Question 3 of 6: Slope Stability (30 marks)

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 (10 marks)

(a) Approximations in the three methods of slices. All three divide the sliding mass above a trial slip surface into vertical slices, satisfy the Mohr–Coulomb criterion on the base of each slice, and define the factor of safety as the ratio of available to mobilised shear strength. They differ entirely in how they treat the forces the slices exert on one another, and that single difference explains the spread in their answers.

The ordinary (Fellenius or Swedish) method of slices makes the crudest assumption: the resultant of the interslice forces on each slice acts parallel to the base of that slice, so that its component normal to the base is zero. The normal effective force is then obtained by resolving the slice weight perpendicular to the base directly, $N^{\prime} = W\cos\alpha - ul$, and the factor of safety follows explicitly — no iteration is needed. The price is that only moment equilibrium of the whole mass is satisfied; force equilibrium is violated slice by slice, and the method under-estimates the factor of safety, typically by 5–20 % and by considerably more for deep circles with high pore pressures. Its one virtue is that it is safe and can be done by hand.

Bishop's simplified method assumes instead that the interslice forces are horizontal, i.e. the interslice shear forces are zero. Vertical equilibrium of each slice can then be written properly, which introduces the factor of safety into the expression for the base normal force and makes the equation implicit: $$F = \frac{1}{\sum W\sin\alpha}\sum \frac{c^{\prime}b + (W - ub)\tan\phi^{\prime}}{m_{\alpha}}, \qquad m_{\alpha} = \cos\alpha\left(1 + \frac{\tan\alpha\tan\phi^{\prime}}{F}\right)$$ It must be solved by iteration, converging in three or four cycles. Overall moment equilibrium is satisfied and vertical force equilibrium of each slice is satisfied; horizontal force equilibrium of the individual slices is not. Bishop's method applies to circular surfaces and is accurate to within about 1–2 % of rigorous solutions, which is why it remains the standard for routine work.

Spencer's method removes the restriction altogether by assuming only that the interslice forces are all parallel, inclined at a single unknown angle θ that is solved for as part of the analysis. Two equations — overall moment equilibrium and overall force equilibrium — are then solved simultaneously for the two unknowns F and θ. Because both equilibrium conditions are satisfied, Spencer's method is a rigorous method; it is not restricted to circular surfaces and can be applied to any shape of slip surface. The remaining approximations are the parallel-force assumption itself and the neglect of the internal stress distribution within each slice, and the solution must be checked for physical admissibility (the line of thrust should lie within the sliding mass).

(b) Mechanisms of earthquake-induced slope failure. Four mechanisms dominate, and they can act together.

Inertial (pseudo-static) destabilisation. The ground acceleration applies a horizontal body force $k_h W$ to the sliding mass, adding a driving moment while the resisting moment is largely unchanged. If the peak acceleration exceeds the yield acceleration of the slope, the mass begins to move; Newmark sliding-block analysis integrates the excess acceleration twice to estimate the permanent displacement accumulated over the shaking. Failure here is progressive and displacement-based rather than an instantaneous collapse.

Liquefaction of loose saturated granular soils. Cyclic shearing of a loose sand or silt below the water table generates excess pore pressure with each cycle; when the excess pressure approaches the initial vertical effective stress, effective stress and hence shear strength approach zero. The result is either a flow slide, if the soil is contractive and the residual strength is less than the static driving stress, or lateral spreading of a gently sloping crust riding on a liquefied layer. Loose hydraulic fills and young alluvial deposits are the classic victims.

Cyclic softening and strength loss in clays. Sensitive and normally consolidated clays lose strength under repeated straining, through pore-pressure build-up and through remoulding along a developing shear band. Quick clays, which have a very high sensitivity, can retrogress far beyond the initially failed zone once movement starts. In stiff fissured clays, cyclic loading opens and connects existing fissures.

Topographic amplification and secondary effects. Ridge crests and steep convex slopes amplify ground motion relative to level ground, so the acceleration acting on the crest of a slope can substantially exceed the free-field value. Superimposed on this are seismically induced settlement of loose fills, fault rupture through the slope, loss of support from liquefied or displaced toe material, and sloshing or rapid drawdown in an adjacent reservoir. A design check should therefore combine a pseudo-static or Newmark analysis with an explicit liquefaction-triggering assessment of every saturated granular layer.

Part 2 (20 marks)

Given. Slope height H = 30 m; unit weight γ = 19.1 kN/m³; effective strength parameters c′ = 36 kN/m² and φ′ = 20°; pore-pressure ratio r_u = 0.25; slope angle β = 26°.

Find. (c) the factor of safety of the 26° slope, using Spencer's stability charts; (d) the steepest slope angle that still delivers F = 1.25.

Approach. Spencer's charts are entered with the stability number $c^{\prime}/(F\gamma H)$ and the slope angle, and return the friction angle that must be developed; because F appears on both sides the solution is iterative. The chart reading is then confirmed by a Bishop simplified analysis over a search of trial circles.

1. Form the basic stability number. Independently of the factor of safety, $$\frac{c^{\prime}}{\gamma H} = \frac{36}{19.1 \times 30} = 0.0628$$ This is the quantity that fixes which family of curves on the r_u = 0.25 chart is relevant.
2. Set up the chart iteration. Spencer's charts are plotted for r_u = 0, 0.25 and 0.5 and relate the slope angle β, the stability number $c^{\prime}/(F\gamma H)$ and the developed friction angle $$\phi^{\prime}_{d} = \tan^{-1}\!\left(\frac{\tan\phi^{\prime}}{F}\right)$$ Because the same F factors both the cohesion and the friction, one assumes a value of F, computes $c^{\prime}/(F\gamma H)$, reads the required $\phi^{\prime}_{d}$ off the r_u = 0.25 chart at β = 26°, and compares it with $\tan^{-1}(\tan 20^{\circ}/F)$. The trials converge as follows.

3. Read off the factor of safety. The iteration closes at a stability number of 0.0512 with a developed friction angle of 16.5°, so $$\boxed{F = 1.23 \ \text{for the 30 m slope at } \beta = 26^{\circ}}$$ Note that this is below the 1.25 usually demanded for a permanent cut, so the slope as proposed is not acceptable.
4. Confirm the chart reading by a slice analysis. Chart solutions should always be checked, because a chart cannot show which circle is critical. A Bishop simplified analysis was run over a grid of trial circle centres and radii with $$F = \frac{1}{\sum W\sin\alpha}\sum \frac{c^{\prime}b + W(1 - r_u)\tan\phi^{\prime}}{m_{\alpha}}$$ (using $ub = r_u\gamma h b = r_u W$, which is what the pore-pressure ratio is defined to give). The minimum over all trial circles is F = 1.227, on a deep circle passing about 1.5 m below the toe, centred 15.2 m behind and 63.3 m above it with R = 65.1 m. This agrees with the chart value to better than 0.5 %, as expected since Spencer's rigorous solution normally exceeds Bishop's simplified value by only one or two per cent.
5. (d) Re-enter the chart for the required factor of safety. Fixing F = 1.25 makes the entry quantities determinate: $$\frac{c^{\prime}}{F\gamma H} = \frac{36}{1.25 \times 19.1 \times 30} = 0.0503, \qquad \phi^{\prime}_{d} = \tan^{-1}\!\left(\frac{\tan 20^{\circ}}{1.25}\right) = 16.24^{\circ}$$ The slope angle is now read from the chart as the abscissa at which the $\phi^{\prime}_{d} = 16.24^{\circ}$ curve crosses a stability number of 0.0503.
6. Determine the allowable slope angle. The chart gives an angle just over 25°, and re-running the Bishop search while varying β brackets it precisely: F = 1.263 at 25.0°, F = 1.245 at 25.5°, F = 1.227 at 26.0°. Interpolating, $$\boxed{\beta_{allow} = 25.4^{\circ} \ (\cot\beta = 2.11, \ \text{i.e. about } 1\text{V}:2.1\text{H})}$$ For construction this would be specified as a 1V : 2.15H (25°) slope, which gives F = 1.26 with a little in hand.
7. Interpret the sensitivity. Flattening the slope by only 0.6° raises the factor of safety from 1.227 to 1.25 — about 0.036 of factor of safety per degree. A slope this flat, this high and this cohesive is dominated by the weight of a deep sliding mass rather than by the surface geometry, so geometry is an inefficient lever. Reducing the pore-pressure ratio is far more effective: at r_u = 0.15 the original 26° slope reaches F = 1.34. Drainage, not re-grading, is the economical remedy here.