Question 5 of 6: Undrained Slope Stability of a Municipal Drainage Ditch
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 May 2017 — 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.
Reference texts. B.M. Das, Principles of Geotechnical
Engineering, 9th ed. (weight-volume relationships, permeability, grain-size
analysis, USCS classification, compaction, slope stability, well hydraulics); R.F.
Craig, Craig's Soil Mechanics, 9th ed. (effective stress, seepage and flow
nets, shear strength); USDA NRCS National Engineering Handbook (compaction
and earthwork field practice).
Question 5: Undrained Slope Stability of a Municipal Drainage Ditch (20 marks)
Find. The factor of safety of the existing 2H:1V cut (a), and the
side slope that would give a minimum FS of 3 (b).
Figure 3 (schematic): the 3 m deep, 2H:1V municipal drainage
ditch in saturated clay.
Approach. For an undrained (φu = 0) clay slope, the
resisting moment along any trial circular slip surface is simply cu
× (arc length) × R, so the factor of safety against a circular failure
reduces to Taylor's stability-number method: search trial circles for the minimum
FS = cuR²θ/(γ·Area·d), where Area and d are
the sliding wedge's cross-sectional area and the horizontal lever arm of its weight
about the circle's centre. The search here was carried out numerically and checked
against Taylor's own published values at β = 90° (Ns = 3.83) and
β = 53.1° (Ns = 5.52) before being applied to the actual slope.
a) Governing failure mode and FS. Because β = 26.6° is
well below 53°, the critical surface here is a deep (base) circle passing below
the toe rather than a toe circle, and Taylor's stability number for this regime
saturates near its asymptotic value Ns ≈ 5.52 — the SAME value
that governs right at the β = 53.1° boundary — because once a slope is
shallow enough, the critical circle no longer even touches the sloping face. The
numerical search confirms this: the minimum-FS circle for the actual 2H:1V cut
converges to
$$FS=\frac{N_s}{\gamma H/c_u}=\frac{5.52}{1.3875}=\boxed{FS\approx3.98}$$
The existing ditch is comfortably stable, with roughly a factor of 4 margin against
shear failure.
b) Side slope for FS = 3. Because the deep-circle FS above is
essentially independent of face angle for any β ≤ 53°, flattening the
ditch beyond 2H:1V would buy almost no extra safety margin. Reducing FS to 3 instead
requires STEEPENING the cut well past β = 53°, into the toe-circle regime,
where Taylor's stability number falls smoothly from 5.52 (at 53.1°) toward 3.83
(at a vertical, 90° face). Repeating the numerical search over toe circles for a
family of steeper slopes and interpolating to FS = 3 gives
$$\cot\beta\approx0.20 \quad(\beta\approx79^\circ), \qquad
FS=\frac{N_s(\beta)}{\gamma H/c_u}\approx\frac{4.16}{1.3875}\approx\boxed{3}$$
i.e. the cut could be made far steeper than the existing 2H:1V — roughly
1V:0.2H, close to a vertical face — and still meet a minimum FS of 3.
Quantity
Value
γH/cu
1.39
Governing failure mode at 2H:1V
Deep (base) circle
FS of the existing 2H:1V ditch
≈ 3.98
Side slope giving FS = 3
≈ 0.2H:1V (β ≈ 79°)
Check
No firm stratum depth is given below the ditch, so the deep-circle search assumes a
thick, homogeneous clay layer (no impermeable/rigid boundary shallower than the
critical circle) — a real firm layer close to the ditch invert would force a
shallower critical circle and could raise or lower FS from the value above.
