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.

Part 1(a) — the slip surface (5 marks)

Given. A homogeneous $c^{\prime}\!-\!\phi^{\prime}$ clay cutting, 18 m deep at 20°, with $\gamma = 19$ kN/m³, $c^{\prime} = 30$ kPa, $\phi^{\prime} = 25^\circ$ and a uniform pore-pressure ratio $r_u = 0.25$.

Find. The shape and position of the slip surface to be expected, and the form of failure.

Expected form of failure. The cutting is in a homogeneous clay with no stronger or weaker stratum to deflect the failure surface and no hard layer at any particular depth. Under those conditions the failure is a rotational slip on a near-circular surface, and because $\phi^{\prime} > 0$ it is a toe circle — the friction component makes deep-seated arcs progressively less efficient, since a deeper arc must mobilise a longer surface through soil whose normal stress is higher but whose driving moment arm has not grown in proportion. Only in a $\phi_u = 0$ analysis, or where a firm base limits the depth, do base circles or midpoint circles become critical.

The circle drawn above is the critical one located by minimising Bishop's simplified factor of safety over the centre coordinates and radius. Its centre lies at (15.8, 46.5) m relative to the toe and its radius is 49.2 m; the arc emerges at the toe and daylights on the crest about 40 m behind it. The corresponding factor of safety is

with $F = 2.40$ if the pore pressures are removed — an 18% penalty for $r_u = 0.25$, and a useful reminder of how much of the stability of a clay cutting is bought by drainage. The expected failure mechanism, then, is a slow rotational movement of a wedge roughly 40 m long and up to 2.7 m deep below the toe, with tension cracking at the crest, heave and bulging at the toe, and back-tilted blocks in between.

Part 1(b) — approximations in the three methods (5 marks)

All three are methods of slices: the sliding mass is divided into vertical slices, equilibrium is written slice by slice, and the results are summed. They differ entirely in how they handle the interslice forces, which are statically indeterminate.

Ordinary (Fellenius, Swedish) method of slices. The approximation is the boldest: the interslice forces are assumed to be equal, opposite and parallel to the base of each slice, so that they cancel and the normal force on the base is simply $N = W\cos\alpha$. Only overall moment equilibrium about the centre of rotation is satisfied; force equilibrium of the individual slice is not. This gives a closed-form, non-iterative expression

which is easy to compute by hand but is known to underestimate $F$, typically by 10–15% and by as much as 60% for deep circles with high pore pressures, because subtracting $u\ell$ from $W\cos\alpha$ can drive the effective normal force negative at the steep ends of the arc.

Bishop's simplified method. The approximation is that the resultant interslice force is horizontal: the interslice shear forces are neglected, but the normal ones are not. Vertical equilibrium of each slice is then satisfied, giving

Because $F$ appears on both sides the solution is iterative, but it converges in three or four cycles and is accurate to about 1% against rigorous methods for circular surfaces. Overall moment equilibrium is satisfied; overall horizontal force equilibrium is not. The method fails ($m_\alpha \to 0$) for steeply inclined slice bases at the tail of the arc, which is why implementations reject trial circles where $m_\alpha$ falls below about 0.2.

Spencer's method. The approximation is only that the inclination of the interslice resultant is constant along the sliding mass, $X/E = \tan\theta$, with $\theta$ unknown. Two equations are then written — one for moment equilibrium, one for force equilibrium — and solved simultaneously for the pair $(F, \theta)$ at which both are satisfied. Spencer's method is therefore rigorous in the sense that complete statics is satisfied, it applies to non-circular as well as circular surfaces, and it needs no assumption about the shape of the slip surface. Its costs are a two-variable iteration that occasionally fails to converge, and a residual physical approximation: a single constant interslice inclination cannot represent a mass in which the interslice forces genuinely rotate, and the solution must be checked afterwards for admissibility (no tension, and the line of thrust inside the sliding mass).

Part 2 — drained factor of safety of the 12 m slope (20 marks)

Given. A slope cut in clayey soil, 12 m high at 26°, with $c^{\prime} = 20$ kPa, $\phi^{\prime} = 30^\circ$ and $\gamma = 17.8$ kN/m³; the drained (long-term, effective-stress) condition is required.

Find. The factor of safety (a) with the water table well below the slope, and (b) with the water table 6 m below the ground surface.

Approach. Both cases are effective-stress analyses with the same strength parameters; only the pore pressures change. Bishop's simplified method is applied over a search of trial circles, with $u = 0$ in case (a) and $u = \gamma_w h_w$ from the phreatic surface in case (b).

1. Set up the geometry and confirm the mechanism. With the toe at the origin, the slope face rises to the crest at $L = H/\tan\beta = 12/\tan 26^\circ = 24.6$ m horizontally. Note first that $\beta = 26^\circ < \phi^{\prime} = 30^\circ$: even with $c^{\prime} = 0$ an infinite slope in this material would stand, so any failure must be a shallow-to-moderate rotational slip in which cohesion carries the ends of the arc.
2. Case (a), water table well below the slope. With $u = 0$, Bishop's simplified expression reduces to $$F = \frac{1}{\sum W\sin\alpha}\sum\frac{c^{\prime}b + W\tan\phi^{\prime}}{m_\alpha}$$

   Minimising over the centre and radius locates a toe circle centred at (5.8, 25.2) m with $R = 25.9$ m, extending only about 0.7 m below toe level — a shallow slip, as anticipated:

   $$\boxed{F_a = 2.34}$$

   As a check on order of magnitude, Taylor's stability number for the mobilised friction angle $\phi_d = \arctan(\tan 30^\circ/2.34) = 13.9^\circ$ at $\beta = 26^\circ$ is about 0.040, and $c_d/(\gamma H) = (20/2.34)/(17.8\times12) = 0.040$ — consistent.

3. Case (b), define the pore-pressure field. "Six metres below the ground surface" is taken as a horizontal water table at 6 m below the crest, i.e. standing 6 m above the toe and daylighting halfway up the slope face; that is the natural reading for a cutting whose original ground surface is the crest, and it is the more severe of the two possible readings. On each slice base the pore pressure is $$u = \gamma_w\left(\min(y_{wt},\,y_{surface}) - y_{base}\right)$$

   which sets $u = 0$ above the phreatic surface and treats the seepage face below it as fully saturated to the slope surface. This is the standard textbook simplification: it uses the vertical distance to the water table rather than the true piezometric head from a flow net, and is slightly conservative because flow lines curve away from the vertical near the face.

4. Case (b), the factor of safety. Repeating the minimisation with that pore-pressure field draws the critical circle downslope and deepens it slightly — centre (7.1, 19.2) m, $R = 20.8$ m, about 1.6 m below toe level — because the wetted lower half of the mass has lost most of its frictional resistance: $$\boxed{F_b = 1.69}$$

   Expressed as an equivalent uniform pore-pressure ratio, this phreatic surface is worth $r_u \approx 0.39$ on that circle, appreciably more severe than the $r_u = 0.25$ a designer might have assumed by eye. If the water table is instead read as following the topography 6 m below the surface everywhere, it barely touches the shallow critical circle and $F$ only falls to 2.26 — a 33% spread in answer that comes entirely from how one sentence is read, which is worth stating on the answer paper.

Figure 5 — Part 2(b): the same slope with the water table 6 m below the crest. The saturated zone is shaded; the critical circle moves downslope and deepens, and F falls to 1.69.
5. Assess the result. Both cases clear the conventional long-term requirement of $F \ge 1.5$ for a permanent slope, so the cutting is stable as designed. The engineering point is the size of the reduction: raising the water table to 6 m below the crest costs 28% of the factor of safety and moves the critical mechanism deeper. On a permanent cutting that is an argument for a toe drain or a counterfort drain rather than for a flatter slope — drainage buys the same margin far more cheaply than earthworks.