18-Env-A3 Geotechnical and Hydrogeological Engineering · May 2017
Question 6 of 6: Plane (Wedge) Slope Stability — Dry and Rain-Saturated Conditions
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2017 — 04-Env-A3 / Geotechnical & Hydrogeological Engineering. 3 hours duration; open book exam, any non-communicating calculator permitted. FIVE (5) questions constitute a complete exam paper (the first five as they appear in the answer book are marked, 20 marks each, 100 marks total); all six printed questions are solved below for completeness.
Reference texts. Braja M. Das, Principles of Geotechnical Engineering (9th ed.) — weight–volume relations, seepage/flow nets, grain-size analysis, consolidation and slope-stability chapters; Craig & Knappett, Craig's Soil Mechanics (8th ed.) — cross-reference for flow-net theory and the Method of Fragments; Freeze & Cherry, Groundwater (1979) — Darcy's law and the Dupuit–Thiem equation for radial flow to a well.
Find. (a) FS against sliding on plane AB, dry; (b) FS after saturation and pore-pressure buildup on AB.
Approach. Treat ABC as a finite wedge on a rigid base: get its weight from the triangle geometry, resolve the weight into components normal and parallel to the assumed failure plane, and compare Mohr–Coulomb shear resistance to the driving shear force.
Fig. Q4 (source) — wedge ABC: A at the toe, failure plane AB at $\theta$, face AC at $\beta$, top BC horizontal.
Wedge geometry. With B and C both at height $H$ above A, $\text{Area}=\tfrac12 H^2(\cot\theta-\cot\beta)=\tfrac12(3)^2(1.1918-0.8391)=1.587\ \text{m}^2/\text{m}$, and the failure-plane length $L=H/\sin\theta=3/\sin40^\circ=4.667$ m.
Saturated unit weight. With no $G_s$ given, use $\gamma_{sat}=\gamma_{dry}+n\gamma_w=18+0.25(9.81)=20.45$ kN/m³, giving $W_{sat}=32.46$ kN/m, $N_{sat}=24.86$ kN/m, $T_{sat}=20.86$ kN/m.
Part (b) — pore pressure on AB. With the water table rising to the crest (ground surface at elevation $H$) during the rain event, the vertical depth of water above a point on AB at elevation $z$ is $H-z$, giving a hydrostatic pore pressure that is maximum at the toe A ($z=0$) and zero at the crest B ($z=H$):
$$u_A=\gamma_w H=9.81(3)=29.43\ \text{kPa},\qquad U=\tfrac12 u_A L=\tfrac12(29.43)(4.667)=68.68\ \text{kN/m}\ \text{(triangular resultant, normal to AB)}.$$
$$N'=N_{sat}-U=24.86-68.68=-43.81\ \text{kN/m},$$
$$S_{sat}=cL+N'\tan\phi=116.68-25.30=91.38\ \text{kN/m},\qquad FS_b=\frac{S_{sat}}{T_{sat}}=\frac{91.38}{20.86}=\boxed{4.38}.$$
Check: the computed pore-pressure resultant $U=68.7$ kN/m exceeds the saturated normal force $N_{sat}=24.9$ kN/m, i.e. $N'$ is formally negative. Physically this means the plane would separate (go into tension) rather than sustain a negative friction contribution; flooring $N'$ at zero instead gives $FS_b=5.59$. Either way the very large cohesion term ($cL=116.7$ kN/m) keeps $FS_b$ well above 1 — the wedge remains stable, but the margin drops sharply (FS 7.04→4.38) once saturation and pore pressure are introduced, which is the point of the exercise.
