

Given data
Quantity	Symbol	Value
Backfill, Soil 1	$\gamma_1,\ \phi'_1,\ c'_1$	20 kN/m³, 36°, 5 kPa
Foundation soil, Soil 2	$\gamma_2,\ \phi'_2,\ c'_2$	18 kN/m³, 30°, 10 kPa
Concrete	$\gamma_c$	24 kN/m³
Height from top of stem to underside of base	$H$	$11.0+1.5=12.5$ m
Base slab width and thickness	$B$, $t$	9.0 m (3.0 m toe side + 6.0 m heel), 1.0 m
Stem thickness at top and at base	—	0.5 m, 1.5 m; back face vertical at 3.0 m from the toe
Depth of soil in front of the toe	$D$	1.5 m to the underside of the base
Groundwater	—	Not shown; drained backfill assumed

Find. (i) the factor of safety against overturning about the toe, and (ii) the factor of safety against sliding along the base.

[Figure not reproduced: Figure 4 (redrawn) — cantilever retaining wall section, showing the vertical plane through the heel on which the Rankine active thrust P_a is taken and the passive block in front of the toe. See the official exam paper.]

Approach. Take the Rankine active pressure on a vertical plane through the heel over the full height $H$, so that the soil wedge above the heel is treated as part of the resisting mass. Neglect the tensile block at the top of the cohesive backfill, compute the active thrust and its moment about the toe, then sum the weights of the concrete and of the soil carried on the heel to obtain the resisting moment. Sliding is checked on the base with a reduced friction angle and adhesion, with and without the passive resistance in front of the toe.

Active earth pressure coefficient for the backfill. With a level backfill and a vertical plane through the heel, Rankine applies with $$K_a=\tan^2\!\left(45^\circ-\frac{36^\circ}{2}\right)=\tan^2 27^\circ=0.2596$$
Active pressure diagram allowing for cohesion. The Rankine active pressure in a $c'$–$\phi'$ soil is $$\sigma'_a=K_a\gamma z-2c'\sqrt{K_a}$$ At the underside of the base, $z=H=12.5$ m, $$\sigma'_a=0.2596(20)(12.5)-2(5)\sqrt{0.2596}=64.90-5.10=59.81\ \text{kPa}$$ The pressure is negative above the depth $$z_c=\frac{2c'}{\gamma\sqrt{K_a}}=\frac{2(5)}{20\sqrt{0.2596}}=0.981\ \text{m}$$ Soil cannot sustain that tension, and a crack will open, so the tensile block is discarded and only the triangle below $z_c$ is carried into the analysis.
Active thrust and its line of action. The remaining diagram is a triangle of height $59.81$ kPa over a depth of $12.5-0.981=11.519$ m: $$P_a=\tfrac12(59.81)(11.519)=344.5\ \text{kN/m}$$ acting horizontally at $$\bar y=\frac{11.519}{3}=3.840\ \text{m}$$ above the underside of the base. The overturning moment about the toe is therefore $$M_o=344.5(3.840)=1322.6\ \text{kN}\cdot\text{m/m}$$
Weights of the wall and the soil it carries. Taking $x$ from the toe (the front edge of the base) and working per metre run:

Resisting weights and their moments about the toe
Component	Area or dimensions	$\gamma$ (kN/m³)	$W$ (kN/m)	$\bar x$ (m)	$W\bar x$ (kN·m/m)
Base slab, 9.0 m × 1.0 m	9.00 m²	24	216.0	4.500	972.0
Stem, rectangular part 0.5 m × 11.5 m	5.75 m²	24	138.0	2.750	379.5
Stem, tapered part (triangle, 1.0 m × 11.5 m)	5.75 m²	24	138.0	2.167	299.0
Backfill on the heel, 6.0 m × 11.5 m	69.00 m²	20	1380.0	6.000	8280.0
Soil over the toe, 0.5 m deep	0.761 m²	18	13.7	0.761	10.4
Totals			1885.7		9940.9

The stem is a trapezium 11.5 m tall (the height of the wall less the 1.0 m base slab), 0.5 m wide at the top and 1.5 m at the bottom, with its back face vertical at $x=3.0$ m. Splitting it into a 0.5 m wide rectangle between $x=2.5$ and $x=3.0$, and a triangle whose corners are $(1.5,\,1.0)$, $(2.5,\,1.0)$ and $(2.5,\,12.5)$, gives the two entries above; the triangle's centroid lies at $\bar x=(1.5+2.5+2.5)/3=2.167$ m. The backfill carried on the heel — 6.0 m wide and 11.5 m deep, from the top of the base slab up to the backfill surface — supplies 1380 kN/m, more than 70% of the total weight and by far the largest resisting moment.

(i) Factor of safety against overturning. Taking moments about the toe, $$FS_{\text{overturning}}=\frac{\sum M_R}{M_o}=\frac{9940.9}{1322.6}=\boxed{FS_{OT}=7.5}$$ This comfortably exceeds the usual requirement of 1.5 to 2.0. It is not surprising: the base is 9.0 m wide against a height of 12.5 m, a ratio of 0.72 that is at the generous end of the usual 0.5 to 0.7, and the 6.0 m heel places nearly the whole of the retained soil wedge on the resisting side of the toe.
Base friction and adhesion. The base bears on Soil 2. Taking the customary reductions for a concrete base cast against soil, $$\delta=\tfrac23\phi'_2=\tfrac23(30^\circ)=20^\circ,\qquad c_a=\tfrac23c'_2=\tfrac23(10)=6.67\ \text{kPa}$$ so the resistance available along the base is $$R=\sum W\tan\delta+c_aB=1885.7\tan20^\circ+6.67(9.0)=686.3+60.0=746.3\ \text{kN/m}$$
Passive resistance in front of the toe. Over the 1.5 m of Soil 2 standing in front of the wall, $$K_p=\tan^2\!\left(45^\circ+\frac{30^\circ}{2}\right)=\tan^2 60^\circ=3.000$$ $$P_p=\tfrac12K_p\gamma_2D^2+2c'_2\sqrt{K_p}\,D=\tfrac12(3.000)(18)(1.5)^2+2(10)(1.732)(1.5)$$ $$P_p=60.75+51.96=112.7\ \text{kN/m}$$
(ii) Factor of safety against sliding. The driving force is the horizontal active thrust, 344.5 kN/m. Ignoring the passive resistance, $$FS_{\text{sliding}}=\frac{746.3}{344.5}=\boxed{FS_{SL}=2.17}$$ and including it, $$FS_{\text{sliding}}=\frac{746.3+112.7}{344.5}=\frac{859.0}{344.5}=2.49$$ Both exceed the usual minimum of 1.5. The lower of the two, 2.17, is the value that should be quoted for design, because the passive block in front of the toe can be removed by a service trench, by scour or by erosion at any time in the life of the wall and should not be relied on unless it is protected.

Question 9 — results
Quantity	Value
Active pressure coefficient, backfill	$K_a=0.2596$
Depth of the tension crack	$z_c=0.98$ m
Active pressure at the underside of the base	59.81 kPa
Active thrust $P_a$ and its height above the base	344.5 kN/m at 3.84 m
Overturning moment about the toe	1322.6 kN·m/m
Total vertical weight $\sum W$	1885.7 kN/m
Resisting moment about the toe	9940.9 kN·m/m
(i) Factor of safety against overturning	$\boxed{FS_{OT}=7.5}$
Base resistance $\sum W\tan\delta+c_aB$	746.3 kN/m
Passive resistance in front of the toe	112.7 kN/m
(ii) Factor of safety against sliding (passive ignored)	$\boxed{FS_{SL}=2.17}$ (2.49 if $P_p$ is included)

Check — reading of Figure 4 and the checks not asked for. The figure gives 11.0 m from the top of the stem to the ground surface in front of the wall and a further 1.5 m to the underside of the base, so the height used for earth pressure on the vertical plane through the heel is $H=12.5$ m. Soil 1 is taken as the backfill retained by the wall and Soil 2 as the natural ground the wall is founded in and cut into; the wavy line on the right of the figure is the boundary between them, and it lies outside the vertical plane through the heel over the full height, so Soil 2 does not enter the active pressure calculation. The small 0.20 m dimension at the heel is a taper on the top of the base slab and has been idealised away by treating the slab as a uniform 1.0 m thick rectangle. Two further checks are required before a wall of these proportions could be signed off but are not asked for here: bearing pressure under the base including the eccentricity of the resultant (the resultant must fall within the middle third, $e\le B/6=1.5$ m), and overall (global) slope stability of the wall and its founding soil as a single mass.

Topic. External stability of a cantilever retaining wall: overturning, sliding, and the Rankine active thrust in a $c'$–$\phi'$ backfill.

Key relations. $K_a=\tan^2(45^\circ-\phi'/2)$; $\sigma'_a=K_a\gamma z-2c'\sqrt{K_a}$; $z_c=2c'/(\gamma\sqrt{K_a})$; $FS_{OT}=\sum M_R/\sum M_O$ about the toe; $FS_{SL}=\left(\sum W\tan\delta+c_aB+P_p\right)/P_a$ with $\delta=\tfrac23\phi'$ and $c_a=\tfrac23c'$; $K_p=\tan^2(45^\circ+\phi'/2)$; $P_p=\tfrac12K_p\gamma D^2+2c'\sqrt{K_p}D$.

Why this works. Rankine's active state applies on a vertical plane where the soil is free to strain laterally. Placing that plane at the back of the heel, rather than on the sloping back of the stem, means the wedge of soil above the heel travels with the wall — so it is counted once, as weight on the resisting side, and the wall–soil friction on the stem never has to be estimated.

Common pitfall. Omitting the weight of the backfill carried on the heel. That single component supplies 8280 of the 9941 kN·m/m of resisting moment here; leaving it out would give $FS_{OT}=1.3$ and a wall that appears to fail. The opposite pitfall is including the full cohesive pressure diagram with its negative block at the top, which reduces $P_a$ to 341.9 kN/m and quietly credits the wall with a tensile bond it does not have.

Source. Das, Principles of Foundation Engineering, Ch. 8 (retaining walls: proportioning, check for overturning, check for sliding along the base, check for bearing capacity failure); Das, Principles of Geotechnical Engineering, Ch. 13 (Rankine active and passive pressure, tension crack in cohesive backfill).

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