Question 9 of 11: Slope Design

Question 4 — Slope Design (20 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.

4.1 — the factor-of-safety method. The factor-of-safety (FS) method expresses stability as the ratio of total resisting force/moment to total driving force/moment along an assumed failure surface, FS = resisting/driving; FS>1 implies stability, FS<1 implies failure, with FS closer to 1 implying progressively less margin. It is appropriate as a simple, industry-standard, easily-audited screening and design tool, but it is a LIMIT-EQUILIBRIUM method — it says nothing about deformation, progressive failure, or the actual probability of failure given real parameter uncertainty. Alternative/complementary methods include probabilistic slope-stability analysis (treating c, φ, γ and geometry as distributions and computing a probability of failure rather than a single FS), numerical stress-deformation modelling (finite-element/finite-difference codes such as Phase2/RS2 or FLAC, which capture progressive yielding and deformation directly), and empirical rock-mass rating methods (Slope Mass Rating, SMR) for structurally-controlled rock slopes. For a NON-PERMANENT (temporary, operating) pit slope the accepted minimum FS is commonly taken as about 1.2–1.3 (lower than the 1.5+ typically required for permanent/civil slopes), reflecting the shorter exposure period and closer operational monitoring of an active mine wall.

Given (4.2/4.5/4.6). Bench height H = 24 m; failure-plane angle β = 17°; slope-face angle α = 35°; cohesion c = 5.51 kPa; friction angle φ = 30°; density ρ = 1922 kg/m³ (γ = 18.85 kN/m³); arc radius R = 55 m; total arc length 107 m; driving weight W1 = 1360 t at lever arm l1 = 20 m; opposing weight W2 = 982 t at lever arm l2 = 27 m; average depth to the plane of weakness h̄ = H/2 = 12 m (the ONE calculation needing this assumption, per the question’s own note).

Find. FS_dry and FS_sat for the simple planar wedge (4.2) and for the Swedish slip circle (4.5/4.6).

Approach (4.2). Model the sliding mass as the triangular wedge bounded by the failure plane, the slope face and the horizontal crest (Hoek & Bray planar-failure form); resolve its weight normal and tangential to the plane; add cohesion over the plane length; for the saturated case, subtract an uplift force from a triangular pore-pressure distribution along the plane.

1. Wedge geometry and weight. Area/unit width = ½H²(cotβ − cotα) = ½(24)²(cot17° − cot35°) = 530.7 m²/m. W = γ×Area = 18.85×530.7 = 10,006 kN/m. Plane length L = H/sinβ = 24/sin17° = 82.1 m.
2. Resolve normal/tangential and compute FS_dry. N = Wcosβ = 9,570 kN/m; T = Wsinβ = 2,926 kN/m. $$FS_{dry}=\dfrac{cL+N\tan\phi}{T}=\dfrac{5.51(82.1)+9570\tan30^\circ}{2926}$$ $$\boxed{FS_{dry}=2.04}$$
3. Saturated case — triangular uplift. With the water table following the SAME upper boundary as the wedge (face, then horizontal crest), the maximum vertical thickness of saturated ground above the plane is z_max = H(1 − cotα·tanβ) = 24(1 − cot35°·tan17°) = 13.5 m, giving a peak normal pore pressure u_max = γ_wz_maxcos²β = 9.81(13.5)cos²17° = 121.4 kPa and, for the triangular distribution along the plane, U = ½u_maxL = ½(121.4)(82.1) = 4,984 kN/m.
4. FS_saturated. $$FS_{sat}=\dfrac{cL+(N-U)\tan\phi}{T}=\dfrac{452.4+(9570-4984)\tan30^\circ}{2926}$$ $$\boxed{FS_{sat}=1.06}$$

4.3 — conclusions. Even a MODEST rise in the water table collapses the factor of safety from a comfortably stable 2.04 (dry) to a marginally-stable 1.06 (saturated) — a drop of nearly half — for exactly the same slope geometry and strength parameters, which is the single most important practical conclusion of a planar-failure analysis: water control (dewatering, drainage) is usually a far cheaper and more effective stability lever than flattening the slope. The uplift force is triangular, not uniform, because the water depth (and hence pore pressure) above the failure plane is ZERO at the toe (where the plane daylights at the ground surface) and increases with distance back into the slope up to the point where the wedge is thickest, mirroring the linearly-increasing overburden thickness above the plane — a uniform-pressure assumption would either under- or over-state the true uplift depending on where it was anchored.

4.4 — method of slices; bank vs. base failure. The method of slices divides the sliding mass above an assumed (usually circular) failure arc into a series of vertical slices, computes the weight, and hence the driving (tangential) and resisting (normal × friction, plus cohesion) force of EACH slice independently along its own small arc segment, then sums all slices’ contributions to get overall driving and resisting moments about the arc centre — allowing the strength and geometry to vary slice-by-slice rather than requiring one uniform assumption for the whole arc (the Swedish/Fellenius circle problem below is the single-slice DEGENERATE case of this same method). “Bank failure” is a slip surface that stays entirely within the slope itself, exiting on the FACE above the toe; “base failure” is a deeper slip surface that extends below the toe elevation and exits beyond the toe on the floor — base failures mobilise a larger volume and are generally associated with weaker, more ductile foundation material beneath the slope.

Approach (4.5/4.6). Effective-stress strength parameters (c, φ) do NOT themselves change with saturation — only the pore pressure acting on the failure surface changes — so the SAME c=5.51 kPa and φ=30° and the SAME driving/opposing moments (W1, W2 unaffected by water) apply to both the dry and saturated case; the only quantity that differs is the shear strength S along the arc, via the pore-pressure term.

1. Driving and opposing moments (same for both cases). Md = W1·l1 = 13,342×20 = 266,832 kN·m. Mr,weight = W2·l2 = 9,632×27 = 260,102 kN·m.
2. Shear strength S, dry. σ = γh̄ = 18.85×12 = 226.2 kPa. S_dry = c + σtanφ = 5.51 + 226.2(0.5774) = 136.1 kPa.
3. FS, dry. $$FS_{dry}=\dfrac{M_{r,weight}+S_{dry}\cdot L_{arc}\cdot R}{M_d}=\dfrac{260{,}102+136.1(107)(55)}{266{,}832}$$ $$\boxed{FS_{dry}=3.98}$$
4. Shear strength S, saturated. Using the same h̄=12 m for the pore pressure, u = γ_wh̄ = 9.81×12 = 117.7 kPa. S_sat = c + (σ−u)tanφ = 5.51 + (226.2−117.7)(0.5774) = 68.2 kPa.
5. FS, saturated. $$FS_{sat}=\dfrac{260{,}102+68.2(107)(55)}{266{,}832}$$ $$\boxed{FS_{sat}=2.48}$$

Both Swedish-circle results are ACCEPTABLE (FS > the 1.2–1.3 non-permanent-slope minimum from 4.1) even saturated, in contrast to the simple planar wedge of 4.2 which drops to a marginal 1.06 when saturated — illustrating why a SINGLE representative failure mode/geometry should never be assumed governing without checking the alternative mechanisms the site structure actually permits.