07-Str-B5 · December 2018

Question 4 of 6: Retaining Structures

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

Notes on this paper

Paper format: National Exams, December 2018 — 07-Str-B5 Foundation Engineering. Three hours, open book, any non-communicating calculator permitted. Six questions of equal value (30 marks each); five constitute a complete paper and only the first five appearing in the answer book are marked. All six are solved here. This subject is pure geotechnical engineering — the cover page names it Foundation Engineering.

Reference texts (the books an open-book candidate should have on the desk for this subject):

Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the Canadian limit-states framework, geotechnical resistance factors, frost depth, pile design.
B. M. Das, Principles of Foundation Engineering, 9th ed. — bearing capacity with shape/depth/inclination factors, retaining walls, driven piles (§12), drilled shafts (§13), pile groups.
R. F. Craig & J. Knappett, Craig's Soil Mechanics, 9th ed. — effective stress, Bishop's simplified method, consolidation settlement.
M. J. Tomlinson & J. Woodward, Pile Design and Construction Practice, 6th ed. — adhesion factors, the equivalent-raft settlement method, block failure of pile groups.
L. C. Reese & M. W. O'Neill, Drilled Shafts: Construction Procedures and Design Methods (FHWA) — the α* = 0.55 shaft rule and the exclusion zones used in Question 5.
D. W. Taylor, Fundamentals of Soil Mechanics; A. W. Bishop & N. Morgenstern, Stability coefficients for earth slopes (Géotechnique, 1960) — the slope-stability charts behind Question 3.

Check — assumptions adopted across this paper. This examination omits at least one parameter that each design step needs. Every adoption is made explicitly here and flagged again at the point of use. (1) Questions 2 and 5 give no groundwater table. Both profiles are clays quoted by undrained shear strength, so they are saturated; the water table is therefore taken at ground level and effective stresses are computed with submerged unit weights. This is the conservative choice for settlement, and the sensitivity is reported in each answer (Question 2 settles 293 mm with the water table at surface against 153 mm with no free water). (2) Question 2 gives no adhesion factor. α is taken from Das's Table 12.6 (Terzaghi, Peck & Mesri, α against c_u/p_a) because that table needs no assumed effective-stress profile; the answer tabulates what Tomlinson's and the API's rules would give instead, and identifies the α at which the group size changes. (3) Question 1 does not say whether the quoted leg loads are factored. They are treated as specified (unfactored) loads, and both a serviceability check against the 200 kPa SLS capacity and an ultimate check at 1.4 × leg load against the 300 kPa factored resistance are carried out; the governing one is stated. (4) Question 5 gives no preconsolidation pressure. The requested Skempton correlation is used on the virgin compression line, and the (much smaller) settlement that follows if the profile's own c_u/σ′_v0 ratio of about 1.1 is honoured as an OCR of 2 is reported alongside. (5) Concrete unit weight is taken as 24 kN/m³ and γ_w as 9.81 kN/m³ throughout. (6) Where a question states only "factor of safety", the gross definition (ultimate bearing capacity over total applied pressure) is used, and the convention is restated in each answer.

Question 4 — Retaining Structures (30 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 of total height 6 m (5 m stem plus 1 m base), 4 m wide at the base, retaining a level granular-A backfill and founded 1 m into a cohesionless foundation soil.

Given data — Question 4
Quantity	Symbol	Value
Total retained height (Rankine plane through the heel)	H	6.0 m
Base width, thickness, embedment	B, t, D_f	4.0 m, 1.0 m, 1.0 m
Toe / stem at base / heel	—	0.6 m / 0.5 m / 2.9 m
Stem thickness, top and bottom	—	0.4 m, 0.5 m
Backfill: friction angle, unit weight	φ′₁, γ₁	40°, 20 kN/m³
Foundation soil: friction angle, cohesion, unit weight	φ′₂, c′, γ₂	35°, 0, 19 kN/m³
Concrete unit weight	γ_c	24 kN/m³
Groundwater	—	4 m below the base (part d); at base level (part e)

Find. The Rankine pressure distribution, then factors of safety against overturning about the toe, sliding on the base, and bearing failure — with the bearing check repeated for a water table risen to founding level.

Figure 6 — the cantilever wall and the triangular Rankine active pressure on the vertical plane through the back of the heel. The 290 kN/m of backfill sitting on the heel supplies 76% of the resisting moment.

Approach. Take a vertical Rankine plane through the back of the heel and treat the soil wedge above the heel as part of the wall. Compute the active thrust on that plane, sum the vertical weights and their moments about the toe, and form the three factors of safety in turn; for bearing, locate the resultant, work out the eccentricity and the base pressure distribution, and compare the maximum pressure with a Meyerhof capacity computed on the effective width with inclination factors.

Part (a) — Rankine pressure distribution. The back face of the stem is vertical and the backfill is level, so Rankine's conditions are met on the vertical plane through the heel, of height $H = 6$ m: $$K_a = \tan^2\!\left(45 - \frac{\phi^{\prime}_1}{2}\right) = \tan^2 25^\circ = 0.2174$$
The active pressure increases linearly with depth, $\sigma_a = K_a\gamma_1 z$, from zero at the surface to
$$\sigma_a(6\ \text{m}) = 0.2174 (20)(6) = 26.09\ \text{kPa}$$
a triangular distribution whose resultant is
$$\boxed{P_a = \tfrac12 K_a \gamma_1 H^2 = 78.28\ \text{kN/m},\ \ \text{acting}\ H/3 = 2.0\ \text{m above the base}}$$
In front, over the 1 m of embedment, the passive pressure with $K_p = \tan^2(45+17.5^\circ) = 3.690$ reaches 70.1 kPa at the base and gives $P_p = 35.06$ kN/m at 0.33 m above the base. It is shown on the figure but excluded from the overturning and bearing checks, and reported separately in the sliding check, because the soil in front of a toe can be excavated for services at any time in the wall's life.

Part (b) — weights and resisting moment. Working per metre run and taking moments about the front bottom edge of the toe:

Vertical forces and their moments about the toe
Component	W (kN/m)	x̄ (m)	M_R (kN·m/m)
Stem, rectangular part (0.4 × 5.0)	48.0	0.900	43.20
Stem, tapered part (0.1 × 5.0 / 2)	6.0	0.633	3.80
Base slab (4.0 × 1.0)	96.0	2.000	192.00
Backfill over the heel (2.9 × 5.0)	290.0	2.550	739.50
Total	440.0	—	978.50

There is no soil over the toe, because the figure puts the top of the base slab at the front ground surface. The overturning moment is the active thrust times its lever arm,

$$M_O = P_a\frac{H}{3} = 78.28 (2.0) = 156.56\ \text{kN}\!\cdot\!\text{m/m}$$ $$\boxed{FS_{overturning} = \frac{978.5}{156.56} = 6.25 \;>\; 2.0}$$

Part (c) — sliding. For concrete cast against a granular foundation soil, the base friction angle is taken as $\delta = \tfrac23\phi^{\prime}_2 = 23.3^\circ$, with $c^{\prime} = 0$ so there is no base adhesion: $$F_R = \left(\sum V\right)\tan\delta = 440.0\tan 23.33^\circ = 189.8\ \text{kN/m}$$ $$\boxed{FS_{sliding} = \frac{189.8}{78.28} = 2.43 \;>\; 1.5}$$
Counting the passive wedge in front of the toe would raise this to $(189.8 + 35.1)/78.28 = 2.87$, but the wall passes comfortably without it, so the conservative value is the one to report.
Part (d) — locate the resultant and find the base pressures. The line of action of the resultant crosses the base at $$\bar{x} = \frac{M_R - M_O}{\sum V} = \frac{978.5 - 156.56}{440.0} = 1.868\ \text{m from the toe}$$ $$e = \frac{B}{2} - \bar{x} = 2.000 - 1.868 = 0.132\ \text{m} \;<\; \frac{B}{6} = 0.667\ \text{m}$$
The resultant is inside the middle third, so the base is everywhere in compression and the linear pressure distribution applies:
$$q_{max,min} = \frac{\sum V}{B}\left(1 \pm \frac{6e}{B}\right) = 110.0(1 \pm 0.198)$$ $$q_{max} = 131.8\ \text{kPa}, \qquad q_{min} = 88.2\ \text{kPa}$$
Part (d) — bearing capacity of the base. The footing is a strip, so shape factors are unity, and it carries an inclined, eccentric load. Using the effective width $B^{\prime} = B - 2e = 3.736$ m with $\phi^{\prime}_2 = 35^\circ$: $$N_q = 33.30, \qquad N_\gamma = (N_q - 1)\tan(1.4\phi^{\prime}) = 37.15$$
The depth factor is $d_q = 1 + 2\tan\phi^{\prime}(1-\sin\phi^{\prime})^2 (D_f/B) = 1.064$, and the load inclination is $\alpha = \arctan(P_a/\sum V) = 10.09^\circ$, giving $i_q = (1-\alpha/90)^2 = 0.788$ and $i_\gamma = (1-\alpha/\phi^{\prime})^2 = 0.507$. With the water table 4 m down — just below the $B^{\prime} = 3.74$ m deep failure zone — the bulk unit weight applies throughout:
$$q_u = qN_q d_q i_q + \tfrac12\gamma_2 B^{\prime}N_\gamma i_\gamma = 530.5 + 668.1$$ $$\boxed{q_u = 1198.6\ \text{kPa}, \qquad FS_{bearing} = \frac{1198.6}{131.8} = 9.1 \;>\; 3.0}$$
Part (e) — water table risen to base level. The water table is now at the underside of the base slab. Three things could change; only one does. The retained soil is still above the water, so $P_a$, $\sum V$, $e$ and $q_{max}$ are unchanged; and the 1 m of soil above founding level in front of the wall is likewise above the water, so the surcharge term $q = \gamma_2 D_f = 19$ kPa is unchanged. What changes is the soil within the failure wedge below the base, which is now submerged: $$\gamma^{\prime}_2 = 19 - 9.81 = 9.19\ \text{kN/m}^3$$
Only the self-weight term of the bearing-capacity equation is affected, and it is cut by the ratio $9.19/19 = 0.484$:
$$q_u = 530.5 + \tfrac12 (9.19)(3.736)(37.15)(0.507) = 530.5 + 323.1$$ $$\boxed{q_u = 853.6\ \text{kPa}, \qquad FS_{bearing} = \frac{853.6}{131.8} = 6.5}$$
The ultimate capacity falls by 29% and the factor of safety from 9.1 to 6.5, still far above the required 3. The wall is insensitive to the rise only because it is so lightly stressed; on a wall proportioned closer to the limit this single change is often what turns an adequate design into an inadequate one, which is why the question is worth 5 marks.

Final results — Question 4
Quantity	Value
(a) Rankine K_a, pressure at the base, thrust	0.2174, 26.09 kPa, P_a = 78.28 kN/m at 2.0 m
Total vertical load / resisting moment	440.0 kN/m / 978.5 kN·m/m
Overturning moment	156.56 kN·m/m
(b) FS against overturning	6.25
(c) FS against sliding (passive ignored)	2.43 (2.87 with passive)
Eccentricity / base pressures	0.132 m; q_max = 131.8, q_min = 88.2 kPa
(d) q_u and FS against bearing failure	1198.6 kPa, FS = 9.1
(e) q_u and FS with GWT at base level	853.6 kPa, FS = 6.5 (29% loss)

Topic. External stability of a cantilever retaining wall: overturning, sliding and bearing, and the effect of the groundwater table on bearing capacity.

Key relations. $K_a = \tan^2(45 - \phi^{\prime}/2)$, $P_a = \tfrac12 K_a\gamma H^2$; $FS_{ot} = \sum M_R/\sum M_O$; $FS_{sl} = \sum V\tan\delta/P_a$; $e = B/2 - (\sum M_R - \sum M_O)/\sum V$; $q_{max} = \frac{\sum V}{B}(1 + 6e/B)$; Meyerhof $q_u = qN_qd_qi_q + \tfrac12\gamma B^{\prime}N_\gamma i_\gamma$.

Why this works. The Rankine analysis of a cantilever wall relies on a trick: the soil sitting on the heel moves with the wall, so the "wall" for equilibrium purposes is the concrete plus that wedge, and the earth pressure is taken on the vertical plane at the back of the heel where Rankine's assumption of no shear on a vertical plane is genuinely satisfied. That soil weight is what makes a cantilever wall work — here it is 290 of the 440 kN/m of vertical load and 76% of the resisting moment. The bearing check must use the maximum edge pressure rather than the average, because a bearing failure initiates under the most heavily loaded edge, and it must use $B^{\prime}$ in the $N_\gamma$ term because only that width of soil is genuinely mobilised.

Common pitfall. Taking the Rankine plane at the back face of the stem instead of the back of the heel, which loses the backfill weight over the heel and roughly halves every factor of safety. A second is forgetting that the base slab is 1 m thick, so the retained height for $P_a$ is 6 m and not the 5 m in the question's opening sentence. A third, in part (e), is submerging the backfill as well: the water table is at the base of the wall, so the retained soil stays dry and the active thrust does not change.

Source. Das, Principles of Foundation Engineering 9th ed., §8.3–8.6 (stability of retaining walls, worked in exactly this order) and §3.6 (water table correction); CFEM 4th ed. §27.

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