Question 5 of 6: Cut-off wall — flow net and effective stress

Question 5: Cut-off wall — flow net and effective stress (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 sheet-pile cut-off wall in a 6 m permeable stratum over an impermeable layer. Upstream water stands 3 m above the ground surface; the downstream water table is at the ground surface, so the head loss is $H=3\ \text{m}$. The wall penetrates 3 m below ground; point A is on the back (upstream) face, 1 m below the ground surface. $\gamma_{sat}=20\ \text{kN/m}^3$, $\gamma_w=9.81\ \text{kN/m}^3$, $k=2.0\times10^{-5}\ \text{m/s}$.

Find. (a) a valid flow net; (b) the effective stress $\sigma'$ at point A.

[Figure not reproduced: Single sheet-pile cut-off wall. Flow passes down the upstream face, around the wall tip, and up to the downstream bed; equipotential lines (dashed) cross the flow lines at right angles. The net is traced from a finite-difference Laplace solution ($N_f=3$, $N_d=6$; the vertical line below the tip is . See the official exam paper.]

Approach. Sketch a curvilinear-square flow net to get $N_f$ and $N_d$; use it for the seepage and to read the total head at A, then combine total stress and pore pressure to get $\sigma'$.

1. (a) Flow net and seepage. A curvilinear-square net for this single wall gives $N_f = 3$ flow channels and $N_d = 6$ equipotential drops, so each drop is $\Delta h = H/N_d = 3/6 = 0.5\ \text{m}$. The seepage per metre run is $$q = k\,H\,\frac{N_f}{N_d} = (2.0\times10^{-5})(3)\tfrac{3}{6} = \boxed{3.0\times10^{-5}\ \text{m}^3/\text{s/m} = 2.59\ \text{m}^3/\text{day/m}}.$$
2. (b) Total stress at A. Above A lie 3 m of upstream water and 1 m of saturated soil: $$\sigma_A = 3\gamma_w + 1\gamma_{sat} = 3(9.81)+1(20) = 49.43\ \text{kPa}.$$
3. Total head at A from the net. Take the datum at the downstream ground surface, so the upstream total head is $+3\ \text{m}$. Point A is only 1 m down the 3 m pile. The first equipotential meets the upstream face about 1.5 m down, so A lies above it, about 0.6 of a drop below the upstream bed. A finite-difference Laplace solve of this geometry gives a head loss of 0.305 m to A, i.e. 0.61 drops. $$h_A = H - n_d\,\Delta h \approx 3 - (0.61)(0.5) = 2.69\ \text{m}.$$
4. Pore pressure at A. With A at elevation $z_A = -1\ \text{m}$, the pressure head is $h_A - z_A = 2.69-(-1)=3.69\ \text{m}$, so $$u_A = (h_A - z_A)\gamma_w = (3.69)(9.81) = 36.20\ \text{kPa}.$$
5. Effective stress at A. $$\sigma_A' = \sigma_A - u_A = 49.43 - 36.20 = \boxed{13.2\ \text{kPa}}.$$