07-Str-B1 · May 2015

Question 9 of 9: Factor of Safety Against Overturning of a Gravity Retaining Wall by Coulomb's Theory

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

Notes on this paper

National Examinations — May 2015 — 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 discussion questions of 7 marks each, of which any FOUR are to be answered; Section B carries four design problems of 24 marks each, of which any THREE are to be answered — a marked total of 100. The paper instructs candidates to state any interpretive assumptions, to identify the source of every design chart and assumed value, and to exercise sound engineering judgment where data are absent. All nine printed questions are worked below, because the set is intended as a study resource.

Reference texts: Das, B.M., Principles of Foundation Engineering (9th ed., Cengage) — general bearing-capacity equation, pile and pile-group capacity, consolidation settlement of footings, retaining walls; Das, B.M., Principles of Geotechnical Engineering (9th ed., Cengage) — lateral earth pressure, effective stress, consolidation theory; Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM, 4th ed., 2006) — Canadian practice for site investigation, SPT/CPT interpretation, pile design and tolerable settlement; Craig, R.F. / Knappett, J.A., Craig's Soil Mechanics (8th ed., CRC Press) — shear strength and earth-pressure theory; Duncan, J.M., Wright, S.G. & Brandon, T.L., Soil Strength and Slope Stability (2nd ed., Wiley) — fully softened and residual strengths for fissured and expansive clays; Fredlund, D.G., Rahardjo, H. & Fredlund, M.D., Unsaturated Soil Mechanics in Engineering Practice (Wiley) — swelling soils and matric suction.

Note — Figure 2 is printed over a coarse halftone. The soil-property annotations inside the photograph-style Figure 2 (Question 8) are printed over a coarse dot screen. The values used below are read from the printed figure and are: upper sand $\gamma = 15\ \text{kN/m}^3$ over 1.5 m, lower sand $\gamma_{sat} = 18\ \text{kN/m}^3$ over 1.5 m, normally consolidated clay 2.5 m thick with $w = 35\%$ and $LL = 48$, over sand; groundwater table at the underside of the footing.

Assumptions declared once, applied throughout. $\gamma_w = 9.81\ \text{kN/m}^3$; reinforced concrete $\gamma_c = 24\ \text{kN/m}^3$; specific gravity of soil solids $G_s = 2.70$ where a void ratio must be back-figured from water content; loads are vertical and concentric unless stated. Every assumption that changes a numerical answer is repeated in the question where it is used.

Question 9: Factor of Safety Against Overturning of a Gravity Retaining Wall by Coulomb's Theory (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, 1.2 m thick base slab, with a tapering stem 8.0 m high (2.5 m wide at the slab, 1.0 m at the top), a 0.5 m toe projection and a 2.0 m heel, retaining level cohesionless backfill to the top of the stem, with 2.0 m of soil in front of the wall (Figure 3).

Given data and assumed values — Question 9
Quantity	Symbol	Value
Total wall height (stem + base)	$H$	8.0 + 1.2 = 9.2 m
Base slab: width × thickness	—	5.0 m × 1.2 m
Stem width at slab / at top	—	2.5 m / 1.0 m
Toe projection / heel projection	—	0.5 m / 2.0 m
Backfill unit weight	$\gamma$	20 kN/m³
Backfill strength	$c',\ \phi'$	0 kPa, 30°
Soil–wall friction angle	$\delta = 0.6\phi'$	18°
Concrete unit weight (assumed)	$\gamma_c$	24 kN/m³

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

[Figure not reproduced: Figure 9.1 — Figure 3 redrawn to scale, with the vertical pressure plane through the back of the heel and the Coulomb active thrust resolved at δ = 18° below the horizontal, acting at H/3 above the base. See the official exam paper.]

Approach. Take the vertical plane through the back of the heel as the pressure plane, compute the Coulomb active thrust on it over the full wall height, resolve it into horizontal and vertical components, take moments about the toe of the overturning thrust and of every stabilising weight (concrete stem, base slab, soil carried on the heel and the vertical component of the thrust), and form the ratio.

Assumptions, stated because they change the number. (i) Pressure plane. The wall has a 2.0 m heel, so the plane on which the active thrust is computed is taken vertically through the back edge of the base, height $H = 9.2$ m; the soil sitting on the heel is then part of the stabilising mass. This is standard practice and avoids the physically impossible alternative of a pressure plane cutting through the concrete. (ii) $\delta = 0.6\phi' = 18^\circ$ is applied on that plane, as the question directs, and the thrust therefore acts $18^\circ$ below the horizontal. (iii) $\gamma_c = 24$ kN/m³ for the concrete; the figure does not give it. (iv) Passive resistance of the 2.0 m of soil in front of the wall is neglected — it could be removed by future excavation or scour, and it needs a wall movement the structure cannot tolerate. (v) The backfill surface is level with the top of the stem and carries no surcharge. (vi) $c' = 0$, so no tension crack or cohesion term arises. (vii) Overturning is taken about the front bottom edge of the base (the toe).

Coulomb active earth-pressure coefficient. With a vertical back face ($\beta = 90^\circ$), a horizontal backfill ($\alpha = 0$), $\phi' = 30^\circ$ and $\delta = 18^\circ$, $$K_a = \frac{\sin^2(\beta+\phi')}{\sin^2\beta\,\sin(\beta-\delta)\left[1 + \sqrt{\dfrac{\sin(\phi'+\delta)\sin(\phi'-\alpha)}{\sin(\beta-\delta)\sin(\alpha+\beta)}}\right]^2}$$ Substituting $\sin 120^\circ = 0.8660$, $\sin 72^\circ = 0.9511$, $\sin 48^\circ = 0.7431$ and $\sin 30^\circ = 0.5000$, $$K_a = \frac{0.7500}{(1)(0.9511)\left[1+\sqrt{0.3907}\right]^2} = \frac{0.7500}{0.9511(2.6408)} = \boxed{0.2986}$$ For comparison, Rankine's $K_a$ with $\delta = 0$ would be 0.3333; wall friction reduces the thrust and, more importantly, tilts it downward.
Active thrust and its components. For a uniform cohesionless backfill, $$P_a = \tfrac{1}{2}\gamma H^2K_a = 0.5(20)(9.2)^2(0.2986) = 252.8\ \text{kN/m}$$ $$P_h = P_a\cos\delta = 252.8\cos 18^\circ = 240.4\ \text{kN/m} \qquad P_v = P_a\sin\delta = 252.8\sin 18^\circ = 78.1\ \text{kN/m}$$ The pressure distribution is triangular, so the thrust acts at $H/3 = 3.067$ m above the base, on the plane 5.0 m from the toe.
Overturning moment about the toe. Only the horizontal component overturns; the vertical component acts on the heel side and resists: $$M_o = P_h\left(\frac{H}{3}\right) = 240.4(3.067) = \boxed{737.2\ \text{kN}\cdot\text{m per metre}}$$
Weight and lever arm of the base slab. Measuring $x$ from the toe, $$W_1 = (5.0)(1.2)(24) = 144.0\ \text{kN/m}, \qquad \bar{x}_1 = 2.50\ \text{m}, \qquad M_1 = 360.0\ \text{kN}\cdot\text{m}$$
Weight and lever arm of the tapering stem. The stem is a trapezoid 8.0 m high, 2.5 m wide at the slab (from $x = 0.5$ to $x = 3.0$) and 1.0 m at the top (from $x = 1.0$ to $x = 2.0$). Splitting it into a central rectangle and two triangles gives areas of 8.0, 2.0 and 4.0 m² with centroids at 1.500, 0.833 and 2.333 m, so $$A = 14.0\ \text{m}^2, \qquad W_2 = 14.0(24) = 336.0\ \text{kN/m}, \qquad \bar{x}_2 = \frac{8.0(1.500)+2.0(0.833)+4.0(2.333)}{14.0} = 1.643\ \text{m}$$ $$M_2 = 336.0(1.643) = 552.0\ \text{kN}\cdot\text{m}$$
Weight and lever arm of the soil carried on the heel. The soil between the battered back face of the stem and the pressure plane is a trapezoid 8.0 m high, 2.0 m wide at the slab and 3.0 m wide at the top: a rectangle of 16.0 m² centred at 4.000 m plus a triangle of 4.0 m² centred at 2.667 m. $$A = 20.0\ \text{m}^2,\qquad W_3 = 20.0(20) = 400.0\ \text{kN/m}, \qquad \bar{x}_3 = \frac{16.0(4.000)+4.0(2.667)}{20.0} = 3.733\ \text{m}$$ $$M_3 = 400.0(3.733) = 1493.3\ \text{kN}\cdot\text{m}$$
Moment of the vertical component of the thrust. $P_v$ acts on the pressure plane, 5.0 m from the toe: $$M_4 = 78.1(5.00) = 390.6\ \text{kN}\cdot\text{m}$$
Total resisting moment and the factor of safety. Summing, $$M_R = 360.0 + 552.0 + 1493.3 + 390.6 = 2795.9\ \text{kN}\cdot\text{m per metre}$$ $$FS_{overturning} = \frac{M_R}{M_o} = \frac{2795.9}{737.2} = \boxed{3.79}$$ This comfortably exceeds the minimum of 2.0 normally required against overturning, and the wall is safe in this mode. (Sliding and bearing pressure would still need separate checks; with $P_h = 240$ kN/m and a total vertical load of 958 kN/m the sliding factor of safety on a base friction angle of $\tfrac{2}{3}\phi'$ is about 1.5, which is marginal and is where this wall would in fact be governed.)

Effect of a rising groundwater table. Read literally, the water rising only to the base of the wall leaves the retained backfill drained: no water pressure acts on the pressure plane, the backfill unit weight is unchanged over the full 9.2 m, and the overturning factor of safety is therefore essentially unaltered at 3.79. What does change is everything that depends on effective stress beneath the base — the soil under the footing becomes buoyant, so the bearing capacity falls by up to a half and the sliding resistance falls with the reduced effective normal stress on the base. Both of those checks become significantly less favourable even though the overturning check does not move.

If the water rises any higher, into the retained fill, the factor of safety falls sharply, and it is worth quantifying the limiting case in which the backfill is fully submerged. The soil then contributes only its buoyant weight, $\gamma' = 20 - 9.81 = 10.19$ kN/m³, while a full hydrostatic thrust is added:

$$P_a' = \tfrac{1}{2}\gamma' H^2K_a = 128.8\ \text{kN/m}, \qquad P_w = \tfrac{1}{2}\gamma_wH^2 = 0.5(9.81)(9.2)^2 = 415.2\ \text{kN/m}$$

The horizontal thrust more than doubles to $122.5 + 415.2 = 537.6$ kN/m, giving $M_o = 1648.8\ \text{kN}\cdot\text{m}$, while the soil on the heel now weighs only $203.8$ kN/m so that $M_R$ falls to $1871.8\ \text{kN}\cdot\text{m}$ and

$$FS_{overturning} = \frac{1871.8}{1648.8} = \boxed{1.14}$$

a collapse from 3.79 to barely above unity. The factor of safety decreases, and the reason is that water is far more damaging than soil: it exerts its full hydrostatic pressure with no $K_a$ reduction and no shear strength, while simultaneously halving the stabilising weight. This is why every retaining wall must be provided with a drainage system — a free-draining granular blanket or geocomposite drain behind the wall, weep holes or a longitudinal perforated collector pipe at the base, and a low-permeability cap at the surface — and why the drainage detail, not the earth-pressure calculation, is what most often distinguishes a wall that survives from one that fails.

Final results — Question 9
Quantity	Symbol	Value
Coulomb active coefficient ($\beta=90^\circ,\alpha=0,\delta=18^\circ$)	$K_a$	0.2986
Active thrust on the pressure plane	$P_a$	252.8 kN/m
Horizontal / vertical components	$P_h$ / $P_v$	240.4 / 78.1 kN/m
Overturning moment about the toe	$M_o$	737.2 kN·m/m
Base slab: weight, arm, moment	$W_1,\bar{x}_1,M_1$	144.0 kN/m, 2.500 m, 360.0 kN·m/m
Stem: weight, arm, moment	$W_2,\bar{x}_2,M_2$	336.0 kN/m, 1.643 m, 552.0 kN·m/m
Soil on heel: weight, arm, moment	$W_3,\bar{x}_3,M_3$	400.0 kN/m, 3.733 m, 1493.3 kN·m/m
Moment of vertical thrust component	$M_4$	390.6 kN·m/m
Total resisting moment	$M_R$	2795.9 kN·m/m
Factor of safety against overturning (dry)	$FS$	3.79
Water at base of wall only	$FS$	3.79 (unchanged; bearing and sliding worsen)
Backfill fully submerged (limiting case)	$FS$	1.14 — a large decrease

Topic. Stability of a gravity retaining wall against overturning, using Coulomb's active earth pressure with wall friction.

Key relations. Coulomb $K_a$ as written in Step 1; $P_a = \tfrac{1}{2}\gamma H^2K_a$ acting at $H/3$ and inclined at $\delta$ to the normal of the pressure plane; $FS_{ot} = \sum M_R / \sum M_o$ with moments taken about the toe; submerged case $P = \tfrac{1}{2}\gamma'H^2K_a + \tfrac{1}{2}\gamma_wH^2$; minimum acceptable $FS_{ot} \ge 2$, $FS_{sliding} \ge 1.5$.

Why this works. Overturning is a rigid-body rotation about the toe, so the check is a straightforward moment balance: everything with a downward force and a lever arm measured from the toe resists, and the horizontal thrust drives. Coulomb's wedge analysis supplies the thrust by finding the trial wedge that maximises the force the wall must provide, and including $\delta$ tilts the resultant downward, which both reduces the horizontal component and adds a stabilising vertical component — so ignoring wall friction is conservative. Water changes the problem qualitatively because it transmits pressure isotropically: the hydrostatic term carries no $K_a$ factor at all.

Common pitfall. Applying Coulomb's $K_a$ to the backfill and then also counting the soil above the heel as an independent surcharge on the same plane — the wedge analysis already accounts for it. The other frequent error is omitting $P_v$, or including it as an overturning rather than a resisting moment; here it contributes 391 of the 2796 kN·m of resistance. Taking moments about the centre of the base instead of the toe is a third.

Source. Das, Principles of Foundation Engineering, 9th ed., §8.3–§8.7 (Retaining walls: proportioning, stability against overturning, sliding and bearing failure) and §7.6 (Coulomb's active pressure); CFEM 4th ed., §26 and §27 (Earth Pressures; Design of Retaining Structures and Drainage).

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