16-Civ-B3 Geotechnical Design · December 2016

Question 8 of 9: Short-term stability of a canal bank on a trial slip circle

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 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 strain-influence method (Ch. 5), retaining walls (Ch. 8), pile foundations (Ch. 11), drilled shafts (Ch. 12).
B. M. Das, Principles of Geotechnical Engineering, 9th ed. — 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 and earth-retaining structures.
R. F. Craig, Craig's Soil Mechanics, 9th ed. — effective stress, undrained versus drained behaviour, earth pressure, and the total-stress circular-arc slope analysis.
M. J. Tomlinson & J. Woodward, Pile Design and Construction Practice, 6th ed. — shaft adhesion in clay and under-reamed (belled) bored piles.
P. K. Robertson & K. L. Cabal, Guide to Cone Penetration Testing, 6th ed. — CPT interpretation and CPT-versus-SPT selection.

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

Adhesion factor α = 0.55 for a drilled shaft in clay (Q6) — O'Neill and Reese (1999), reproduced in Das, Principles of Foundation Engineering, 9th ed., Section 12.9; the exclusion of the top 1.5 m and of one shaft diameter above the bell comes from the same source.
Bearing-capacity factor N_c^* = 9 (Q6) — Skempton (1951), as tabulated in Das, Section 11.11.
Overburden correction C_N = √(p_a/σ'_o) (Q7) — Liao and Whitman (1986), Das Principles of Geotechnical Engineering, Section 17.6.
SPT-to-friction-angle correlation (Q7) — Peck, Hanson and Thornburn (1974) as fitted by Wolff (1989); cross-checked against Kulhawy and Mayne (1990). Both are tabulated in Das, Principles of Foundation Engineering, 9th ed., Section 2.9.
Bearing-capacity factors and shape/depth factors (Q7) — Vesic (1973) and De Beer (1970), Das Sections 3.6 and 3.7.
Strain-influence diagram and the C₁, C₂ factors (Q7) — Schmertmann, Hartman and Brown (1978), Das Section 5.6; the modulus correlation E_s = 500(N₆₀ + 15) kPa is Bowles (1996), reproduced in the same section.
Rankine active coefficient for an inclined backfill (Q9) — Das, Principles of Geotechnical Engineering, 9th ed., Eq. (13.35); the base friction and adhesion reductions k₁ = k₂ = 2/3 are Das Section 8.4.
Unit weight of the submerged backfill (Q9) — assumed equal to the printed moist unit weight, 18 kN/m³, in the absence of a saturated value; the consequence of that assumption is bounded in the Q9 callout.

Section A

Question 8: Short-term stability of a canal bank on a trial slip circle (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
Radius of the trial circle	R	15.75 m
Angle subtended by the arc AED	θ	95.5°
Area of the sliding mass ABCDE	A	155 m²
Horizontal offset of the centroid G from the vertical through O	x̄	3.2 m
Bulk unit weight of the saturated clay	γ	20 kN/m³
Undrained shear strength	c_u	40 kN/m²
Slope angle / height	β / H	30° / 7.5 m
Tension crack CD (given on the figure)	—	2.92 m

Find. The short-term (undrained) factor of safety on the trial circle with the canal empty, and then the same factor of safety with the canal full to the level of the top of the bank.

Figure 8.1 — The trial slip circle of Figure 2. The arc runs from the toe A through E to D; the vertical face CD is the tension crack, along which no shear can be mobilised, so the resisting arc stops at D. The disturbing moment is the weight of ABCDE acting at the lever arm 3.2 m measured from the vertical through O.

Approach. This is a total-stress (φ_u = 0) circular-arc analysis. With no friction to mobilise, every element of the arc contributes the same shear stress c_u, and because every element is at the same radius R from O the whole resisting moment can be written down in one line without slices. The factor of safety is the ratio of that resisting moment to the moment of the weight of the sliding mass about O.

Compute the length of the failure arc. The arc runs from the toe A through the lowest point E to the foot of the tension crack D, subtending 95.5°: $$L_a = R\,\theta_{\text{rad}} = 15.75 \times \frac{95.5\pi}{180} = 15.75 \times 1.66684 = 26.25\ \text{m}$$ It stops at D and does not continue up to C, because the tension crack is an open fissure that transmits no shear. Recognising that is the whole point of the note attached to the question.
Compute the weight of the sliding mass. Per metre run of the canal, $$W = A\,\gamma = 155 \times 20 = 3100\ \text{kN/m}$$
Compute the resisting moment about O. The shear resistance available per unit length of arc is $c_u = 40$ kPa, it acts everywhere tangentially, and its lever arm about O is the constant radius R: $$M_R = c_u\,L_a\,R = 40 \times 26.25 \times 15.75 = 16\,539\ \text{kN}\cdot\text{m/m}$$
Compute the disturbing moment about O. The only disturbing agency with the canal empty is the self weight of the mass, acting vertically through the centroid G, whose lever arm about O is the horizontal offset of G from the vertical through O: $$M_D = W\,\bar{x} = 3100 \times 3.2 = 9920\ \text{kN}\cdot\text{m/m}$$ Only the horizontal offset is a lever arm; the vertical position of G is irrelevant, which is why the figure dimensions it horizontally.
Factor of safety, canal empty. $$FS = \frac{M_R}{M_D} = \frac{16\,539}{9920} = 1.667$$ $$\boxed{FS_{\text{empty}} = 1.67}$$ This is comfortably above the 1.3 to 1.5 usually required for a permanent slope in the short term, on this trial circle.
Set up the canal-full case. Raising the water to the top of the bank submerges the entire sliding mass. Three things change and one does not. The pore pressures on the whole boundary of the mass — on the slope face AB, in the tension crack, and on the slip surface itself — now form a closed hydrostatic system, and their resultant is exactly the buoyancy of the mass. The clay is still loaded undrained, so c_u is unchanged: an undrained strength is fixed by the water content that existed before the water rose, and a few days of external flooding does not alter it.
Apply buoyancy to the disturbing moment. Because the water level is the same on both sides of every boundary, the water pressures can be replaced exactly by using the submerged (buoyant) unit weight for the whole mass: $$\gamma' = \gamma - \gamma_w = 20 - 9.81 = 10.19\ \text{kN/m}^3$$ $$W' = A\,\gamma' = 155 \times 10.19 = 1579.5\ \text{kN/m}$$ $$M_D' = W'\,\bar{x} = 1579.5 \times 3.2 = 5054\ \text{kN}\cdot\text{m/m}$$ The centroid is unmoved, so the lever arm is unchanged.
Factor of safety, canal full. The resisting moment is untouched at 16 539 kN·m/m, so $$FS = \frac{16\,539}{5054} = 3.272$$ $$\boxed{FS_{\text{full}} = 3.27}$$ Filling the canal roughly doubles the factor of safety, from 1.67 to 3.27, an increase of 96 per cent. The water acts as a stabilising counterweight against the bank — and this is precisely why the dangerous condition for a canal or reservoir slope is not the full state but rapid drawdown, in which the external water is removed while the pore pressures inside the bank have not yet fallen. The empty case analysed in the first part is the long-standing empty condition; a rapid drawdown from full to empty would be more severe still, because the clay would then also carry positive excess pore pressures locked in from the loaded state.

Quantity	Value
Arc length AED, L_a	26.25 m
Weight of mass ABCDE (canal empty)	3100 kN/m
Resisting moment about O (both cases)	16 539 kN·m/m
Disturbing moment, canal empty	9920 kN·m/m
Factor of safety, canal empty	1.67
Buoyant weight of mass, canal full	1579.5 kN/m
Disturbing moment, canal full	5054 kN·m/m
Factor of safety, canal full	3.27
Effect of filling the canal	FS increases by 96 per cent

Check: the printed figure is over-dimensioned and mildly self-inconsistent. Figure 2 supplies four quantities where three would fix the problem, and they do not all close. Reconstructing the circle from the printed radius (15.75 m), the printed 3.0 m offset from the vertical through O to the crest break B, and the printed 30° slope of height 7.5 m puts O at 9.99 m horizontally and 12.18 m vertically from the toe A. Sweeping the printed 95.5° from A then places D at 4.10 m below the crest, so the reconstructed tension crack is CD = 4.10 m, not the 2.92 m printed — and 4.10 m is within 3 per cent of the theoretical tension-crack depth z₀ = 2c_u/γ = 2(40)/20 = 4.00 m, which strongly suggests the printed 2.92 m is the value in error. The same reconstruction gives an area of 168.4 m² against the printed 155 m² and a centroid lever arm of 2.88 m against the printed 3.2 m.

The reassuring part is that the two errors act in opposite directions, so the answer is robust: the fully reconstructed geometry gives FS = 16 539 / (168.4 × 20 × 2.88) = 1.70 against the 1.67 obtained from the printed values — a 2 per cent difference. The printed values are used for the answer, because they are what the candidate is instructed to work from and what a marker expects; the reconstruction is reported because Note 6 requires assumed and derived values to be identified, and because the check is what licenses confidence in the result. Reconstructed values for the flooded case give FS = 3.34 against 3.27, and the conclusion — that filling the canal roughly doubles the factor of safety — is unaffected.

Other assumptions. (i) γ_w is taken as 9.81 kN/m³; using 10.0 kN/m³ gives FS_full = 3.33 instead of 3.27. (ii) The tension crack is assumed dry in the empty case; if it filled with rainwater over the printed 2.92 m the hydrostatic thrust would add about 41.8 kN/m at a lever arm of about 6.6 m about O, reducing FS_empty to about 1.62, and over the theoretical z₀ = 4.0 m it would add 78.5 kN/m at about 7.3 m, reducing it to about 1.58 — a real and unfavourable effect that justifies sealing or draining the crest. (iii) Only one trial circle has been analysed. The true factor of safety is the minimum over all circles, and finding it requires a search; the value reported here is an upper bound on the minimum for this slope.

Topic. Total-stress (φ_u = 0) circular-arc slope analysis, including the tension crack and the effect of submergence.

Key relations. $FS = \dfrac{M_R}{M_D} = \dfrac{c_u L_a R}{W\bar{x}}$ with $L_a = R\theta$; tension-crack depth $z_0 = \dfrac{2c_u}{\gamma}$ (the depth at which the active total stress $\gamma z - 2c_u$ falls to zero); submerged unit weight $\gamma' = \gamma - \gamma_w$.

Why this works. When φ_u = 0 the available shear stress does not depend on the normal stress, so the distribution of normal stress around the arc — the thing that makes a general slope analysis require slices — simply does not enter. Every element of arc offers c_u at the same lever arm R, so the resisting moment is a product of three numbers. Submergence is handled by the same argument that underlies Archimedes: a boundary water pressure field in equilibrium with a single external water level contributes a resultant equal and opposite to the weight of displaced water, so replacing γ by γ' accounts for every one of those pressures at once, including the pressure inside the tension crack.

Common pitfall. Three recur. Running the arc all the way to C instead of stopping at D, which inflates L_a and the factor of safety. Using the straight-line distance OG, or the vertical offset of G, as the lever arm instead of the horizontal offset. And reducing c_u when the canal fills — the strength is undrained and does not know the canal is full; only the driving weight changes.

Source. Craig, Craig's Soil Mechanics, 9th ed., Section 9.3 (the φ_u = 0 analysis, tension cracks, and submerged slopes); Das, Principles of Geotechnical Engineering, 9th ed., Sections 15.5 and 15.6.

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