16-Civ-A4 Geotechnical Materials and Analysis · December 2017

Question 6 of 6: Seepage, Effective Stress and Exit Gradient (Flow Net)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 6: Seepage, Effective Stress and Exit Gradient (Flow Net) (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. A concrete dam, with its base embedded 1.6 m below the ground, on a permeable foundation 17.2 m deep over an impermeable stratum. The upstream water surface is 6.3 m above the ground; the downstream water level is at the ground surface (the ∇ symbol sits on the ground line). Counting the printed net gives N_f = 3 flow channels and N_d = 10 equipotential drops. Point A lies 9.4 m below the ground surface, downstream of the dam toe, on the last full equipotential before the exit.

Find. (i) The seepage q per metre of dam; (ii) the effective stress at A; (iii) the maximum exit gradient and its consequence relative to the critical gradient.

Approach. Use the flow-net discharge formula for q; obtain the total head at A from the number of potential drops, convert to pore pressure via the pressure head, and subtract from the total vertical stress; then compare the exit gradient (drop over the shortest exit field) with the critical gradient γ′/γ_w.

1. Seepage (i). The downstream water is at ground level, so the head lost across the net is the full upstream height, H = 6.3 m (the 1.6 m is the dam’s embedment, not a tailwater depth). The head loss per drop is Δh = H/N_d = 6.3/10 = 0.63 m, and $$q=k\,H\,\frac{N_f}{N_d}=25\times10^{-6}\times6.3\times\frac{3}{10}=4.73\times10^{-5}\ \text{m}^3/\text{s per m}.$$ Per day: $q=4.73\times10^{-5}\times86400=\boxed{4.1\ \text{m}^3/\text{day per m}}.$
2. Total head at A. Take the downstream water surface (the ground) as datum. A is 8 drops from the upstream face, so 2 drops remain to the exit and its total head above datum is $$h_A=6.3-8\times0.63=2\times0.63=1.26\ \text{m}.$$
3. Pore pressure at A. A lies 9.4 m below the datum (z_A = −9.4 m), so the pressure head is h_p = h_A − z_A = 1.26 + 9.4 = 10.66 m and $$u_A=\gamma_w\,h_p=9.81\times10.66=104.6\ \text{kPa}.$$
4. Total and effective stress at A. No free water stands on the downstream ground, so only the 9.4 m of saturated soil acts above A: $$\sigma_A=9.4\times19=178.6\ \text{kPa},$$ $$\sigma_A'=\sigma_A-u_A=178.6-104.6=\boxed{74\ \text{kPa}}.$$ Check: with no flow, σ′ would be γ′z = 9.19 × 9.4 = 86.4 kPa. The flow at A is upward (toward the exit), which cuts σ′ by the 2 remaining drops, 2 × 0.63 × 9.81 = 12.4 kPa, giving 74.0 kPa.
5. Critical gradient (iii). $i_{cr}=\dfrac{\gamma'}{\gamma_w}=\dfrac{19-9.81}{9.81}=\boxed{0.94}.$
6. Maximum exit gradient. The largest exit gradient is in the smallest last field, the one against the dam’s downstream face. It is bounded by the short equipotential leaving the dam face, the first interior flow line and the ground. Scaled from the net (17.2 m = full depth), its flow-path length is about 1.0 m at the dam face and 1.5 m at the flow line, so l ≈ 1.25 m and $$i_{exit}=\frac{\Delta h}{l}\approx\frac{0.63}{1.25}\approx0.50,\qquad \text{FS}=\frac{i_{cr}}{i_{exit}}\approx1.9.$$ The exit gradient is below critical, so the dam as drawn does not boil. Still, a factor of safety of about 1.9 is below the 3–4 usually recommended against piping, so an exit filter is advisable. If the exit gradient exceeded i_cr, the effective stress at the downstream surface would fall to zero: the sand would boil (a quick condition), and progressive backward erosion (piping) beneath the dam would undermine and could ultimately fail it. The standard remedies are a downstream loaded filter / inverted graded filter, a longer or deeper cut-off, and relief wells to bleed off the head.