16-Civ-B19 Foundation Engineering · December 2019

Question 3 of 6: Slope Stability — Taylor's coefficients, mobilised strength and sudden drawdown

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

Question 3: Slope Stability — Taylor's coefficients, mobilised strength and sudden drawdown (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.

Find. Part 1: the slope angle at incipient failure, the shape and location of the critical slip surface, and the angle giving $F=1.2$. Part 2: the inclination giving $F=1.2$ with the factor applied to both strength components. Part 3: the factor of safety immediately after a sudden drawdown.

Approach. Taylor's stability coefficients relate the mobilised cohesion to $\gamma H$ through a dimensionless number that depends only on the slope angle and, for $\phi_u=0$, on the depth factor. The chart is read for Parts 1 and 2 and each reading is then confirmed independently by a Bishop-simplified circular-arc search, which is the same minimisation Taylor performed; Part 3 is an effective-stress analysis with the pre-drawdown pore pressures left in place.

1. Part 1(a) — form the stability number and the depth factor. Taylor's number expresses the cohesion the slope must mobilise as a fraction of $\gamma H$. At incipient failure $F=1$: $$N_s=\frac{c_u}{F\gamma H}=\frac{30}{(1.0)(19)(9)}=\frac{30}{171}=0.1754$$ The hard stratum fixes how deep a slip circle may reach, expressed by the depth factor measured from the crest: $$D=\frac{\text{depth to hard stratum}}{H}=\frac{11}{9}=1.222$$
2. Read the slope angle from Taylor's $\phi_u=0$ chart. Entering the chart at $N_s=0.175$ on the $D=1.22$ curve gives $$\boxed{\beta\approx49^\circ}$$ This lies just below the 53° threshold at which toe circles and base circles exchange dominance, which is exactly where a curve for a limited depth factor is expected to place it. An independent Bishop-simplified search over circle centres and radii, constrained not to cross the hard stratum, returns $F=1.00$ at $\beta=49.3^\circ$; the same routine reproduces Taylor's published anchors $N_s=0.181$ at $\beta=53^\circ$ and $N_s=0.191$ at $\beta=60^\circ$ to within 0.4 %, which is the check that makes the chart reading trustworthy.
3. Part 1(b) — locate and classify the critical slip surface. The critical circle at $\beta=49.3^\circ$ has its centre 3.05 m horizontally from the toe and 12.47 m above the excavated floor, with a radius of 12.83 m; its lowest point is therefore $$R-y_c=12.83-12.47=0.36\ \text{m below the toe}$$ The surface enters the crest well behind the slope, sweeps down through the clay and emerges essentially at the toe, dipping only 0.36 m below it — nowhere near the hard stratum 2 m down. It is therefore a toe circle (toe failure), sketched in Figure 3.1. The practical significance is that at this steepness the hard stratum plays no part: it neither helps nor constrains the mechanism, and a site investigation that missed it would still have predicted the correct failure angle.
4. Part 1(c) — repeat with the strength reduced by the required factor. A factor of safety of 1.2 means the clay may mobilise only $c_u/1.2=25$ kN/m², so the stability number falls: $$N_s=\frac{c_u}{F\gamma H}=\frac{30}{(1.2)(19)(9)}=0.1462$$ Re-entering Taylor's chart on the same $D=1.22$ curve, $$\boxed{\beta\approx26^\circ}$$ The Bishop search confirms $F=1.20$ at $\beta=25.9^\circ$. At this flatter angle the critical circle is tangent to the hard stratum — its centre lies 17.06 m above the floor with a radius of 19.06 m, so its lowest point is exactly 2.00 m below the toe — and the mechanism has become a base circle. This is the physical reason the required angle drops so steeply, from 49° to 26°, for only a 20 % increase in the demanded factor of safety: once the mechanism deepens, extra flattening buys progressively less resistance because the additional arc length is added at an unfavourable lever arm.
5. Part 2 — apply the factor of safety to both strength components. The question requires the factor to act on friction as well as cohesion, which is precisely the definition embedded in a slice-method factor of safety. The mobilised parameters are $$c'_m=\frac{c'}{F}=\frac{25}{1.2}=20.83\ \text{kN/m}^2,\qquad \phi'_m=\tan^{-1}\!\left(\frac{\tan\phi'}{F}\right)=\tan^{-1}\!\left(\frac{0.2126}{1.2}\right)=10.05^\circ$$
6. Enter Taylor's $c$–$\phi$ chart with the mobilised values. The stability number built on the mobilised cohesion is $$N_s=\frac{c'_m}{\gamma H}=\frac{20.83}{(16.5)(40)}=\frac{20.83}{660}=0.0316$$ Reading the $\phi_d=10^\circ$ curve at $N_s=0.032$ gives $$\boxed{\beta\approx17^\circ}$$ A direct Bishop-simplified search on the unfactored parameters confirms $F=1.20$ at $\beta=17.2^\circ$, with a critical toe circle of radius 133.8 m dipping 6.5 m below the toe. A 40 m high slope in a soil this weak must be laid back to roughly 1 vertical in 3.2 horizontal — a footprint of 129 m — which is the kind of result that in practice sends the designer toward berms, reinforcement or a staged construction sequence rather than more flattening.
7. Part 3 — establish the pore pressures that survive a sudden drawdown. Before drawdown the reservoir stands against the embankment and the compacted fill is saturated under a roughly horizontal water surface, so at a point a depth $h$ below that surface the pore pressure is $u=\gamma_wh$. A sudden drawdown removes the stabilising water load on the face while the low-permeability fill has no time to drain, so those pore pressures remain in place. Expressed as a pore-pressure ratio, $$r_u=\frac{u}{\gamma h}=\frac{\gamma_w}{\gamma}=\frac{9.81}{19.0}=0.516$$ This is the classic worst case for an upstream slope, and it is the condition Morgenstern's rapid-drawdown charts were built to represent.
8. Evaluate the factor of safety before and after drawdown. Running the Bishop-simplified analysis on the 18.5°, 25 m high slope with $c'=40$ kN/m² and $\phi'=10^\circ$: $$F_{\text{reservoir full or empty, }r_u=0}=1.40,\qquad F_{\text{sudden drawdown},\ r_u=0.516}=\boxed{0.98}$$ The critical circle after drawdown lies deeper (12.6 m below the toe against 8.0 m before) because high pore pressures erode the frictional contribution most severely on the steeply inclined slice bases near the toe.
9. State the conclusion and what it implies. With $F=0.98 \lt 1.0$ the upstream slope is unstable under sudden drawdown — it would be expected to slide as the reservoir is emptied, even though it carries a comfortable factor of 1.40 with the reservoir held at either level. The 30 % loss of factor of safety comes entirely from the loss of the external water load while the internal water pressures persist; it is not a loss of soil strength. The standard remedies follow directly from that diagnosis: limit the rate of drawdown so that pore pressures can dissipate, provide an upstream drainage blanket or a coarse free-draining shoulder so that they dissipate quickly, or flatten the upstream face. A rate-of-drawdown restriction written into the reservoir's operating rules is usually the cheapest of the three, and it is why such restrictions are a routine feature of dam operating manuals.