18-Env-A3 Geotechnical and Hydrogeological Engineering · Undated paper
Question 3 of 6: Seepage Beneath a Retaining Wall
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.
Given. Wall height 5 m, base width 3 m, top width 1 m; sand layer 5 m thick beneath the wall over an impervious stratum; $G_s=2.65$, $n=20\%$, $k=1.5\times10^{-4}$ cm/s; water table upstream at the top of the wall (5 m above the base); downstream ground is dry, at the elevation of the wall base; wall length 100 m.
Find. (a) Total seepage rate under the wall. (b) Piezometric head at Point A. (c) Factor of safety against heave at the downstream toe.
Fig. 1 — Retaining wall on a permeable sand layer, underlain by an impervious stratum. Water floods the upstream (back) side to the top of the wall; the downstream toe is dry. Point A sits at the base of the sand layer directly beneath the downstream toe.
Approach. The wall's own base is the only impervious "floor" the water must travel beneath (the sand above and behind the wall is not itself impervious), so this reduces to the single-fragment Method of Fragments: treat the seepage path as 1-D horizontal flow of length equal to the wall's base width, driven by the full head difference between the upstream water table and the dry downstream grade.
(a) Seepage rate. Sand properties: $e=n/(1-n)=0.20/0.80=0.25$, $\gamma_{sat}=(G_s+e)\gamma_w/(1+e)=(2.65+0.25)/1.25\times9.81=22.76\text{ kN/m}^3$, $\gamma_b=\gamma_{sat}-\gamma_w=12.95\text{ kN/m}^3$. Convert $k$: $1.5\times10^{-4}\text{ cm/s}\times36 = 5.4\times10^{-3}\text{ m/hr}$ (since $1\text{ cm/s}=36\text{ m/hr}$). With floor width $B=3\text{ m}$ (the wall's base), layer thickness $a=5\text{ m}$, and head loss $\Delta H = 5\text{ m}$ (upstream WT at the wall top down to the dry downstream grade at the wall base): $$q = \dfrac{k\,a\,\Delta H}{B} = \dfrac{0.0054\times5\times5}{3} = 0.0450\text{ m}^3/\text{hr per m}$$ Over the 100 m wall length: $$Q = 0.0450\times100 = \boxed{4.50\text{ m}^3/\text{hr}}$$
(b) Piezometric head at A. In this single-fragment (1-D horizontal-flow) idealisation the total head at any point is uniform across the depth of the layer and varies only along the flow direction, from $\Delta H$ above the exit datum at the entry (upstream edge of the wall base) down to the downstream boundary condition at the exit (the dry, atmospheric ground surface at the toe, elevation $=a=5\text{ m}$ above the base of the layer). Point A sits exactly at that exit boundary, so its total head, referenced to the bottom of the sand layer, equals the downstream grade elevation: $$\boxed{\text{head at A} = 5\text{ m}}$$ i.e. a piezometer at A would show water rising 5 m above A, to the level of the downstream ground surface.
(c) Factor of safety against heave. The critical (upward) hydraulic gradient for this cohesionless sand is $$i_{cr} = \dfrac{\gamma_b}{\gamma_w} = \dfrac{G_s-1}{1+e} = \dfrac{1.65}{1.25} = 1.32$$ Absent a full flow net, the exit gradient beneath the floor is approximated by the same average horizontal gradient used to find $q$ in part (a): $$i_{exit} \approx \dfrac{\Delta H}{B} = \dfrac{5}{3} = 1.667$$ $$FS_{heave} = \dfrac{i_{cr}}{i_{exit}} = \dfrac{1.32}{1.667} = \boxed{0.79}$$
Check: part (c)'s exit gradient is approximated as the average gradient under the floor ($\Delta H/B$), since the figure gives no flow net to read a true local gradient at the toe. A rigorous flow net typically shows the gradient concentrating (even singular, per the corner-effect noted for cutoff-wall tips elsewhere in this subject) right at the toe corner, which would only lower this FS further — so $FS_{heave}=0.79$ is, if anything, an optimistic (upper-bound) estimate. Either way $FS<1$: the toe is at real risk of heave/piping and would need a filter, weighted apron, or an extended cutoff in practice.
