16-Civ-B3 Geotechnical Design · December 2014

Question 10 of 10: Factor of safety against overturning of a gravity 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 2014 — 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 five design questions of 24 marks each (answer any three); the examinable total is 4 × 7 + 3 × 24 = 100 marks. All ten 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. — bearing capacity (Ch. 3), stress increase in a soil mass (Ch. 6), retaining walls (Ch. 8), pile foundations (Ch. 11).
B. M. Das, Principles of Geotechnical Engineering, 9th ed. — lateral earth pressure (Ch. 13), shear strength (Ch. 12), slope stability (Ch. 15), subsurface exploration (Ch. 17).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for site investigation, bearing resistance, deep foundations and earth-retaining structures.
R. F. Craig, Craig's Soil Mechanics, 8th ed. — earth pressure theory and slope stability.
D. P. Coduto, Foundation Design: Principles and Practices, 2nd ed. — in-situ testing and SPT correlations.
J. E. Bowles, Foundation Analysis and Design, 5th ed. — bearing capacity factors and retaining-wall stability tables.

Sources of charts and assumed values (page-1 Note 6). Note 6 of this paper requires the candidate to identify the source of every design chart and every assumed value. Each chart reading and each assumption below is therefore named where it is used, and the values assumed in the absence of data are collected here:

Q6 — adhesion factor α from Das, Principles of Foundation Engineering, Table 11.6 (Terzaghi, Peck & Mesri form, α against $c_u/p_a$); $\lambda$ from Vijayvergiya & Focht (1972) as tabulated by Das, Table 11.7.
Q7 — overburden correction $C_N$ from Liao & Whitman (1986); $\phi'$ from Wolff (1989) and from Hatanaka & Uchida (1996), both reproduced in Das, Ch. 2; settlement-controlled bearing pressure from Meyerhof (1965) as given by Das, Ch. 5, used only as a serviceability check because the question forbids direct correlations of bearing capacity to penetration index. Table I prints the blow counts as field values $N_f$; with no hammer data they are converted as $N_{60} = N_f$, i.e. a safety hammer at the reference 60 per cent energy ratio with borehole, sampler and rod-length factors of 1 (Das, Ch. 2, hammer-efficiency and correction-factor tables).
Q8 — embankment influence factor from Osterberg (1957), reproduced as Das Fig. 6.24; the closed form of that chart is used so the reading carries no chart-scaling error.
Q9 — Meyerhof general bearing-capacity equation with the shape factors of De Beer (1970) and the depth factors of Hansen (1970), as set out in Das, Ch. 3.
Q10 — Coulomb active earth-pressure coefficient, Das Eq. 13.31; unit weight of the mass-concrete wall assumed $\gamma_c = 24\ \text{kN/m}^3$ (CFEM 4th ed., normal-density concrete), the only value the figure does not supply.

Question 10: Factor of safety against overturning 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. A mass-concrete gravity wall on a 5.0 m wide by 1.2 m thick base, with a stem 7.5 m high that tapers from 2.5 m at its foot to 1.0 m at the top, retaining a level granular backfill to the top of the wall.

Given data — Question 10 (dimensions read from Figure 2)
Quantity	Symbol	Value
Base slab	$B \times t$	5.0 m wide by 1.2 m thick
Base segments, toe to heel	—	0.5 + 0.5 + 1.0 + 1.0 + 2.0 = 5.0 m
Stem height above the base	$H_s$	7.5 m
Stem width, top / bottom	—	1.0 m / 2.5 m
Total wall height	$H$	7.5 + 1.2 = 8.7 m
Embedment at the toe	$D$	1.5 m
Backfill	$\gamma$, $c'$, $\phi'$	20 kN/m³, 0, 36°
Wall friction	$\delta$	$\tfrac{2}{3}\phi' = 24^\circ$
Concrete unit weight (assumed, CFEM)	$\gamma_c$	24 kN/m³

Find. The factor of safety against overturning about the toe, $FS_{OT} = \sum M_R / \sum M_O$.

Figure 10.1 — Gravity wall geometry from Figure 2 of the paper. The base is 5.0 m wide in five dimensioned segments; the stem batters 0.5 m on the front face and 1.0 m on the back over its 7.5 m height, giving a back-face inclination of $\beta = 7.60^\circ$ from the vertical. Moments are taken about the toe O.

Approach. Take the free body as the wall together with the soil standing on the heel, with the back boundary the plane of the stem's inclined back face projected to founding level. Compute Coulomb's active thrust on that plane, resolve it into horizontal and vertical components, then take moments of everything about the toe.

Back-face inclination. The stem's rear face runs 1.0 m horizontally over its 7.5 m height, so measured from the vertical $$\beta = \tan^{-1}\!\left(\frac{1.0}{7.5}\right) = 7.595^\circ$$ The backfill surface is level, so the surcharge slope is $\alpha = 0$, and the wall friction is $\delta = \tfrac{2}{3}(36^\circ) = 24^\circ$ as specified.
Coulomb active coefficient. Using Das Eq. 13.31, $$K_a = \frac{\cos^2\!\left(\phi' - \beta\right)} {\cos^2\beta\,\cos\!\left(\delta + \beta\right) \left[1 + \sqrt{\dfrac{\sin\!\left(\phi' + \delta\right)\sin\!\left(\phi' - \alpha\right)} {\cos\!\left(\delta + \beta\right)\cos\!\left(\alpha - \beta\right)}}\,\right]^2}$$ Substituting $\phi' = 36^\circ$, $\beta = 7.595^\circ$, $\delta = 24^\circ$ and $\alpha = 0$ gives $$\boxed{K_a = 0.2929}$$ As a check on the algebra, setting $\beta = \delta = 0$ in the same expression returns 0.2596, which is exactly the Rankine value $\tan^2(45^\circ - 18^\circ) = 0.2596$. The back face leans back under the fill (its foot is 1.0 m further into the backfill than its top), so soil rests on it and the sliding wedge is heavier; an independent trial-wedge search over failure-plane angles returns the same $K_a = 0.2929$.
Active thrust and its components. With the vertical height of the back face taken to founding level, $H = 7.5 + 1.2 = 8.7\ \text{m}$, $$P_a = \tfrac{1}{2}\gamma H^2 K_a = \tfrac{1}{2}(20)(8.7)^2(0.2929) = 221.7\ \text{kN/m}$$ Coulomb's thrust acts at $\delta$ to the normal of the back face, and since that normal is itself inclined $\beta$ to the horizontal, the resultant makes an angle $\delta + \beta = 31.59^\circ$ with the horizontal: $$P_h = P_a\cos\!\left(\delta + \beta\right) = 221.7\cos 31.59^\circ = 188.9\ \text{kN/m}$$ $$P_v = P_a\sin\!\left(\delta + \beta\right) = 221.7\sin 31.59^\circ = 116.2\ \text{kN/m}$$ Both act at $H/3 = 2.90\ \text{m}$ above the base, and at that elevation the back-face plane lies $2.773\ \text{m}$ from the toe.
Weights of the free body. The wall is decomposed into the base slab and three stem components, and the retained soil into the block standing on the 2.0 m heel plus the triangular wedge that sits on the battered back face. Areas are per metre run, concrete at 24 kN/m³ and soil at 20 kN/m³, with lever arms measured from the toe: $$W_1 = (5.0)(1.2)(24) = 144.0\ \text{kN/m} \ \text{at}\ 2.500\ \text{m}$$ $$W_2 = \tfrac{1}{2}(0.5)(7.5)(24) = 45.0\ \text{kN/m} \ \text{at}\ 0.833\ \text{m}$$ $$W_3 = (1.0)(7.5)(24) = 180.0\ \text{kN/m} \ \text{at}\ 1.500\ \text{m}$$ $$W_4 = \tfrac{1}{2}(1.0)(7.5)(24) = 90.0\ \text{kN/m} \ \text{at}\ 2.333\ \text{m}$$ $$W_5 = (2.0)(7.5)(20) = 300.0\ \text{kN/m} \ \text{at}\ 4.000\ \text{m}$$ $$W_6 = \tfrac{1}{2}(1.0)(7.5)(20) = 75.0\ \text{kN/m} \ \text{at}\ 2.667\ \text{m}$$ so that $\sum W = 834.0\ \text{kN/m}$.
Resisting moment about the toe. Summing the weight moments, $$\sum W x = 360.0 + 37.5 + 270.0 + 210.0 + 1200.0 + 200.0 = 2277.5\ \text{kN}\!\cdot\!\text{m/m}$$ and adding the stabilising moment of the vertical component of the thrust, $P_v x_P = 116.2(2.773) = 322.2\ \text{kN}\!\cdot\!\text{m/m}$, $$\boxed{\sum M_R = 2277.5 + 322.2 = 2599.7\ \text{kN}\!\cdot\!\text{m/m}}$$ Note that the soil standing on the heel supplies 1200 of the 2278 kN·m of weight moment — more than all the concrete together. That is why a gravity wall of this proportion works.
Overturning moment about the toe. Only the horizontal component of the thrust overturns, and its lever arm is its height above the base: $$\sum M_O = P_h \frac{H}{3} = 188.9(2.90) = 547.7\ \text{kN}\!\cdot\!\text{m/m}$$ Passive resistance from the 1.5 m of soil in front of the toe is deliberately neglected, which is conservative and is normal practice because that soil can be excavated or eroded.
Factor of safety against overturning. $$\boxed{FS_{OT} = \frac{\sum M_R}{\sum M_O} = \frac{2599.7}{547.7} = 4.75}$$ This comfortably exceeds the value of 2 normally required against overturning for a gravity wall, and exceeds 1.5 even if the vertical component of the thrust is discarded entirely (which would give 2277.5/547.7 = 4.16).
Complete the stability picture. Overturning is rarely the governing check for a wall of this shape, so the two companion checks are worth recording. The total vertical force on the base is $V = \sum W + P_v = 834.0 + 116.2 = 950.2\ \text{kN/m}$, and with a base friction angle of $\tfrac{2}{3}\phi' = 24^\circ$ and no passive contribution, $$FS_{sliding} = \frac{V\tan 24^\circ}{P_h} = \frac{950.2(0.4452)}{188.9} = 2.24$$ The eccentricity of the resultant is $e = B/2 - \left(\sum M_R - \sum M_O\right)/V = 2.5 - 2052.0/950.2 = 0.340\ \text{m}$, comfortably inside the middle third ($B/6 = 0.833\ \text{m}$), so the whole base stays in compression with $$q_{max,min} = \frac{V}{B}\left(1 \pm \frac{6e}{B}\right) = 190.0(1 \pm 0.408) = 268\ \text{and}\ 112\ \text{kPa}$$ Both checks are satisfied, and the bearing pressure of 268 kPa would then be compared with the allowable bearing pressure of the founding soil.

Final results — Question 10
Quantity	Symbol	Value
Back-face inclination from vertical	$\beta$	7.60°
Wall friction angle	$\delta$	24°
Coulomb active coefficient	$K_a$	0.2929
Active thrust	$P_a$	221.7 kN/m
Horizontal / vertical components	$P_h$ / $P_v$	188.9 / 116.2 kN/m
Total weight of the free body	$\sum W$	834.0 kN/m
Resisting moment about the toe	$\sum M_R$	2599.7 kN·m/m
Overturning moment about the toe	$\sum M_O$	547.7 kN·m/m
Factor of safety against overturning	$FS_{OT}$	4.75
Factor of safety against sliding (no passive)	$FS_{SL}$	2.24
Eccentricity / base pressures	$e$ / $q_{max}$	0.340 m / 268 kPa

Check: two assumptions are recorded under page-1 Notes 1 and 6. First, the unit weight of the wall is not given and is taken as $\gamma_c = 24\ \text{kN/m}^3$ for normal-density mass concrete (CFEM 4th ed.); at 23 kN/m³ the factor of safety would fall only to 4.68, so the answer is insensitive to this choice. Second, and much more consequential, is the choice of free body. The calculation above follows the conventional examination treatment for a gravity wall: Coulomb's thrust on the stem's inclined back face projected to founding level, with the whole soil column over the 2.0 m heel counted as resisting weight. That pairing is not strictly self-consistent — the soil over the heel lies on the far side of the pressure plane — so it is worth recording where the two fully consistent free bodies land. Taking a vertical plane through the back edge of the heel with Rankine's coefficient ($K_a = 0.2596$, $P_a = 196.5\ \text{kN/m}$, no vertical component) and the same 834 kN/m of weight gives $FS_{OT} = 4.00$. Taking Coulomb's thrust on the plane from the top of the back face down to the heel of the base, inclined $\beta = 19.03^\circ$ ($K_a = 0.4055$, $P_a = 306.9\ \text{kN/m}$), with the free body then reduced to the wall plus only the 119 kN/m soil wedge that lies inboard of that plane, gives $FS_{OT} = 3.22$. The three treatments therefore bracket the answer between about 3.2 and 4.8, and every one of them clears the value of 2 normally required, so the conclusion that this wall is safe against overturning does not depend on the choice. The lowest of the three, 3.22, is the number to carry forward if a conservative design value is wanted.

Topic: stability of a gravity retaining wall against overturning, using Coulomb active earth pressure on a battered back face.

Key relations: $K_a = \dfrac{\cos^2(\phi' - \beta)}{\cos^2\beta\cos(\delta+\beta) \left[1 + \sqrt{\dfrac{\sin(\phi'+\delta)\sin(\phi'-\alpha)} {\cos(\delta+\beta)\cos(\alpha-\beta)}}\right]^2}$; $P_a = \tfrac{1}{2}\gamma H^2 K_a$; $P_h = P_a\cos(\delta+\beta)$, $P_v = P_a\sin(\delta+\beta)$; $FS_{OT} = \dfrac{\sum M_R}{\sum M_O}$ about the toe.

Why this works: Coulomb's theory finds the plane failure wedge that maximises the thrust on the wall, and it can handle a battered face, a sloping backfill and wall friction, all of which Rankine's theory cannot. Wall friction is what gives the thrust a vertical component, and that component acts a long way from the toe, so it is genuinely stabilising — on this wall it contributes an eighth of the resisting moment.

Common pitfall: counting the soil on the heel as resisting weight and applying the earth thrust on a vertical plane behind the heel. The two are alternative free bodies, not additive: choose one boundary, and apply the wedge thrust on that boundary only. A second frequent error is to include $P_v$ in the resisting moment while also taking the pressure plane vertical with $\delta$ against soil rather than against concrete, which double-counts wall friction.

Source: B. M. Das, Principles of Foundation Engineering, 9th ed., Ch. 8 (Retaining walls — stability checks) and Principles of Geotechnical Engineering, 9th ed., Ch. 13 (Coulomb's active pressure, Eq. 13.31); CFEM 4th ed., Ch. 27; J. E. Bowles, Foundation Analysis and Design, 5th ed., Ch. 12.

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