Question 2 of 6: Flow Net and Uplift Force on a Tunnel in an Embankment

Question 2: Flow Net and Uplift Force on a Tunnel in an Embankment (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. The embankment cross-section (Figure 1): a 30 m tall, 80 m wide (toe-to-toe) trapezoidal fill with a 25 m wide crest, carrying a 20 m wide by 10 m tall tunnel centred in the base (10 m below crest, resting on rock at mid-height). A river holds the upstream face wetted to $h=25$ m and the downstream face wetted to $h=15$ m (both measured above the impermeable rock base).

Find. (a) the flow net for the worst-case (highest sustained differential-head) condition; (b) the uplift force per unit length of tunnel acting on its base.

[Figure not reproduced: Figure 1 (redrawn) with the solved flow net superimposed — equipotentials (blue, head in m) and flow lines (red). See the official exam paper.]

Approach. "Worst case" means treating the embankment as fully saturated up to crest level rather than relying on an uncertain internal phreatic surface — the conservative, maximum-uplift condition. Laplace's equation $\nabla^2h=0$ then governs head $h(x,z)$ in the soil, with $h=25$ m fixed on the wetted upstream face, $h=15$ m fixed on the wetted downstream face, and no-flow (zero head-gradient) on the rock base, the dry crest/slope surfaces, and the impermeable tunnel lining. Rather than hand-sketch curvilinear squares, the same boundary-value problem is solved numerically (finite-difference relaxation on a 0.5 m grid) — this is exactly a flow net, just read off a computed head field instead of a ruler; $N_f/N_d$ and the uplift both fall out of it directly.

1. Set up the seepage domain and boundary conditions. With the base spanning $x=0$ (upstream toe) to $x=80$ m (downstream toe) and the crest centred at $x=27.5$–$52.5$ m, the tunnel occupies $x=30$–$50$ m, $z=10$–$20$ m. Fixed heads of 25 m and 15 m apply on the wetted portions of the two side slopes (below their respective water lines); every other boundary (rock base, dry slope/crest, tunnel lining) is no-flow.
2. Solve Laplace's equation for the head field. Gauss–Seidel relaxation on the masked grid (dry region and tunnel void excluded) converges to the head $h(x,z)$ shown by the blue equipotentials above; the red flow lines are the corresponding orthogonal streamlines. Reading off the number of flow channels and equipotential drops gives an equivalent $N_f/N_d\approx1.12$, so $Q\approx k\,\Delta H\,(N_f/N_d)\approx8\times10^{-7}\times10\times1.12=8.9\times10^{-6}\ \text{m}^3/\text{s per m}$ — a useful cross-check that the mesh has converged to a physically sensible flow net.
3. Read the head along the tunnel base. Sampling $h$ along $z=10$ m, $x=30$–$50$ m gives a head that falls from about 23.0 m (upstream side) to 18.7 m (downstream side), averaging $\bar h=\boxed{20.8\ \text{m}}$.
4. Convert head to pore pressure and integrate for the uplift force. At each point on the base, $u=\gamma_w(h-z)$ with $z=10$ m and $\gamma_w=9.81$ kN/m³. Integrating $u$ across the 20 m base width, $$U=\int_{30}^{50}\gamma_w\big(h(x,10)-10\big)\,dx=\boxed{2113\ \text{kN per metre of tunnel}}.$$