22-Agric-A2 Soil Physics and Mechanics · December 2019
Question 5 of 6: Slope Stability by the Ordinary Method of Slices
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 04-Agric-A2 Soil Physics & Mechanics,
National Exams December 2019 — a three-hour open-book examination;
any non-communicating calculator is permitted. The cover page states that five (5)
questions constitute a complete exam paper and that only the first five as they appear
in the answer book are marked, that each question is of equal value, and that some
questions require a written answer whose clarity and organization matter for marks. All
six printed questions are worked here, because the set is a study resource rather than a
timed attempt; on exam day a candidate submits only the first five, in order.
the overall
slope height (7 m + 1 m = 8 m) and the two horizontal slice widths (6 m, 4 m) are legible, but
the exact shape of the printed circular failure arc between the toe and the crest is not. The slip circle used below is reconstructed: a circular arc through the toe
(0, 0) and the point where the arc daylights at the crest (10, 8) — consistent with the
legible 6 m / 4 m slice split — with radius R = 12 m chosen as a representative toe-circle
radius for a slope of this height. The Ordinary Method of Slices mechanics below are exact for
whatever circle is used; only the specific slice weights/angles (and hence the numeric FS)
depend on this reconstruction, and should be checked against the original exam figure if a
clean copy becomes available.
Given.
Unit weight γ = 17 kN/m³; friction angle φ′ = 30°; c′ = 0
(dry, non-cohesive); slope height 8 m; slice 2 width 6 m (toe side); slice 1 width 4 m (crest
side).
Find. Factor of safety against slope shear failure (a); three stabilization
methods (b).
Reconstructed slip circle (toe at origin, daylights 10 m along at the
crest, R = 12 m) with the two given slices: Slice 2 (0–6 m, toe side, under the sloped
face) and Slice 1 (6–10 m, crest side, under the flat crest).
Approach. With c′ = 0, only the frictional term of the Ordinary
Method of Slices survives; compute each slice's weight (from its average height above the
arc) and the inclination of the arc at its centre, then sum.
Slice geometry (from the reconstructed circle, centre (−1.34, 11.92), R =
12 m). At each slice's mid-width, the ground surface height minus the arc height
gives the slice's representative height, and the arc's local slope gives its base angle
α:
$$\begin{aligned}
\text{Slice 2 (}x_c&=3\text{ m):}\quad h_2=4.00-0.74=3.26\ \text{m}, \quad \alpha_2=21.2^\circ\\
W_2&=\gamma h_2(6)=17(3.26)(6)=332.8\ \text{kN/m}
\end{aligned}$$
$$\begin{aligned}
\text{Slice 1 (}x_c&=8\text{ m):}\quad h_1=8.00-4.39=3.61\ \text{m}, \quad \alpha_1=51.1^\circ\\
W_1&=\gamma h_1(4)=17(3.61)(4)=245.4\ \text{kN/m}
\end{aligned}$$
Ordinary Method of Slices, c′ = 0. Resolving each slice weight into
components normal and tangential to its own base:
$$FS=\frac{\sum W_i\cos\alpha_i\tan\phi'}{\sum W_i\sin\alpha_i}
=\frac{\tan(30^\circ)\left[W_1\cos\alpha_1+W_2\cos\alpha_2\right]}
{W_1\sin\alpha_1+W_2\sin\alpha_2}$$
$$\begin{aligned}
&=\frac{0.5774\left[245.4(0.628)+332.8(0.933)\right]}{245.4(0.778)+332.8(0.362)}\\
&=\frac{0.5774(464.3)}{190.9+120.3}=\frac{268.1}{311.2}=\boxed{0.86}
\end{aligned}$$
b) Three design methods to improve stability. (i) Flatten the
slope — reducing the overall slope angle directly reduces the driving (tangential)
component of weight on every trial surface, the most reliable fix for a purely frictional
soil where FS = tanφ′/tanβ on an infinite slope. (ii) Provide a toe
buttress or berm — adding mass (a counterweight fill or rock berm) at the toe
increases the resisting normal force exactly where the failure surface exits, without
changing the geometry above. (iii) Install drainage or soil reinforcement —
even though this slope is specified dry, subsurface drains prevent any future pore-pressure
buildup from reducing effective stress and hence shear strength, while soil nails, geogrid
reinforcement, or a retaining structure add an external resisting force/moment directly to
the limit-equilibrium calculation.
Quantity
Value
a) Factor of safety, Ordinary Method of Slices
0.86 (reconstructed circle — see the check note)
b) Stabilization methods
flatten slope; toe buttress/berm; drainage or reinforcement
