Question 3 of 6: Slope Stability

Question 3 — Slope Stability (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. Three independent slopes, each in a homogeneous soil, are to be assessed by limit equilibrium on circular slip surfaces.

Find. The factor of safety of the 30 m clay slope; the inclination that gives F = 1.2 for the 40 m slope with the factor applied to both c′ and tanφ′; and the drawdown factor of safety of the 25 m reservoir embankment.

Approach. All three parts are Taylor stability-number problems. Because a chart cannot be read here, each chart value is re-derived numerically from the theory the chart encodes: a search over trial circles using the exact $\phi_u = 0$ Swedish-circle solution in Part 1 and the Bishop simplified method in Parts 2 and 3. The routine is first validated against the two stability numbers every textbook prints for $\phi_u = 0$, and only then applied to the questions.

Part 1 — factor of safety of the 30 m clay slope (10 marks)

1. Classify the problem and find the depth factor. With $\phi_u = 0$ and $\beta = 30^{\circ} < 53^{\circ}$ the critical surface is a base circle whose depth is limited only by the firm stratum. Taylor's depth factor measures the depth to that stratum from the crest: $$D = \frac{H + 45}{H} = \frac{30+45}{30} = 2.5$$
2. Set up the exact moment equilibrium for a purely cohesive soil. For $\phi_u = 0$ the normal stress on the arc does no work in the moment balance about the centre, so the factor of safety is exact and independent of any slice assumption: $$F = \frac{c_u\,R\,L_{arc}}{\sum W_i x_i}$$ where $L_{arc}$ is the length of the slip arc, $R$ the radius, $W_i$ the weight of slice $i$ and $x_i$ its horizontal lever arm from the centre.
3. Minimise over trial circles subject to the rock stratum. Searching over the centre coordinates and radius, with the constraint that no arc may pass below 45 m under the toe, the critical circle is found at $x_c = 26.09$ m, $y_c = 61.44$ m, $R = 105.98$ m (origin at the toe), whose lowest point lies 44.5 m below the toe — essentially tangent to the rock. It gives $$\boxed{F = 1.06}$$
4. Express the result as a Taylor stability number and sense-check it. Rearranging, $$m = \frac{c_u}{F\gamma H} = \frac{96}{1.0645 \times 17.0 \times 30} = 0.177$$ which sits, as it should, just below the deep-base limit of 0.181 that applies when $D \to \infty$: the rock at $D = 2.5$ stops the circle a little short of its unrestricted optimum, so the slope is slightly safer than the limiting case. Had the rock been absent, F would fall to 1.04.
5. Interpret. A factor of safety of 1.06 means the slope is marginally stable at best. For a permanent slope in clay the usual requirement is 1.3 to 1.5 on undrained strength, so this profile would not be accepted: a small strength reduction from softening, fissuring or a rise in pore pressure would take it below unity. It is also worth noting that a 30 m high clay slope is far taller than the critical height $H_c = c_u/(0.181\gamma) \approx 31$ m for this soil, which is exactly why F comes out so close to 1.

Part 2 — inclination for a factor of safety of 1.2 on a 40 m slope (10 marks)

1. Recognise what "the factor of safety applies to friction as well as cohesion" means. It defines $F$ as the factor by which both strength components must be divided to bring the slope to limiting equilibrium: $$c^{\prime}_d = \frac{c^{\prime}}{F},\qquad \tan\phi^{\prime}_d = \frac{\tan\phi^{\prime}}{F}$$ This is precisely the definition embedded in the Bishop simplified factor of safety, so no separate manipulation is needed — a search that returns Bishop's F already answers the question.
2. Compute the developed strengths and the target stability number. For $F = 1.2$: $$c^{\prime}_d = \frac{25}{1.2} = 20.83\ \text{kPa},\qquad \phi^{\prime}_d = \arctan\left(\frac{\tan 12^{\circ}}{1.2}\right) = 10.04^{\circ}$$ $$m = \frac{c^{\prime}_d}{\gamma H} = \frac{20.83}{16.5 \times 40} = 0.0316$$ This pair, $m = 0.0316$ with $\phi^{\prime}_d \approx 10^{\circ}$, is what would be taken into Taylor's chart to read off the required angle.
3. Search for the inclination that returns F = 1.2. Running the Bishop simplified search at a series of slope angles and interpolating gives $$F(\beta): \quad 14^{\circ} \to 1.394,\quad 16^{\circ} \to 1.263,\quad 18^{\circ} \to 1.158,\quad 20^{\circ} \to 1.073$$ so F = 1.2 falls at $\beta = 17.15^{\circ}$. Rounding to a constructible angle, $$\boxed{\beta = 17^{\circ}\ \ (\text{about } 3.3\text{H}:1\text{V}), \qquad F = 1.21}$$ The critical circle at that inclination is a deep one, centred well behind the crest at $x_c = 41.1$ m, $y_c = 127.9$ m with $R = 134.6$ m.
4. Comment on the answer. A 40 m high cutting in a clay with only 25 kPa of effective cohesion and 12° of friction has to be laid back to roughly 1 vertical in 3.3 horizontal, which means a slope width of about 131 m and a very large volume of excavation. In practice a designer would first ask whether berms, a toe buttress or drainage could raise F at a steeper angle, since the flat slope required here is driven by the sheer height: the same soil at 20 m height would stand at about 28° for the same factor of safety.

Part 3 — sudden drawdown of the reservoir embankment (10 marks)

1. Model the sudden-drawdown pore pressures. Before drawdown the embankment is submerged and steady seepage has brought the pore pressures into equilibrium with the reservoir. When the level falls faster than the fill can drain, the water pressures inside the embankment stay at their pre-drawdown values while the stabilising external water load disappears. For a compacted fill of low permeability the standard idealisation sets the pore-pressure ratio to $$r_u = \frac{u}{\gamma h} = \frac{\gamma_w}{\gamma} = \frac{9.81}{19.0} = 0.516$$
2. Write the Bishop simplified factor of safety. With effective-stress strengths and pore pressure $u = r_u\gamma h$ on each slice base, $$F = \frac{\displaystyle\sum \frac{c^{\prime}b + (W - ub)\tan\phi^{\prime}}{m_{\alpha}}}{\displaystyle\sum W\sin\alpha},\qquad m_{\alpha} = \cos\alpha\left(1 + \frac{\tan\alpha\tan\phi^{\prime}}{F}\right)$$ which is solved by iteration because F appears on both sides.
3. Evaluate the fully drained (pre-drawdown, long-term) case first, as a datum. Setting $r_u = 0$ and searching over circles gives a critical circle at $x_c = 27.9$ m, $y_c = 62.7$ m, $R = 70.5$ m and $$F_{drained} = 1.39$$ which is comfortably stable and is the condition the embankment would normally be checked for.
4. Evaluate the sudden-drawdown case. Repeating the search with $r_u = 0.516$ throughout the fill, the critical circle moves slightly (centre $x_c = 31.1$ m, $y_c = 57.2$ m, $R = 70.0$ m) and the factor of safety falls to $$\boxed{F_{drawdown} = 0.97}$$
5. Interpret and recommend. The embankment is unstable under sudden drawdown: a factor of safety below 1.0 means a slip would develop as the reservoir was emptied, and even the 1.2 to 1.3 usually required for this transient case is far out of reach. The physical reason is that the buoyant weight of the submerged fill was carrying part of the load through water pressure, and removing the reservoir removes the external support while leaving the internal pore pressures intact — halving the effective normal stress on the slip surface and therefore halving the frictional component of strength. Because $\phi^{\prime}$ is only 10° the frictional contribution is small to begin with, so what remains is nearly all cohesion, and 40 kPa is not enough over a 70 m arc. The remedies are to flatten the upstream slope, to place a free-draining upstream zone or blanket drain on the upstream face so that pore pressures fall with the reservoir, or to impose a maximum rate of drawdown that the fill can follow.