Question 5 of 6: Flow net and effective stress at a cut-off wall

Question 5: Flow net and effective stress at a cut-off wall (Value: 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) a valid flow net for the section, and (b) the effective stress at $A$.

(a) Constructing the flow net

Approach. Fix the boundary conditions first, sketch a net of curvilinear squares between them, then audit the resulting shape factor against a closed-form solution so that the drawing becomes a checkable result rather than an opinion.

1. Identify the boundaries, because they fix the net. Four boundaries bound the flow region, and each is either a flow line or an equipotential. The upstream ground surface, beneath 3 m of standing water, is a surface of constant total head and is therefore the first equipotential, $h = 3$ m. The downstream ground surface, with the tailwater at grade, is the last equipotential, $h = 0$. The two faces and the tip of the sheet pile form the first flow line, and the top of the impermeable layer forms the last flow line.
2. Apply the drawing rules. Flow lines and equipotentials must intersect at right angles everywhere; each figure enclosed between two adjacent flow lines and two adjacent equipotentials must be a curvilinear square, so that the width $b$ and the length $l$ of every field are equal; the net must meet the impermeable base tangentially and the constant-head surfaces perpendicularly; and it must be drawn to true scale, since a distorted section destroys the squareness on which the arithmetic rests.
3. Choose the number of channels. The pile is driven to exactly half the depth of the stratum, $$\frac{D}{T} = \frac{3}{6} = 0.500,$$ which is the geometry for which a square net closes with four flow channels and eight potential drops. Three interior flow lines are drawn, nested around the tip, and seven interior equipotentials cross them; the equipotential at $h = H/2$ is the pile itself continued vertically to the base, by antisymmetry.
4. Audit the net rather than trusting the sketch. Two independent checks are available and both should be quoted. Harr’s conformal solution for a single sheet pile in a stratum of finite depth gives the exact shape factor $$\frac{q}{kH} = \frac{K(m')}{2K(m)}, \qquad m = \sin\frac{\pi D}{2T}, \quad m' = \sqrt{1-m^2},$$ and at $D/T = 1/2$ both moduli equal $\cos 45^\circ$, the two complete elliptic integrals cancel and the factor is exactly $0.500$. A direct finite-difference solution of Laplace’s equation on the same section returns $0.495$. The drawn net gives $$\frac{N_f}{N_d} = \frac{4}{8} = 0.500,$$ so all three agree and the net is confirmed. A plausible-looking five-and-fourteen net would give 0.357 and be 29 % low.
5. Quantify the seepage. With the net established, $$q = k H \frac{N_f}{N_d} = (2.0\times10^{-5})(3)\left(\frac{4}{8}\right) = 3.00\times10^{-5}\ \text{m}^3/\text{s per m},$$ $$\boxed{q = 3.00\times10^{-5}\ \text{m}^3\!/\text{s per m} = 2.59\ \text{m}^3\!/\text{day per m width}}$$
6. Record the head lost per drop, for use in part (b). The whole head is lost in $N_d = 8$ equal steps, so $$\Delta h = \frac{H}{N_d} = \frac{3}{8} = 0.375\ \text{m per drop}.$$

(b) Effective stress at point A

Approach. Compute the total stress from the column of water and soil above $A$, obtain the pore pressure from the total head that the flow net assigns to $A$, and subtract. Because there is seepage, the pore pressure is not hydrostatic and must come from the net.

1. Fix a datum. Take the datum at the downstream water level, which is the downstream ground surface. Then the upstream water surface is at total head $h = +3$ m, the downstream surface at $h = 0$, and point $A$ — 1 m below ground on the upstream face of the pile — has elevation $z_A = -1.0$ m.
2. Locate $A$ on the net. $A$ lies on the upstream face of the pile, one metre into a three-metre embedment, i.e. one third of the way down the face. Counting the equipotentials down that face puts it about $n_d = 0.8$ of a drop below the first equipotential; solving Laplace’s equation numerically on the same section and reading the head on the upstream face at a depth of 1 m gives $n_d = 0.82$, so the count off the drawing is sound.
3. Total head at A. $$h_A = H - n_d\,\Delta h = 3 - (0.8)(0.375) = 3 - 0.30,$$ $$\boxed{h_A = 2.70\ \text{m}}$$ The numerical solution gives 2.69 m, a difference of 0.3 %.
4. Pore-water pressure at A. Using the relation printed with the question, $u = (h-z)\gamma_w$, with the pressure head being the total head less the elevation head, $$u_A = (h_A - z_A)\gamma_w = \big(2.70 - (-1.00)\big)(9.81) = (3.70)(9.81),$$ $$u_A = 36.30\ \text{kPa}.$$
5. Total vertical stress at A. Above $A$ stand 3 m of free water and 1 m of saturated soil: $$\sigma_A = H\gamma_w + (1.0)\gamma_{sat} = (3)(9.81) + (1.0)(20) = 29.43 + 20.00,$$ $$\sigma_A = 49.43\ \text{kPa}.$$
6. Effective stress at A. $$\sigma_A' = \sigma_A - u_A = 49.43 - 36.30,$$ $$\boxed{\sigma_A' = 13.13\ \text{kPa}\ \ \text{(upstream face, as drawn)}}$$
7. Check the sign of the seepage effect, and give the other face. With no flow at all the pore pressure at $A$ would be hydrostatic, $u = (4.00)(9.81) = 39.24$ kPa, and the effective stress would be the buoyant weight of 1 m of soil, $\sigma' = (1.0)(20 - 9.81) = 10.19$ kPa. The computed 13.13 kPa is larger, as it must be: on the upstream side the flow is downward, the seepage force acts with gravity and adds to the effective stress. On the downstream face at the same elevation the picture reverses. By the antisymmetry of the section, $h = 3 - 2.70 = 0.30$ m there, so $u = (0.30 + 1.00)(9.81) = 12.75$ kPa against a total stress of only $(1.0)(20) = 20.00$ kPa, giving $\sigma' = 7.25$ kPa — below the static 10.19 kPa, because the upward flow lifts part of the soil weight.

One consequence of the downstream figures deserves a line, because it is what the flow net is usually drawn for. The reduced effective stress on the downstream side is the piping condition, and it is checked on Terzaghi’s exit prism — a block $D = 3$ m deep and $D/2 = 1.5$ m wide against the downstream face. The average total head on its base is 0.98 m, so the average upward gradient through it is $0.98/3 = 0.33$ against a critical gradient $i_c = (\gamma_{sat}-\gamma_w)/\gamma_w = 1.04$, and the factor of safety against heave is $1.04/0.33 = 3.2$. That is comfortable, but it is the number a designer would quote, not the effective stress itself.