16-Civ-B3 Geotechnical Design · May 2015

Question 9 of 9: Factor of Safety of a Gravity Retaining Wall against Overturning

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

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada National Examinations, May 2015 — 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. — bearing capacity (Ch. 3), stress increase in a soil mass (Ch. 6), retaining walls (Ch. 8), pile foundations (Ch. 11), subsurface exploration (Ch. 2).
B. M. Das, Principles of Geotechnical Engineering, 9th ed. — consolidation (Ch. 11), shear strength (Ch. 12), lateral earth pressure (Ch. 13).
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, 9th ed. — earth pressure theory, effective stress and slope stability.
D. P. Coduto, Foundation Design: Principles and Practices, 3rd ed. — in-situ testing, settlement of shallow and deep foundations.
M. J. Tomlinson & J. Woodward, Pile Design and Construction Practice, 6th ed. — shaft adhesion in clay, pile-group behaviour, load testing.

Sources of design charts and assumed values (page-1 Note 6). Note 6 of this paper requires the candidate to identify the source of every design chart used and of every value assumed in the absence of data. They are named where used and collected here:

Q6 — bearing capacity factors $N_c$, $N_q$, $N_{\gamma}$ from Das, Principles of Foundation Engineering, Table 3.3 (Prandtl–Reissner $N_q$, Vesic $N_{\gamma}=2(N_q+1)\tan\phi'$); shape factors after De Beer (1970) and depth factors after Hansen (1970), Das Table 3.4. Assumed: unit weight of water $\gamma_w = 9.81$ kN/m$^3$; general shear failure; the sand extends at least $2B$ below the base.
Q7 — adhesion factor $\alpha = 1.0$ for soft clay ($c_u \le 50$ kPa) from NAVFAC DM-7.2 Fig. 1 and Tomlinson & Woodward Table 4.6; $\lambda = 0.24$ at an embedded length of 12 m from Vijayvergiya & Focht (1972) as tabulated by Das, Table 11.7; bearing factor $N_c^{*}=9$ for $L/D \ge 4$ (Skempton). Assumed: pile spacing $s = 3d = 1.5$ m centre to centre, driven closed-end concrete piles, clay $\gamma_{sat} = 17$ kN/m$^3$ with the water table at ground level.
Q8 — compression index from the Skempton correlation $C_c = 0.009\,(LL-10)$, Das Principles of Geotechnical Engineering Eq. (11.42); $2{:}1$ stress distribution, Das Eq. (6.31); Boussinesq rectangular influence factor (Das Table 6.6) used as the cross-check.
Q9 — Coulomb active pressure coefficient, Das Principles of Foundation Engineering Eq. (8.13) with the wall friction angle prescribed by the question, $\delta = 0.6\phi' = 18^{\circ}$. Assumed: unit weight of reinforced concrete $\gamma_c = 24$ kN/m$^3$ (CSA A23.3 nominal); passive resistance in front of the toe neglected; the backfill surface is horizontal and carries no surcharge.

Question 9: Factor of Safety of a Gravity Retaining Wall against Overturning (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	$B_b$ / $t_b$	5.0 m / 1.2 m
Toe projection / stem base width / heel projection	—	0.5 m / 2.5 m / 2.0 m
Stem height / top width	$h_s$ / $b_t$	8.0 m / 1.0 m
Total height above base underside	$H$	1.2 + 8.0 = 9.2 m
Backfill unit weight, cohesion, friction angle	$\gamma,\ c',\ \phi'$	20 kN/m$^3$, 0, $30^{\circ}$
Soil-wall friction angle	$\delta = 0.6\phi'$	$18^{\circ}$
Concrete unit weight (assumed)	$\gamma_c$	24 kN/m$^3$
Depth of soil in front of the wall	$D$	2.0 m

Find. The factor of safety against overturning about the toe, and the direction in which it moves if the groundwater table rises to the base of the wall.

[Figure not reproduced: Figure 3 (redrawn) — gravity retaining wall, all dimensions in metres. The back face of the stem is battered 1.0 m over 8.0 m, giving $\beta = 7.13^{\circ}$ from the vertical; overturning is taken about the toe at the base underside. See the official exam paper.]

Approach. Compute Coulomb's active coefficient on the battered back face of the stem, resolve the thrust into horizontal and vertical components, take moments about the toe of the base, and divide the sum of the resisting moments by the overturning moment; then repeat with the water table raised to see which way the answer moves.

Set up the geometry and the batter of the pressure face. Take the origin at the toe of the base underside, $x$ positive toward the heel and $y$ positive upward. The base occupies $0 \le x \le 5.0$, $0 \le y \le 1.2$; the stem is the trapezium with corners $(0.5, 1.2)$, $(3.0, 1.2)$, $(2.0, 9.2)$ and $(1.0, 9.2)$. The back face therefore runs from $(3.0,\,1.2)$ to $(2.0,\,9.2)$, a batter of 1.0 m over 8.0 m, so its inclination from the vertical is $$\beta = \tan^{-1}\!\left(\frac{1.0}{8.0}\right) = 7.13^{\circ}$$ The backfill surface is horizontal at the top of the wall, so the surcharge angle is $\alpha = 0$ and the total height for the pressure calculation is $H = 9.2$ m.
Coulomb active earth pressure coefficient. With $\phi' = 30^{\circ}$, $\delta = 0.6\phi' = 18^{\circ}$, $\beta = 7.13^{\circ}$ and $\alpha = 0$, $$K_a = \frac{\cos^2(\phi'-\beta)}{\cos^2\beta\,\cos(\delta+\beta)\left[1+\sqrt{\dfrac{\sin(\delta+\phi')\sin(\phi'-\alpha)}{\cos(\delta+\beta)\cos(\beta-\alpha)}}\right]^2}$$ Evaluating the parts, $\cos^2(22.87^{\circ}) = 0.8488$, $\cos^2\beta = 0.9846$, $\cos(25.13^{\circ}) = 0.9054$, and the bracket is $[1+\sqrt{(0.7431)(0.5)/(0.9054 \times 0.9923)}]^2 = (1.6431)^2 = 2.6998$, so $$K_a = \frac{0.8488}{0.9846 \times 0.9054 \times 2.6998} = 0.3526$$ The battered face raises $K_a$ above the vertical-face Coulomb value of 0.2986, because a face that leans away from the backfill at the top must support a larger wedge.
Active thrust and its components. The thrust on the full height is $$P_a = \tfrac{1}{2}\gamma H^2 K_a = \tfrac{1}{2}(20)(9.2)^2(0.3526) = 298.5\ \text{kN/m}$$ acting at $\delta$ to the normal of the back face, that is at $(\delta+\beta) = 25.13^{\circ}$ below the horizontal, at $H/3 = 3.067$ m above the base underside. Resolving, $$P_{ah} = P_a\cos(25.13^{\circ}) = 270.3\ \text{kN/m}, \qquad P_{av} = P_a\sin(25.13^{\circ}) = 126.8\ \text{kN/m}$$
Overturning moment about the toe. Only the horizontal component overturns; the vertical component and every weight resist. Hence $$M_O = P_{ah}\times\frac{H}{3} = 270.3 \times 3.067 = 828.9\ \text{kN}\cdot\text{m/m}$$
Weights and their lever arms. The base slab has area $5.0 \times 1.2 = 6.0$ m$^2$, so $W_1 = 6.0(24) = 144.0$ kN/m at $\bar x_1 = 2.50$ m. The stem is a trapezium of area $\tfrac{1}{2}(2.5+1.0)(8.0) = 14.0$ m$^2$, giving $W_2 = 14.0(24) = 336.0$ kN/m; decomposing it into a central rectangle (area 8.0 at $x = 1.50$), a front triangle (area 2.0 at $x = 0.833$) and a rear triangle (area 4.0 at $x = 2.333$) places its centroid at $\bar x_2 = 23.0/14.0 = 1.643$ m. The soil resting on the heel occupies the rectangle $3.0 \le x \le 5.0$ over the full 8.0 m (area 16.0 at $x = 4.00$) plus the triangle between the battered face and the vertical through $x = 3.0$ (area 4.0 at $x = 2.667$), a total of 20.0 m$^2$, so $W_3 = 20.0(20) = 400.0$ kN/m at $\bar x_3 = 74.67/20.0 = 3.733$ m.
Resisting moment about the toe. The vertical component of the thrust acts on the back-face plane, whose $x$-ordinate at the level of application ($y = 3.067$ m) is $x = 3.0 - (3.067-1.2)/8.0 = 2.767$ m. Summing, $$M_R = 144.0(2.50) + 336.0(1.643) + 400.0(3.733) + 126.8(2.767)$$ which gives $M_R = 360.0 + 552.0 + 1493.3 + 350.7 = 2756.0$ kN·m/m. The passive resistance of the 2.0 m of soil in front of the toe has deliberately been ignored, since it can be removed by a service trench at any time.
Factor of safety against overturning. $$\boxed{FS_{OT} = \frac{M_R}{M_O} = \frac{2756.0}{828.9} = 3.32}$$ This comfortably exceeds the value of 2.0 normally required against overturning (CFEM Ch. 27 and Das §8.4), so the wall is stable in this mode.
Effect of the water table rising to the base of the wall. The factor of safety decreases. Raising the water table to base level does not change the active thrust, because the backfill above the base remains drained, but it puts the underside of the base into contact with water and introduces an uplift force. Taking the piezometric level at the top of the base slab, the uniform uplift is $u = 9.81(1.2) = 11.8$ kPa, giving $U = 11.8(5.0) = 58.9$ kN/m at $x = 2.5$ m and reducing $M_R$ by 147.2 kN·m/m, so $$FS_{OT} = \frac{2756.0-147.2}{828.9} = 3.15$$ The decrease is modest only because the water has not yet entered the retained soil. If the water continues to rise the deterioration is severe: with the backfill saturated to the surface, the soil thrust falls to $\tfrac{1}{2}(20-9.81)(9.2)^2(0.3526) = 152.1$ kN/m but a hydrostatic thrust $\tfrac{1}{2}(9.81)(9.2)^2 = 415.2$ kN/m is added, the stabilising soil weight over the heel drops from 400 to 203.8 kN/m, and a trapezoidal uplift of 274.7 kN/m acts under the base; the resisting moment collapses to about 1018 kN·m/m against an overturning moment of 1695 kN·m/m, giving $FS_{OT} \approx 0.60$ — outright failure. The wall must therefore be drained, by a granular filter blanket or geocomposite drain behind the stem discharging through weep holes or a longitudinal collector at the base.

Quantity	Value
Back-face batter from vertical, $\beta$	$7.13^{\circ}$
Coulomb active coefficient, $K_a$	0.3526
Active thrust, $P_a$	298.5 kN/m
Horizontal / vertical components	270.3 / 126.8 kN/m
Overturning moment, $M_O$	828.9 kN·m/m
Weight of base / stem / soil on heel	144.0 / 336.0 / 400.0 kN/m
Resisting moment, $M_R$	2756.0 kN·m/m
Factor of safety against overturning	3.32
With water table at base (uplift on base)	3.15 — decreases
With backfill saturated to the surface	0.60 — fails

Check — the pressure plane is a modelling choice, and it matters. This wall has a 2.0 m heel that projects well beyond the foot of the battered back face, so the soil above the heel lies partly outside the plane on which the Coulomb thrust was computed. Three defensible treatments give a wide spread, and all three are reported here as page-1 Note 6 requires:

Coulomb on the stem's back face ($\beta = 7.13^{\circ}$, $K_a = 0.3526$), with the whole soil column over the heel counted as resisting weight — the conventional examination treatment, and the one shipped above: $FS_{OT} = 3.32$.
Coulomb on the vertical plane through the heel ($\beta = 0$, $K_a = 0.2986$), with $P_{av}$ acting at $x = 5.0$ m — fully self-consistent, since the free body is then the wall plus exactly the soil inside the plane: $FS_{OT} = 3.79$.
Das's gravity-wall procedure, Coulomb on the plane from the top of the back face to the heel of the base ($\beta = 18.06^{\circ}$, $K_a = 0.4558$), with the free body reduced to the wall plus the 128.7 kN/m wedge inboard of that plane: $FS_{OT} = 2.33$.

The lowest value, 2.33, is the one to carry into design; every treatment exceeds 2.0, so the conclusion that the wall is safe against overturning is robust. Other assumptions: $\gamma_c = 24$ kN/m$^3$ for reinforced concrete; no surcharge on the backfill; passive resistance in front of the toe ignored; sliding and bearing-pressure checks not requested here but required before the wall could be accepted — on these dimensions sliding, not overturning, is the mode most likely to govern.

Topic. Overturning stability of a gravity retaining wall using Coulomb's active earth pressure theory.

Key relations. Coulomb $K_a = \cos^2(\phi'-\beta)\big/\{\cos^2\beta\cos(\delta+\beta)[1+\sqrt{\sin(\delta+\phi')\sin(\phi'-\alpha)/(\cos(\delta+\beta)\cos(\beta-\alpha))}]^2\}$; $P_a = \tfrac{1}{2}\gamma H^2K_a$ acting at $H/3$ and at $(\delta+\beta)$ to the horizontal; $FS_{OT} = \sum M_R/\sum M_O$ about the toe, with $FS \ge 2$ required.

Why this works. Coulomb's theory finds the trial wedge that maximises the thrust on the wall, allowing for friction on the wall face and for a face that is not vertical. Wall friction is genuinely beneficial: it tilts the resultant downward, which both reduces the horizontal component that overturns the wall and adds a vertical component that resists. The batter of the back face works the other way here — because the top of the face leans toward the toe, the wedge it must support is larger, so $K_a$ rises from 0.299 to 0.353. Taking moments about the toe rather than about the base centroid is the standard convention because the toe is the physical hinge about which a wall would rotate.

Common pitfall. Measuring $\beta$ from the horizontal instead of from the vertical, which turns $K_a = 0.35$ into a nonsense value; omitting $P_{av}$ from the resisting moment (conservative but wasteful, here worth 15 per cent of $FS$); and computing the thrust over the stem height of 8.0 m instead of the full 9.2 m to the base underside, which understates $P_a$ by a quarter because the thrust varies with $H^2$.

Source. Das, Principles of Foundation Engineering 9th ed. §8.2–8.4 and Eq. (8.13); Das, Principles of Geotechnical Engineering 9th ed. §13.7; CFEM 4th ed. Ch. 27.

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