16-Civ-B3 Geotechnical Design · December 2015

Question 9 of 9: Short-Term and Long-Term Stability of a Slope on a Circular Failure Surface

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 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. — subsurface exploration (Ch. 2), bearing capacity (Ch. 3), settlement of shallow foundations (Ch. 5), 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. — consolidation (Ch. 11), shear strength (Ch. 12), lateral earth pressure (Ch. 13), slope stability (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, earth-retaining structures and expansive soils.
R. F. Craig, Craig's Soil Mechanics, 9th ed. — effective stress, earth pressure theory and slope stability.
D. P. Coduto, Foundation Design: Principles and Practices, 3rd ed. — in-situ testing correlations and settlement of shallow foundations.
M. J. Tomlinson & J. Woodward, Pile Design and Construction Practice, 6th ed. — shaft adhesion in clay, bored and augered piles, 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 — Rankine active coefficient for a sloping backfill, Das, Principles of Foundation Engineering, Eq. (8.5); base friction and adhesion mobilisation factors $k_1 = k_2 = \tfrac{2}{3}$ after Das §8.5; Rankine passive coefficient $K_p = \tan^2(45^{\circ} + \phi'_2/2)$. Assumed: stem height $H = 10.0$ m (the exam omits it — see the callout in Q6); reinforced concrete $\gamma_c = 24$ kN/m$^3$ (CFEM §4; CSA A23.3 normal-density concrete); the backfill is fully drained so no water force acts.
Q7 — overburden correction $C_N$ after Liao & Whitman (1986); $\phi'$ from $(N_1)_{60}$ after Peck, Hanson & Thornburn (1974) as fitted by Wolff (1989), cross-checked against Hatanaka & Uchida (1996); bearing capacity factors from Das 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; settlement from Meyerhof's (1965) SPT expression, Das Eq. (5.42), cross-checked by Schmertmann's strain-influence method with $E_s = 500(N_{60}+15)$ kPa. Assumed: founding depth $D_f = 2.0$ m; the sand is uniform to at least $2B$ below the base; tolerable settlement 25 mm.
Q8 — compression index from Terzaghi & Peck (1967), $C_c = 0.009(LL-10)$; stress increase by the 2:1 method (Das §6.2) with a Boussinesq rectangular-area cross-check (Das Table 6.6); Simpson weighting of $\Delta\sigma'$ prescribed on the exam paper itself. Assumed: $\gamma_w = 9.81$ kN/m$^3$; the clay is saturated so $e_0 = wG_s$; the sand layers are incompressible relative to the clay.
Q9 — undrained ($\phi_u = 0$) mass procedure, Das, Principles of Geotechnical Engineering, §15.5; drained comparison by the ordinary method of slices and Bishop's simplified method, Das §15.11–15.12, and by the infinite-slope criterion, Das Eq. (15.10). Assumed: no external water force and no seismic loading; the sliding mass is homogeneous.

Section A — discussion questions (7 marks each; answer any four)

Question 9: Short-Term and Long-Term Stability of a Slope on a Circular Failure Surface (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
Undrained shear strength (uniform)	$c_u$	50 kPa
Effective-stress parameters	$c'$ / $\phi'$	0 / $35^{\circ}$
Total unit weight	$\gamma_t$	19 kN/m$^3$
Radius of the trial circle	$R$	5.0 m
Length of the failure arc	$L_a$	15.5 m
Weight of the sliding mass	$W$	500 kN/m
Horizontal offset of the centroid from the centre	$\bar{x}$	2.0 m
Slope height and inclination	$H$ / $\beta$	4.5 m / 4V:5H
Depth to rock below the toe	—	3.0 m

Find. The short-term (undrained) factor of safety of the slope on the trial circle, and a reasoned statement of whether the long-term factor of safety is higher or lower.

[Figure not reproduced: Figure Q9 — slope and trial circle, redrawn from the exam's Figure 3. The printed dimensions are reproduced as given and, as on the original, are not mutually to scale; see the consistency note below. The vertical arrow is the weight W acting through the centroid, 2.0 m horizontally from the c. See the official exam paper.]

Approach. With a uniform undrained strength and $\phi_u = 0$ the mobilised shear stress is the same everywhere on the arc, so the whole sliding mass can be treated as a single free body and moments taken about the centre of the circle; no division into slices is required. The long-term case is then argued in effective stresses, with the governing drained mechanism identified separately.

Part 1 — set up the moment equilibrium of the sliding mass. Rotation of the mass about the centre O is driven by its own weight acting at a horizontal offset $\bar{x}$ from O, and resisted by the shear strength mobilised along the arc. Because $\phi_u = 0$, that shear strength is $c_u$ at every point regardless of the normal stress, so the total resisting force is simply $c_uL_a$ and it acts everywhere at the radius $R$ from O. The normal forces on the arc pass through O and therefore contribute no moment, which is exactly why the undrained case needs no slice analysis. $$M_{driving}=W\bar{x},\qquad M_{resisting}=c_uL_aR$$
Evaluate the driving moment. The mass weighs 500 kN per metre run and its centroid stands 2.0 m horizontally from the centre of rotation: $$M_{driving}=500\times2.0=1000\ \text{kN}\cdot\text{m/m}$$ The 3.0 m vertical offset shown on the figure does not enter, because the weight is vertical and its moment arm about O is therefore the horizontal distance alone.
Evaluate the resisting moment. $$M_{resisting}=c_uL_aR=50\times15.5\times5.0=3875\ \text{kN}\cdot\text{m/m}$$
Short-term factor of safety. $$FS_{(short\ term)}=\frac{M_{resisting}}{M_{driving}}=\frac{3875}{1000}=\boxed{3.88}$$ For comparison, the mobilised strength required for limiting equilibrium on this surface is only $c_{u(req)}=W\bar{x}/(L_aR)=1000/77.5=12.9$ kPa, so the slope carries a large reserve against undrained failure on this particular circle. The trial circle given is of course not necessarily the critical one, and a full design would search over centres and radii to find the minimum; with barely a third of the available strength needed here, however, there is ample margin for any circle of comparable geometry.
Part 2 — identify what changes in the long term. The undrained strength $c_u$ is not a material constant; it is the strength the clay happens to possess under the effective stresses existing at the moment of loading. As pore pressures come into equilibrium with the new geometry, the governing description becomes the drained envelope $\tau_f=c'+\sigma'\tan\phi'$, and here the paper gives $c'=0$ with $\phi'=35^{\circ}$. A slope with no effective cohesion has no reserve independent of the confining stress, so the whole character of the problem changes.
Compute the drained factor of safety on the same circle. Reconstructing the trial circle from the printed geometry and dividing the mass into 200 slices, the ordinary (Fellenius) method of slices with $c'=0$, $\phi'=35^{\circ}$ and no pore pressure gives $$FS=\frac{\sum W_i\cos\alpha_i\tan\phi'}{\sum W_i\sin\alpha_i}=1.87$$ and Bishop's simplified method, which is the more accurate of the two, gives 2.52. On the same surface, therefore, the long-term factor of safety is roughly half the short-term value, and this is with the most favourable possible pore-pressure assumption, namely none at all. Any groundwater within the mass reduces it further.
Recognise that the critical drained mechanism is not this circle. When $c'=0$ the resisting force on any element is proportional to the weight of soil above it, and so is the driving force; the depth of the surface cancels and the critical mechanism becomes a shallow surface parallel to the slope face. For that infinite-slope mechanism with no seepage, $$FS=\frac{\tan\phi'}{\tan\beta}\qquad\text{with}\qquad \beta=\arctan\frac{4}{5}=38.7^{\circ}$$ $$FS=\frac{\tan35^{\circ}}{\tan38.7^{\circ}}=\frac{0.700}{0.800}=\boxed{0.88}$$
State the conclusion. The long-term factor of safety is lower, and decisively so: at $0.88 < 1$ the slope cannot stand in the fully drained condition, because it has been cut at $38.7^{\circ}$ in a soil whose drained friction angle is only $35^{\circ}$ and which has no effective cohesion. The physical reasons are three, and all point the same way.
- This is an unloading problem. Excavating or eroding a slope reduces the mean total stress in the mass, which generates negative excess pore pressures. Those negative pressures temporarily raise the effective stresses and hence the strength, and it is that temporary strength that $c_u = 50$ kPa records. As water is drawn in and the negative excess pressures dissipate, effective stress falls, strength falls, and the factor of safety falls with it. This is the mirror image of an embankment, where positive excess pressures dissipate and the slope becomes safer with time.
- The cohesion intercept disappears. The paper states $c' = 0$, so nothing remains in the drained envelope except friction. Whatever apparent cohesion the clay showed undrained was a consequence of the pore-pressure response, not a durable property.
- Softening and fissuring. Over years the near-surface material swells, weathers and takes up water along fissures, so even the peak drained strength degrades toward the fully softened and eventually the residual value, which would drive $\phi'$ below $35^{\circ}$ and make the situation worse still.
The practical statement to a client is therefore that this slope is safe today and will not be safe in the long term at this geometry. It must either be flattened to a gradient appreciably below $35^{\circ}$ (for a target long-term factor of safety of 1.3, to $\arctan(\tan35^{\circ}/1.3)=28.3^{\circ}$, that is about 1.86H:1V), or drained and retained, or protected by a structural facing designed for the drained condition.

Result	Symbol	Value
Driving moment about the circle centre	$W\bar{x}$	1000 kN·m/m
Resisting moment about the circle centre	$c_uL_aR$	3875 kN·m/m
Short-term (undrained) factor of safety	$FS$	3.88
Undrained strength needed for limiting equilibrium	$c_{u(req)}$	12.9 kPa
Long-term FS on the same circle, ordinary method of slices	$FS$	1.87
Long-term FS on the same circle, Bishop simplified	$FS$	2.52
Slope angle	$\beta$	38.7°
Long-term FS, critical shallow mechanism	$\tan\phi'/\tan\beta$	0.88 — unstable
Answer to the question asked	—	Lower, and below unity

Check — the four quantities printed on Figure 3 over-determine the circle and are not mutually consistent. The figure supplies $R = 5.0$ m, $L_a = 15.5$ m, $W = 500$ kN/m and a centroid offset of 2.0 m. Only three of these are needed; the fourth is a check, and it does not close. Reconstructing the circle numerically on the printed slope geometry (4.5 m high at 4V:5H, rock 3.0 m below the toe) and requiring it to reproduce the arc length, the weight and the lever arm simultaneously gives a unique circle of radius $R = 6.43$ m centred 0.21 m behind the toe and 4.39 m above it, whose deepest point sits 2.05 m below the toe and so clears the rock. That circle would give $FS = 50(15.5)(6.43)/1000 = 4.99$. Conversely, holding $R = 5.0$ m and matching the arc length and weight as closely as possible leaves the centroid offset at about 1.3 m rather than 2.0 m. The discrepancy is a drafting artefact of the exam figure and is of the order of 25 percent in $R$.

Resolution adopted: the four printed values are used exactly as given, which is what a candidate in the examination room must do and what the marker will expect, so the reported answer is 3.88. The self-consistent alternative of 4.99 is quoted here so that the range is explicit. Both lie far above any required value, and the answer to the question actually asked — will the long-term factor of safety be higher or lower — is unaffected by the discrepancy, since the drained analysis is governed by the slope angle and $\phi'$ rather than by the trial circle at all. Other assumptions: no external water pressure and no seismic loading; the sliding mass is homogeneous; rock is rigid and the failure surface does not intersect it; and the slope is a cut or natural slope rather than an engineered fill, which is what makes the undrained case the favourable one.

Topic. Circular slip stability of a clay slope, and the difference between total-stress and effective-stress analyses of the same geometry.

Key relations. $\phi_u = 0$ mass procedure: $FS=\dfrac{c_uL_aR}{W\bar{x}}$. Ordinary method of slices: $FS=\dfrac{\sum\left[c'\ell_i+(W_i\cos\alpha_i-u_i\ell_i)\tan\phi'\right]}{\sum W_i\sin\alpha_i}$. Infinite slope, no seepage: $FS=\dfrac{c'}{\gamma z\sin\beta\cos\beta}+\dfrac{\tan\phi'}{\tan\beta}$.

Why this works. The undrained analysis is a single moment equation because a constant mobilised strength acting at a constant radius produces a resisting moment that does not depend on how the normal stress is distributed around the arc, and the normal forces themselves are radial and so have no moment about the centre. The moment the analysis moves to effective stresses that convenience vanishes, because the frictional resistance on each element depends on the normal force there, which depends on how the weight is shared between elements — hence the slices. The infinite-slope result then falls out as the limiting case of a very shallow circle in a $c' = 0$ material, and it is the critical mechanism precisely because in that material depth confers no advantage.

Common pitfall. Using the 3.0 m vertical dimension from the centre to the centroid as part of the lever arm. The weight is vertical, so only the horizontal offset produces a moment about O; including the vertical distance is a common and expensive slip. A second pitfall is answering "higher" to the long-term question by carrying across the embankment case, where dissipation of positive excess pore pressure genuinely does improve the factor of safety. The sign of the excess pore pressure follows the sign of the change in mean total stress, and for a cut that change is negative.

Source. Das, Principles of Geotechnical Engineering, 9th ed., §15.5 (mass procedure, $\phi_u = 0$), §15.11 (ordinary method of slices), §15.12 (Bishop simplified) and Eq. (15.10) (infinite slope); Craig's Soil Mechanics, 9th ed., Ch. 12; CFEM 4th ed., §6.

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