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16-Civ-B3 Geotechnical Design · Undated paper

Question 8 of 9: Overturning and sliding of a cantilever retaining wall

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

Notes on this paper

Paper format. National Examinations (Engineers Canada / EGBC), 16-Civ-B3 Geotechnical Design — 3 hours, open book, any non-communicating calculator permitted (the candidate must write its make and model on the left-hand sheet). The paper prints nine questions in two sections: Section A holds five short questions of 7 marks and asks for any four; Section B holds four design questions of 24 marks and asks for any three. Only the first four of Section A and the first three of Section B are marked, so a complete paper is 4 × 7 + 3 × 24 = 100 marks. Note 1 urges the candidate to state any assumptions made, Note 6 requires the source of every design chart to be identified, and Note 7 permits assumed values provided the source is stated. All nine questions are solved below, because the set is a study resource rather than a sitting.

Reference texts. B. M. Das, Principles of Foundation Engineering, 8th ed. (bearing capacity ch. 3, settlement of shallow foundations ch. 5, drilled shafts ch. 12, retaining walls ch. 8, sheet pile walls ch. 9); B. M. Das, Principles of Geotechnical Engineering, 9th ed. (shear strength, lateral earth pressure, slope stability); R. F. Craig and J. A. Knappett, Craig’s Soil Mechanics, 8th ed. (effective stress, undrained strength, anchored walls); D. P. Coduto, Foundation Design: Principles and Practices, 2nd ed.; and in the Canadian frame the Canadian Foundation Engineering Manual (CFEM), 4th ed., Canadian Geotechnical Society — ch. 4 for site investigation and in-situ testing, ch. 10 for shallow foundations, ch. 18 for deep foundations and ch. 25 for earth retaining structures. Test standards are quoted as ASTM/CSA where the CFEM adopts them (SPT: ASTM D1586; CPT: ASTM D5778; field vane: ASTM D2573).

Source-quality note.

Question 8: Overturning and sliding 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.

Question 8 — geometry and soil data read from Figure 3
QuantityValue
Stem: height above base slab, top width, base width6.1 m, 0.40 m, 1.04 m
Base slab: thickness, toe projection, heel projection0.9 m, 0.76 m, 3.0 m (total width 4.80 m)
Backfill surface slope8 degrees
Uniform surcharge on the backfill$q=20$ kPa
Backfill$\gamma=18$ kN/m3, $\phi'=32$ degrees
Concrete$\gamma_c=23.5$ kN/m3
Foundation soil$\gamma=16.8$ kN/m3, $\delta'=15$ degrees, $c'=30$ kPa
Depth of soil in front of the toe1.0 m

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

q = 20 kPa8 deg1.80 m3.0 m6.1 m0.9 m0.40 mgamma_c = 23.5gamma = 18 kN/m3phi' = 32 deggamma = 16.8 kN/m3base: delta' = 15 deg, c' = 30 kPaP_a at 8 degRankine planeW_soilT_basecantilever wall: Rankine active thrust on the vertical plane through the heel
The wall of Figure 3. The Rankine active thrust is taken on the vertical plane through the heel and acts parallel to the ground surface; the soil wedge between that plane and the back of the stem is counted as part of the stabilising weight.

Approach. Take the Rankine active thrust on the vertical plane through the heel, inclined at the backfill slope; resolve it; sum the stabilising weights and their moments about the toe; then form the two factors of safety. The surcharge is treated both as an additional active pressure and, over the heel, as an additional vertical load, because it is the same load.

  1. Height of the Rankine plane. The vertical plane rises from the heel of the base to the sloping ground surface. The ground rises at 8 degrees from the top of the stem across the 3.0 m heel, so $$\begin{aligned}H&=0.9+6.1+3.0\tan 8^{\circ} \\ &=0.9+6.1+0.422 \\ &=\boxed{7.422\ \text{m}}\end{aligned}$$
  2. Rankine active coefficient for a sloping backfill. With $\beta=8^{\circ}$ and $\phi'=32^{\circ}$, $$\begin{aligned}K_a&=\cos\beta\,\frac{\cos\beta-\sqrt{\cos^{2}\beta-\cos^{2}\phi'}}{\cos\beta+\sqrt{\cos^{2}\beta-\cos^{2}\phi'}} \\ &=0.9903\times\frac{0.9903-0.5113}{0.9903+0.5113} \\ &=0.3159\end{aligned}$$ For comparison the horizontal-backfill value would be $\tan^{2}(45^{\circ}-16^{\circ})=0.3073$, so the 8 degree slope costs about 3 per cent.
  3. Active thrust and its components. The soil and the surcharge are taken separately because their resultants act at different heights, and both act at $\beta$ to the horizontal: $$\begin{aligned}P_{a,soil}&=\tfrac{1}{2}K_a\gamma H^{2} \\ &=\tfrac{1}{2}(0.3159)(18)(7.422)^{2} \\ &=156.6\ \text{kN/m}\end{aligned}$$ $$\begin{aligned}P_{a,q}&=K_a qH \\ &=(0.3159)(20)(7.422) \\ &=46.9\ \text{kN/m}\end{aligned}$$ $$\begin{aligned}P_a&=203.5\ \text{kN/m} \\ P_h&=P_a\cos 8^{\circ}=201.5\ \text{kN/m} \\ P_v&=P_a\sin 8^{\circ}=28.3\ \text{kN/m}\end{aligned}$$
  4. Overturning moment about the toe. The soil triangle acts at $H/3$ and the surcharge rectangle at $H/2$ above the base: $$\begin{aligned}\sum M_o&=156.6\cos 8^{\circ}\left(\frac{7.422}{3}\right)+46.9\cos 8^{\circ}\left(\frac{7.422}{2}\right) \\ &=383.6+172.3 \\ &=\boxed{556\ \text{kN}\cdot\text{m/m}}\end{aligned}$$
  5. Stabilising weights and their moments. Distances are measured from the toe, the front bottom corner of the base slab. The stem is split into the 0.40 m prism and the battered triangle behind it, and the backfill on the heel into the block below the top-of-stem level and the wedge above it.
Question 8 — vertical forces and their moments about the toe
#ComponentWeight, kN/mArm, mMoment, kN·m/m
1Base slab, $4.80\times 0.9\times 23.5$101.52.400243.7
2Stem prism, $0.40\times 6.1\times 23.5$57.31.60091.7
3Stem batter, $\tfrac{1}{2}(0.64)(6.1)(23.5)$45.91.18754.4
4Backfill block, $3.0\times 6.1\times 18$329.43.3001087.0
5Backfill wedge, $\tfrac{1}{2}(3.0)(0.422)(18)$11.43.80043.3
6Surcharge over the heel, $20\times 3.0$60.03.300198.0
7Soil over the toe, $0.76\times 0.1\times 16.8$1.30.3800.5
8Vertical component $P_v$ at the heel28.34.800135.9
Totals635.1—1854.5
  1. Factor of safety against overturning. $$\begin{aligned}\mathrm{FS}_{overturning}&=\frac{\sum M_R}{\sum M_o} \\ &=\frac{1854.5}{555.9} \\ &=\boxed{3.34}\end{aligned}$$ comfortably above the usual requirement of 2.0.
  2. Factor of safety against sliding. The base interface parameters are given directly on the figure as $\delta'=15^{\circ}$ and $c'=30\ \text{kPa}$, so no further reduction factor is applied to them: $$\begin{aligned}\mathrm{FS}_{sliding}&=\frac{\sum V\tan\delta'+Bc'}{P_h} \\ &=\frac{635.1\tan 15^{\circ}+4.80(30)}{201.5} \\ &=\frac{170.2+144.0}{201.5} \\ &=\boxed{1.56}\end{aligned}$$ which meets the usual requirement of 1.5, but only just, and note that base adhesion supplies 46 per cent of the resistance.
  3. Passive resistance in front, if it is relied upon. The 1.0 m of soil over the toe has been ignored above. Taking the foundation soil parameters at face value, $K_p=\tan^{2}(45^{\circ}+7.5^{\circ})=1.698$ and $$\begin{aligned}P_p&=\tfrac{1}{2}K_p\gamma D^{2}+2c'\sqrt{K_p}D \\ &=14.3+78.2 \\ &=92.5\ \text{kN/m}\end{aligned}$$ which would raise the sliding factor of safety to 2.02.
  4. Check that the resultant stays in the middle third. The line of action of the resultant meets the base at $\bar{x}=(\sum M_R-\sum M_o)/\sum V=(1854.5-555.9)/635.1=2.045\ \text{m}$ from the toe, so the eccentricity is $e=B/2-\bar{x}=2.40-2.045=0.355\ \text{m}$, well inside $B/6=0.80\ \text{m}$. The base pressures are then $$q_{max,min}=\frac{\sum V}{B}\left(1\pm\frac{6e}{B}\right)=\boxed{191\ \text{kPa and }74\ \text{kPa}}$$ with no tension anywhere under the base.

Check: two judgement calls that move the sliding answer. First, passive resistance in front of the toe has deliberately been excluded from the reported factor of safety, because that 1.0 m of cover can be removed by a service trench, by scour or by future landscaping, and because it requires wall movement to mobilise. Including it gives 2.02. Second, many codes mobilise only one half to two thirds of the base cohesion, since full adhesion of a cast-in-place slab to the soil beneath cannot be guaranteed; at $c_a=\tfrac{2}{3}c'=20\ \text{kPa}$ the sliding factor of safety falls to 1.32, which is below the usual 1.5. That sensitivity is the governing engineering issue on this wall: it should carry a shear key under the base, or a wider base, rather than depend on full base adhesion. The overturning result is not sensitive to either assumption.

Question 8 — results
QuantitySymbolValue
Height of the Rankine plane$H$7.422 m
Active earth pressure coefficient$K_a$0.3159
Total active thrust$P_a$203.5 kN/m
Horizontal component$P_h$201.5 kN/m
Vertical component$P_v$28.3 kN/m
Overturning moment about the toe$\sum M_o$556 kN·m/m
Resisting moment about the toe$\sum M_R$1854 kN·m/m
Total vertical force$\sum V$635 kN/m
Factor of safety, overturning$\mathrm{FS}_{ot}$3.34
Factor of safety, sliding (no passive)$\mathrm{FS}_{sl}$1.56
Factor of safety, sliding, with $P_p$—2.02
Eccentricity of the base resultant$e$0.355 m ($ \lt B/6=0.80$ m)
Base pressures$q_{max},\ q_{min}$191 kPa, 74 kPa