16-Civ-B3 Geotechnical Design · May 2016

Question 9 of 9: Excavation over a weak inclined seam: planar sliding (24 marks)

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 2016 — 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. — subsurface exploration (Ch. 2), bearing capacity of shallow foundations (Ch. 3), settlement of shallow foundations including Schmertmann's method (Ch. 5), retaining walls (Ch. 8), pile foundations (Ch. 11).
B. M. Das, Principles of Geotechnical Engineering, 9th ed. — shear strength (Ch. 12), lateral earth pressure (Ch. 13), slope stability including the planar-surface analysis (Ch. 15).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for site investigation, in-situ testing, bearing resistance, deep foundations and earth-retaining structures.
R. F. Craig, Craig's Soil Mechanics, 9th ed. — effective stress, undrained versus drained behaviour, earth pressure and slope stability.
D. P. Coduto, Foundation Design: Principles and Practices, 3rd ed. — CPT correlations and settlement of shallow foundations on sand.
M. J. Tomlinson & J. Woodward, Pile Design and Construction Practice, 6th ed. — shaft adhesion in clay and driven displacement piles in sand.

Sources of design charts and assumed values (page-1 Note 6). Note 6 requires the candidate to identify the source of every design chart used and of every value assumed where the paper gives none. They are named at the point of use and collected here:

Strain-influence diagram and the C₁, C₂ correction factors (Q6) — Schmertmann, Hartman and Brown (1978), as tabulated in Das, Principles of Foundation Engineering, 9th ed., Section 5.6.
Rankine active coefficient for an inclined backfill (Q7) — Das, Principles of Geotechnical Engineering, 9th ed., Eq. (13.35).
Bearing-capacity factors N₢, Nᵤ, Nγ and the depth and load- inclination factors (Q7) — Vesic / Meyerhof as tabulated in Das, Principles of Foundation Engineering, 9th ed., Tables 3.3 and 3.4, applied to a retaining-wall base in Section 8.6.
Meyerhof bearing-capacity factor Nᵤ* for a driven pile point and the limiting point resistance (Q8) — Das, 9th ed., Section 11.9 and its interpolated Nᵤ* table; Nᵤ* = 143 at φ′ = 35°.
Adhesion factor α against cu/p₀ (Q8) — Das, 9th ed., Table 11.6 (after Terzaghi, Peck and Mesri); α = 0.68 at cu/p₀ = 0.5.
Earth-pressure coefficient K and interface friction angle δ′ for a driven high-displacement pile (Q8) — Das, 9th ed., Section 11.11: K ≈ 1.4K₀ and δ′ ≈ 0.8φ′ are assumed, and the critical-depth rule L′ = 15D is Das Eq. (11.42).
Unit weight of water γᵣ = 9.81 kN/m³ and g = 9.81 m/s² throughout; atmospheric pressure p₀ = 100 kPa.

Section A — discussion questions (7 marks each)

Question 9 — Excavation over a weak inclined seam: planar sliding (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
Excavation depth	H	5.00 m
Inclination of the weak seam	α	30°
Undrained strength of the seam	c_u	20 kPa
Friction angle of the seam	φ_u	0°
Unit weight of the stiff clay	γ	19.00 kN/m³
Groundwater table	—	deeper than H — no water pressure on the seam
Seam thickness	—	neglected
Excavation face angle	β	85° in part (a); unknown in part (b)

Find. (a) the factor of safety against sliding of the wedge on the seam when the face is cut at 85°; (b) the face angles β that would give factors of safety of 1.3 and 1.5 on the same seam.

Figure Q9. Free body for the planar analysis. The wedge is bounded by the excavation face (angle β), the original ground surface at the crest, and the weak seam (angle α) which daylights at the toe of the cut. Because φ_u = 0 the resistance is the cohesion on the seam alone and is independent of the normal force.

Approach. Treat the block of clay between the face and the seam as a rigid wedge sliding on a plane. With $\phi_u = 0$ the resistance is $c_u$ times the length of the seam and does not depend on the weight, so the factor of safety reduces to a closed-form expression in $\alpha$ and $\beta$ that can be inverted directly for part (b).

Part (a) — geometry of the sliding wedge. The seam daylights at the toe of the excavation and reaches the original ground surface at the crest level, so the wedge is a triangle bounded by the face, the crest and the seam. Its area per metre run is $$A = \tfrac{1}{2}H^{2}\left(\cot\alpha - \cot\beta\right).$$ With $H = 5.00$ m, $\cot 30^\circ = 1.7321$ and $\cot 85^\circ = 0.08749$, $$A = \tfrac{1}{2}(5.00)^{2}\left(1.7321 - 0.08749\right) = 20.557\ \text{m}^2 .$$
Weight of the wedge and length of the failure surface. $$W = \gamma A = 19.00 \times 20.557 = 390.6\ \text{kN/m},$$ $$L_a = \frac{H}{\sin\alpha} = \frac{5.00}{\sin 30^\circ} = 10.00\ \text{m}.$$
Resolve along the seam and form the factor of safety. The driving force is the component of the weight down the plane, $W\sin\alpha$; the resisting force is the cohesion mobilised over the whole plane, $c_uL_a$, with no frictional contribution because $\phi_u = 0$. Hence $$FS = \frac{c_u L_a}{W\sin\alpha} = \frac{20 \times 10.00}{390.6 \times \sin 30^\circ} = \frac{200.0}{195.3} = \boxed{FS = 1.02}.$$ The near-vertical cut is on the point of failure — nominally stable, but with a margin of only two per cent it must be regarded as unsafe.
Reduce the result to a formula in $\alpha$ and $\beta$. Substituting $A$ and $L_a$ into the definition and cancelling $H$ gives a form that is much more useful than the arithmetic above, because it isolates $\beta$: $$FS = \frac{c_u\left(H/\sin\alpha\right)} {\gamma\,\tfrac{1}{2}H^{2}\left(\cot\alpha - \cot\beta\right)\sin\alpha} = \frac{2c_u}{\gamma H \sin^{2}\alpha\left(\cot\alpha - \cot\beta\right)} .$$ Checking against part (a): $2(20)/\left[19(5)(0.25)(1.6446)\right] = 40/39.06 = 1.024$, which reproduces the answer above. Note what the formula says: the factor of safety does not depend on the strength of the stiff clay at all, only on the strength of the seam and on the geometry.
Part (b) — invert the formula for $\beta$. Rearranging, $$\cot\alpha - \cot\beta = \frac{2c_u}{\gamma H\sin^{2}\alpha\;FS}, \qquad\text{so}\qquad \cot\beta = \cot\alpha - \frac{2c_u}{\gamma H \sin^{2}\alpha\;FS}.$$ The constant group is $2c_u/\left(\gamma H\sin^{2}\alpha\right) = 40/\left(19 \times 5 \times 0.25\right) = 1.6842$, so $\cot\beta = 1.7321 - 1.6842/FS$.
Solve for FS = 1.3. $$\cot\beta = 1.7321 - \frac{1.6842}{1.3} = 1.7321 - 1.2955 = 0.43650,$$ $$\beta = \tan^{-1}\!\left(\frac{1}{0.43650}\right) = \boxed{\beta = 66.4^\circ}.$$
Solve for FS = 1.5. $$\cot\beta = 1.7321 - \frac{1.6842}{1.5} = 1.7321 - 1.1228 = 0.60924,$$ $$\beta = \tan^{-1}\!\left(\frac{1}{0.60924}\right) = \boxed{\beta = 58.6^\circ}.$$
Check the trend and the limits. Flattening the face removes material from the top of the wedge without changing the length of the seam, so the driving weight falls while the resistance stays fixed and the factor of safety rises: 1.02 at 85°, 1.30 at 66.4°, 1.50 at 58.6°. The limit is $\beta \to \alpha = 30^\circ$, at which the wedge vanishes and $FS \to \infty$; conversely $FS = 1.0$ is reached at $\cot\beta = 1.7321 - 1.6842 = 0.0479$, that is $\beta = 87.3^\circ$, so the excavation could in principle be cut to within three degrees of vertical before collapse — which is exactly why the 85° face in part (a) is so marginal.
Engineering conclusion. To achieve a routine design factor of safety of 1.5 on a temporary excavation with a weak seam of known orientation, the face must be laid back to about 58.6°, roughly 1V:0.6H. If the site is too confined for that, the alternatives are to support the face (soldier piles and lagging, soil nails or ground anchors taken through and beyond the seam so that they add a resisting force on the plane), to dewater and drain so that no water pressure can develop in a tension crack behind the crest, or to remove the seam locally. Because the mechanism is controlled by a single plane of low strength rather than by the mass of the clay, its orientation must be confirmed by mapping and by additional boreholes before the design angle is fixed — a seam dipping a few degrees more steeply would change the answer substantially.

Quantity	Value
Length of the failure surface, L_a	10.00 m
Wedge area at β = 85°	20.56 m² per metre run
Wedge weight at β = 85°, W	390.6 kN/m
Driving force, W sinα	195.3 kN/m
Resisting force, c_uL_a	200.0 kN/m
(a) Factor of safety at β = 85°	1.02
(b) Face angle for FS = 1.3	β = 66.4°
(b) Face angle for FS = 1.5	β = 58.6°
Face angle at incipient failure (FS = 1.0)	β = 87.3°

Check: wording and assumptions, recorded under page-1 Note 1.

"Soil–rock interface" in part (b). Figure 4 shows no rock; the only discontinuity in the problem is the thin clayey seam, and part (b) is plainly the continuation of part (a). The phrase has been read as referring to the same seam, and the answer is given on that basis. A candidate should state the interpretation, as Note 1 invites.
No water pressure. The Note places the water table deeper than H, so no uplift acts on the seam and no water force acts in a tension crack. This matters: a water-filled tension crack of depth z₀ = 2c_u/γ = 2.11 m at the crest would add a horizontal force of about 22 kN/m. Counted only as an extra driving thrust on the full wedge it drops the part (a) factor of safety to about 0.93; with the crack placed where it meets the seam, so that it also truncates the wedge and the resisting length, the factor of safety falls to about 0.65.
φ_u = 0 on the seam. Because the seam has no friction, the normal force on the plane is irrelevant and the answer is independent of γ except through the wedge weight. This also means the analysis is a short-term, undrained one; in the long term the seam would be governed by c′ and φ′, and if c′ is small the drained factor of safety on a plane at 30° would be approximately tanφ′/tan30°, which for a residual φ′ of 20° in a weak clayey seam is 0.63 — that is, the cut could not stand at all in the long term at any face angle. The excavation must therefore be treated as strictly temporary unless the seam is removed or the face is permanently supported.
Two-dimensional analysis. Plane strain per metre run, with no end effects and no arching from the ends of the cut; that is conservative for a short trench and correct for a long one.

Topic. Stability of a slope with a predefined planar failure surface (the "plane failure" or single-wedge analysis) under undrained conditions.

Key relations. Wedge area $A = \tfrac{1}{2}H^{2}\left(\cot\alpha - \cot\beta\right)$; failure-plane length $L_a = H/\sin\alpha$; general factor of safety on a plane $FS = \dfrac{c\,L_a + \left(W\cos\alpha - u L_a\right)\tan\phi}{W\sin\alpha}$, which for $\phi_u = 0$ and no water reduces to $FS = c_uL_a/\left(W\sin\alpha\right)$ and hence to the closed form $FS = \dfrac{2c_u}{\gamma H\sin^{2}\alpha\left(\cot\alpha - \cot\beta\right)}$.

Why this works. Ordinary slope-stability analysis has to search for the critical surface because the soil is homogeneous and the failure surface could be anywhere. Here the geology has already chosen it: a continuous seam of much lower strength than the surrounding clay is the failure surface, and the problem collapses to the statics of a single rigid block. That is what makes an exact algebraic answer possible and what makes the inversion in part (b) trivial — a circular-arc method would have required iteration. The same reasoning governs planar failures in rock slopes along a bedding plane or joint set.

Common pitfall. Using $\cot\beta - \cot\alpha$ instead of $\cot\alpha - \cot\beta$, which returns a negative area; the seam is always flatter than the face, so $\cot\alpha$ is the larger. The second is to include the strength of the stiff clay somewhere in the calculation — it plays no part, since the wedge is assumed rigid and fails only on the seam. The third is to forget that a $\phi_u = 0$ analysis is an undrained, short-term statement, and to present it as though it settled the permanent stability of the cut.

Source. Das, Principles of Geotechnical Engineering, 9th ed., Ch. 15 — finite slopes with plane failure surfaces (Culmann's analysis and its $\phi = 0$ specialisation); Craig, Craig's Soil Mechanics, 9th ed., Ch. 12 (plane translational slides).

← Question 8 Back to the paper →