Question 5 of 6: Seepage Under a Dam With a Partial Cutoff (Sheet Pile)

Question 5: Seepage Under a Dam With a Partial Cutoff (Sheet Pile) (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.

Find. (a) unit-width seepage rate $q$ under the dam; (b) unit-width uplift force on the dam base.

Approach. The cutoff wall sits INSIDE the dam's footprint (not at either end), so this is not a textbook end-cutoff case; solve the 2-D Laplace seepage equation directly on the actual geometry — Dirichlet heads at the exposed upstream/downstream ground, no-flow along the impervious dam base and stratum floor, and the sheet pile modelled as a zero-thickness no-flow barrier for the top 7 m — with a finite-difference grid (this is exactly a flow net, just solved numerically instead of sketched by hand). Then read off the flow rate as the flux through any vertical section under the dam, and the uplift as the integral of the computed pressure head along the base.

1. Set up the flow domain. Take $y=0$ at the top of the silty sand (ground surface) and $x=0$ at B. Boundary conditions: $h=10$ m for $x<0$ (upstream bed, under the reservoir); $h=0$ for $x>33$ m (downstream bed, at the tailwater datum); no vertical flow ($\partial h/\partial y=0$) along $0\le x\le33$ m at $y=0$ (impervious dam base) and along $y=-20.3$ m (impervious clay contact); no horizontal flow across the sheet pile ($x=2.4$ m, $-7\le y\le0$).
2. Solve numerically (finite-difference / equivalent flow net). Discretising the domain (square cells, grid-refinement checked to <1% change in $q$) and solving Laplace's equation for head $h(x,y)$ gives a smooth potentiometric field; head drops gently across the base upstream of the pile, THEN drops sharply across the sheet pile itself (from ≈9.45 m to ≈5.9 m — about 35% of the total head loss is consumed right at the cutoff), then declines gradually to zero at D. Integrating the resulting horizontal flux $q_x=-k\,\partial h/\partial x$ over the stratum depth, at several sections under the dam, gives a consistent $$q=\boxed{0.603\ \text{m}^3/\text{day per m width}}\ \left(\approx6.98\times10^{-6}\ \text{m}^3/\text{s per m}\right).$$ As a cross-check, this corresponds to an equivalent flow-net ratio $N_f/N_d=q/(kH)=0.603/(0.1728\times10)\approx0.35$, a plausible value for a single mid-length cutoff of this relative depth ($d/T=0.34$).
3. Part (b) — uplift force from the computed pressure-head profile. Because the dam base is horizontal ($z=0$ along B–D), the pressure head at every point on the base equals the total head computed there, $u(x)/\gamma_w=h(x,0)$. Integrating $\gamma_w\,h(x)$ along the base (numerically, from the same solved field) gives the resultant uplift force per unit width: $$U=\int_{0}^{33}\gamma_w\,h(x)\,dx=\boxed{1{,}375\ \text{kN per m width}}.$$ (Grid-refinement checks place this integral in the 1,360–1,390 kN/m range — a ≈2% numerical band, typical of the corner-singularity in head gradient right at B and D.)