Question 4 of 5: Plane failure with a water-filled tension crack

Question 4: Plane failure with a water-filled tension crack (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.

Given. Slope height H = 30 m; face AB at 60° to horizontal; horizontal crest distance BC = b = 8.66 m; fracture plane AD dips 30°; c = 20 kPa, φ = 30°; γ_rock = 25 kN/m³, γ_water = 10 kN/m³.

Find. (a) general FoS(z_w) equation; (b) FoS with the tension crack dry; (c) FoS with the tension crack full of water, and whether it explains the observed winter failure.

Approach. Fix the block geometry (A, B, C, D) from the given angles and dimensions, express the block weight and the two water forces (uplift on AD, thrust on CD) in terms of the unknown crack water depth z_w, then assemble the limit-equilibrium factor of safety as resisting force over driving force along the sliding plane.

1. Part (a) — Fix the block geometry. Taking A as the origin with the face rising at 60°: B = (H/tan60°, H) = (17.32, 30) m. The crest surface runs horizontally a further b = 8.66 m to C = (25.98, 30) m. The tension crack is vertical, so D lies on plane AD directly below C; solving the plane AD (dip 30° from A) for the same x-coordinate as C gives D = (25.98, 15.00) m. This gives a sliding-plane length and a tension-crack depth of $$L_{AD}=|AD|=30.0\ \text{m},\qquad z=|CD|=H-D_y=15.0\ \text{m}$$ i.e. the tension crack reaches exactly to mid-height of the slope — a useful internal check on the figure's stated dimensions. Simplifying assumptions: the slide is a rigid 2-D block on a planar surface of unit thickness (plane-strain, per metre of slope length); the tension crack is vertical and dry above any water level in it; when water is present it fills the crack to depth z_w and seeps along AD, dissipating linearly from a peak of γ_wz_w at D to zero at the toe A (a standard triangular pressure distribution); no seismic or external load is applied; shear strength on AD is Mohr-Coulomb (c, φ) with no scale effect.
2. Weight of the block. The cross-sectional area of quadrilateral ABCD (shoelace formula on A, B, C, D) is 324.8 m² per metre of slope length, so $$W=\gamma_{rock}\times\text{Area}=25\times324.8=8119\ \text{kN/m}$$
3. Water forces as a function of z_w. The triangular uplift pressure along AD (peak γ_wz_w at D, zero at A) resultant is $$U=\tfrac12\gamma_w z_w L_{AD}$$ and the triangular horizontal thrust on the vertical crack face CD (peak γ_wz_w at D, zero at the water surface) resultant is $$V=\tfrac12\gamma_w z_w^2$$
4. Part (a) — Assemble the factor of safety. Resolving forces parallel and normal to plane AD (dip ψ_p=30°), the driving component is the down-dip weight component plus the down-dip component of V; the resisting component is cohesion over the full plane area plus friction on the effective (uplift-reduced) normal force: $$\boxed{FoS(z_w)=\frac{cL_{AD}+\big[W\cos\psi_p-U-V\sin\psi_p\big]\tan\phi}{W\sin\psi_p+V\cos\psi_p}}$$ with U and V from the previous step, both functions of the single unknown z_w.
5. Part (b) — Dry conditions (mid-summer). With z_w=0, U=V=0 and the equation collapses to the simple dry form: $$FoS_{dry}=\frac{cL_{AD}+W\cos\psi_p\tan\phi}{W\sin\psi_p}=\frac{(20)(30)+8119\cos30^\circ\tan30^\circ}{8119\sin30^\circ}$$ $$\boxed{FoS_{dry} = 1.15}$$ The slope was therefore stable (FoS > 1) when it was excavated in dry mid-summer conditions, consistent with no failure being reported at that time.
6. Part (c) — Saturated tension crack (winter). Heavy winter rain is taken as the critical, worst-case condition of the crack filling completely to the surface, z_w = z = 15.0 m (the tension-crack depth found in step 1), giving U = ½(10)(15.0)(30.0) = 2250 kN/m and V = ½(10)(15.0)² = 1125 kN/m. Substituting into the part (a) equation: $$FoS_{wet}=\frac{(20)(30)+\big[8119\cos30^\circ-2250-1125\sin30^\circ\big]\tan30^\circ}{8119\sin30^\circ+1125\cos30^\circ}$$ $$\boxed{FoS_{wet} = 0.60}$$ Since FoS_wet = 0.60 < 1.0, the computed factor of safety does predict failure under a saturated tension crack — the uplift force U and crack thrust V together remove enough of the plane's normal confinement and add enough down-dip driving force that the same geometry and material that was stable dry (FoS 1.15) becomes unstable when the tension crack fills with water (FoS 0.60). This is fully consistent with the slope standing through dry mid-summer and then failing after the winter rain saturated the crack.