18-Env-A3 Geotechnical and Hydrogeological Engineering · Undated paper

Question 6 of 6: Slope Stability — Ordinary Method of Slices

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — 18-Env-A3, Geotechnical and Hydrogeological Engineering — 3 hours, open book. Six questions, each 20 marks, equal value; all six are solved below.

Reference texts: Das, Principles of Geotechnical Engineering, 9th ed.; Craig & Knappett, Craig's Soil Mechanics, 8th ed.; Freeze & Cherry, Groundwater (1979).

Question 6: Slope Stability — Ordinary Method of Slices (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 7 m; slope face runs from the toe up to the crest break; failure surface passes through the toe and daylights on the flat crest; the two slices are divided at the crest-break vertical, with the figure's own dimension line marking slice widths 6 m (slice 2, nearer the toe) and 4 m (slice 1, nearer the daylight point). $\gamma_{dry}=16\text{ kN/m}^3$, $\phi'=30^\circ$, $c'=0$ (non-cohesive), $n=30\%$.

Find. (a) $FS$ against shear failure, dry. (b) $FS$ against shear failure, fully saturated.

[Figure not reproduced: Fig. 4 — Slope with two labelled slices; the exact circular slip-surface radius/centre could not be recovered from the source figure, so the failure surface is approximated by the straight chord shown (see the check note below). See the official exam paper.]

Check — failure-surface reconstruction: the source figure only labels the slope height (7 m) and the two slice widths (6 m, 4 m); it does not give the failure circle's radius or centre, and the purpose of the dimension line is not specified. Per the OMS convention of approximating each slice base as a straight chord of the true arc, the whole two-slice failure surface is modelled here as the single straight chord from the toe (0,0) to the point where the surface daylights on the crest (10 m out, 7 m up) — giving one shared base angle $\alpha=\arctan(7/10)=35.0^\circ$ for both slices. This is a stated engineering assumption, not a recovered measurement: a true circular arc would generally give a shallower angle near the toe and a steeper one near the crest, which would raise the computed $FS$ somewhat. The qualitative conclusion below (saturation collapses an already-marginal margin of safety) is robust to this approximation even though the exact numbers are not.

Approach. With one shared base angle, Ordinary-Method-of-Slices weights, normal/shear components and pore pressures are computed slice-by-slice from the geometry, then summed per Fellenius' equation $FS=\Sigma(N'\tan\phi')/\Sigma T$.

Slice geometry. Base angle $\alpha=\arctan(7/10)=35.0^\circ$ for both slices. Slice 2 (toe side, width 6 m) and slice 1 (crest side, width 4 m) each have a trapezoidal cross-section between the ground surface and the chord; working out the trapezoid areas from the given geometry gives an average height of 1.4 m for both slices, so: $$A_2 = 1.4\times6 = 8.4\text{ m}^2,\qquad A_1 = 1.4\times4 = 5.6\text{ m}^2$$ Base (chord) lengths: $l_2=6/\cos35^\circ=7.33\text{ m}$, $l_1=4/\cos35^\circ=4.88\text{ m}$.
(a) Dry case. $W_2=\gamma A_2=16\times8.4=134.4\text{ kN/m}$, $W_1=16\times5.6=89.6\text{ kN/m}$. With $c'=0$ and no pore pressure, $N_i'=W_i\cos\alpha$ and $T_i=W_i\sin\alpha$; because both slices share the same $\alpha$, the sum collapses to the familiar infinite-slope form: $$FS_a = \dfrac{\Sigma(W_i\cos\alpha)\tan\phi'}{\Sigma(W_i\sin\alpha)} = \dfrac{\tan\phi'}{\tan\alpha} = \dfrac{\tan30^\circ}{\tan35.0^\circ} = \dfrac{0.577}{0.700} = \boxed{0.82}$$
(b) Saturated case. $\gamma_{sat}=\gamma_{dry}+n\gamma_w=16+0.30\times9.81=18.94\text{ kN/m}^3$. $W_{2,sat}=18.94\times8.4=159.1\text{ kN/m}$, $W_{1,sat}=18.94\times5.6=106.1\text{ kN/m}$. With the phreatic surface at the ground surface, pore pressure at each base is $u=\gamma_w h_{avg}=9.81\times1.4=13.73\text{ kPa}$, giving pore forces $U_2=u\,l_2=100.6\text{ kN/m}$, $U_1=u\,l_1=67.1\text{ kN/m}$. $$N_2'=W_{2,sat}\cos\alpha-U_2=130.3-100.6=29.7\text{ kN/m},\quad N_1'=W_{1,sat}\cos\alpha-U_1=86.9-67.1=19.8\text{ kN/m}$$ $$T_2=W_{2,sat}\sin\alpha=91.3\text{ kN/m},\quad T_1=W_{1,sat}\sin\alpha=60.9\text{ kN/m}$$ $$FS_b = \dfrac{(N_2'+N_1')\tan\phi'}{T_2+T_1} = \dfrac{49.6\times0.577}{152.1} = \boxed{0.19}$$

Final results — Question 6
Quantity	Value
(a) $FS$, dry	0.82
(b) $FS$, fully saturated	0.19

Topic: Slope stability — Ordinary (Fellenius) Method of Slices for a cohesionless soil.

Key relations: $N_i'=W_i\cos\alpha_i-u_il_i$; $T_i=W_i\sin\alpha_i$; $FS=\Sigma(N_i'\tan\phi_i')/\Sigma T_i$; $\gamma_{sat}=\gamma_{dry}+n\gamma_w$.

Why this works: The Ordinary Method of Slices resolves each slice's weight into components normal and tangential to its own base, sums the frictional resistance available on every base and compares it to the total driving shear — for a purely frictional soil with every slice sharing one base angle this collapses exactly to the classical infinite-slope result $FS=\tan\phi'/\tan\alpha$, which is why saturation (adding pore pressure $u$ without changing $\tan\phi'$ or $\alpha$) is so damaging: it directly reduces the effective normal force $N'$ that friction acts on, while the driving term $T$ (built from total weight) barely changes.

Common pitfall: Using the saturated unit weight but forgetting to subtract the pore-pressure force $U$ from $N'$ (double counting water's benefit and its detriment), or applying $\gamma_{sat}$ to the driving side without applying the corresponding pore pressure to the resisting side.

Source: Das, Principles of Geotechnical Engineering, Ch. 15 (Slope Stability); Craig & Knappett, Soil Mechanics, Ch. 9.

← Question 5 Back to the paper →