Flow Net, Seepage and Stability of a Cutoff Wall: 07-Str-A3, May 2013, Q5

Question 5: Flow Net, Seepage and Stability of a Cutoff Wall (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 cutoff wall driven through a permeable stratum that rests on an impermeable layer, with the headwater ponded above the upstream bed and the tailwater level with the downstream bed.

Given data — Question 5 (from Figure 4)
Quantity	Symbol	Value
Depth of water above the upstream bed	H	3.0 m
Tailwater level	—	at the downstream bed (no free water)
Sheet-pile penetration below bed level	D	3.0 m
Permeable stratum below the pile tip	—	3.0 m (total thickness T = 6.0 m)
Coefficient of permeability	k	2.0 × 10⁻⁵ m/s
Saturated unit weight of the soil	γ_sat	20 kN/m³
Position of point A on the piling	—	1.0 m below bed level, 2.0 m above the pile tip
Unit weight of water	γ_w	9.81 kN/m³

Find. (a) a valid flow net; (b) the seepage quantity per metre run; (c) the effective stress at A; (d) the factor of safety of the cutoff wall against heave.

Figure 4 with the flow net drawn on it — four flow channels (N_f = 4, solid) and eight equipotential drops (N_d = 8, dashed). The bed surfaces are equipotential boundaries; the sheet pile and the top of the impermeable layer are flow lines.

Approach. Sketch a curvilinear-square flow net between the two equipotential boundaries (the two bed surfaces) and the two flow boundaries (the sheet pile and the impermeable base), read the shape factor $N_f/N_d$ from it, then apply Darcy for the discharge, the equipotential drop for the pore pressure at A, and Terzaghi’s heave criterion for the factor of safety.

Set the boundary conditions (part a). Take the datum at tailwater level, which here coincides with the downstream bed. The upstream bed is a boundary equipotential at total head $h = H = 3.0\ \text{m}$; the downstream bed is a boundary equipotential at $h = 0$. The surface of the sheet pile and the top of the impermeable layer are impervious, so both are flow lines. Sketch flow lines that begin normal to the upstream bed, pass around the pile tip and end normal to the downstream bed, then insert equipotentials so that every field is a curvilinear square (equal average length and breadth, all intersections at right angles).
Count the net. The net that satisfies those rules for this geometry has $$N_f = 4 \qquad N_d = 8 \qquad \frac{N_f}{N_d} = 0.5$$ The shape factor of 0.5 is not a coincidence: for a sheet pile driven to exactly half the depth of the permeable stratum ($D/T = 3/6 = 0.5$) the closed-form conformal-mapping solution also gives a form factor of approximately 0.5, so the sketched net is a good one.
Compute the seepage quantity (part b). For a flow net of curvilinear squares, Darcy’s law integrated over the net gives $$q = k\,H\,\frac{N_f}{N_d} = (2.0 \times 10^{-5})(3.0)\left(\frac{4}{8}\right)$$ $$\boxed{\;q = 3.0 \times 10^{-5}\ \text{m}^3/\text{s per metre run} = 2.59\ \text{m}^3/\text{day per metre}\;}$$ For a 100 m long cofferdam that is about 259 m³/day, which sizes the dewatering pumps.
Fix the head loss per drop. The total head loss is shared equally between the equipotential drops, so $$\Delta h = \frac{H}{N_d} = \frac{3.0}{8} = 0.375\ \text{m per drop}$$
Locate point A on the net (part c). Point A lies on the upstream face of the piling, 1.0 m below bed level. Reading the net there, A sits between the upstream boundary equipotential and the first interior equipotential, about eight tenths of the way to it, so the number of drops consumed at A is $n_d \approx 0.8$ and the total head is $$h_A = H - n_d\,\Delta h = 3.0 - 0.8(0.375) = 2.70\ \text{m}$$
Convert total head to pore pressure. With the datum at tailwater level the elevation head of A is $z_A = -1.0\ \text{m}$, so the pressure head is $h_A - z_A = 2.70 + 1.00 = 3.70\ \text{m}$ and $$u_A = \gamma_w (h_A - z_A) = 9.81 \times 3.70 = 36.30\ \text{kPa}$$
Compute the total vertical stress at A. Above A there is 3.0 m of free water and 1.0 m of saturated soil, hence $$\sigma_A = \gamma_w H + \gamma_{sat}(1.0) = 9.81(3.0) + 20(1.0) = 29.43 + 20.00 = 49.43\ \text{kPa}$$
Take the difference. By Terzaghi’s principle, $$\sigma_A' = \sigma_A - u_A = 49.43 - 36.30$$ $$\boxed{\;\sigma_A' = 13.13\ \text{kPa}\;}$$ Compare this with the no-flow case: hydrostatic pore pressure would be $9.81(4.0) = 39.24\ \text{kPa}$ and the effective stress would be only $\gamma' \times 1.0 = 10.19\ \text{kPa}$. Seepage on the upstream face is downward, so the seepage force acts with gravity and raises the effective stress by about 29 %. On the downstream face at the same depth the head is $h = 0.30\ \text{m}$, the pore pressure is $9.81(1.30) = 12.76\ \text{kPa}$ and the effective stress falls to $20.00 - 12.76 = 7.24\ \text{kPa}$, because there the seepage is upward and opposes gravity. That contrast is the whole story of part (d).
Set up the heave check (part d). The critical mechanism for a cutoff wall is heave (boiling) of the soil prism hanging on the downstream face of the piling. Terzaghi’s prism is $D = 3.0\ \text{m}$ deep and $D/2 = 1.5\ \text{m}$ wide. Its buoyant weight is $$W' = \gamma' D \left(\frac{D}{2}\right) = (20 - 9.81)(3.0)(1.5) = 10.19 \times 4.5 = 45.86\ \text{kN per metre}$$
Compute the uplift on the base of the prism. Reading the excess head along the prism base (at pile-tip level, from the wall out to 1.5 m) off the net gives values falling from about 1.3 m at the wall to about 0.8 m at the outer edge, an average of $h_{avg} \approx 1.00\ \text{m}$, so $$U = \gamma_w\,h_{avg}\left(\frac{D}{2}\right) = 9.81 \times 1.00 \times 1.5 = 14.72\ \text{kN per metre}$$
Form the factor of safety. Dividing the stabilising buoyant weight by the destabilising uplift, $$FS = \frac{W'}{U} = \frac{45.86}{14.72}$$ $$\boxed{\;FS \approx 3.1\ \text{against heave}\;}$$ The same answer follows from the gradient form, since the prism width cancels: the critical hydraulic gradient is $i_{cr} = \gamma'/\gamma_w = 10.19/9.81 = 1.039$ and the average exit gradient is $i_{avg} = h_{avg}/D = 1.00/3.0 = 0.333$, so $FS = 1.039/0.333 = 3.1$.
Judge the result. Canadian practice (CFEM, 4th ed.) looks for a factor of safety of at least 1.5, and commonly 2.0, against heave and piping in a temporary excavation. At $FS \approx 3.1$ the cutoff wall is comfortably stable against heave, and the exit gradient of 0.33 is well below the 0.5 value at which a filter blanket would normally be specified. Note that this is a hydraulic-stability check only; the structural adequacy of the sheet pile (bending, and passive resistance of the 3 m embedment against the unbalanced water thrust) is a separate calculation.

Question 5 — results
Part	Quantity	Result
a	Flow net	N_f = 4, N_d = 8, shape factor 0.5; Δh = 0.375 m per drop
b	Seepage quantity, q	3.0 × 10⁻⁵ m³/s per m (2.59 m³/day per m)
c	Total stress at A, σ_A	49.43 kPa
c	Pore pressure at A, u_A	36.30 kPa
c	Effective stress at A, σ′_A (upstream face)	13.13 kPa
c	Effective stress on the downstream face at the same depth	7.24 kPa
d	Critical gradient, i_cr	1.039
d	Average exit gradient, i_avg	0.333
d	Factor of safety against heave	FS ≈ 3.1 (> 1.5 required)

Check: the stem calls point A the “back of the piling”, while the dot in Figure 4 is drawn on the upstream (water-retaining) face 1 m below bed level. The upstream face is solved as the primary answer because that is where the figure puts the point; the downstream value at the same depth (7.24 kPa) is reported alongside it, so the answer stands whichever face the examiner intended. Parts (b) to (d) depend on the net the candidate draws: a net with N_f = 5 and N_d = 10 has the same shape factor and gives exactly the same q, and locates A within about 0.3 kPa of the value above.

Topic. Two-dimensional confined seepage, flow nets and hydraulic stability.

Key relations. Laplace, $k_x\dfrac{\partial^2 h}{\partial x^2} + k_z\dfrac{\partial^2 h}{\partial z^2} = 0$; discharge $q = kH N_f/N_d$; drop $\Delta h = H/N_d$; pore pressure $u = \gamma_w(h - z)$; critical gradient $i_{cr} = \gamma'/\gamma_w$; Terzaghi heave $FS = \gamma' D / (\gamma_w h_{avg})$.

Why this works. Steady flow through a saturated soil obeys Laplace’s equation, whose two families of solutions — the equipotentials and the stream lines — are orthogonal. Drawing them as curvilinear squares makes the discharge through every flow channel equal and the head drop across every field equal, which is why the discharge collapses to the single ratio $N_f/N_d$ and why a hand sketch, accurate to about five per cent, is good enough for design. The heave check then recognises that seepage exiting upward exerts a body force $i\gamma_w$ per unit volume on the grains; when that force reaches the buoyant unit weight $\gamma'$, the effective stress at the exit vanishes and the soil boils.

Common pitfall. Measuring the pressure head from the wrong datum, or forgetting that the pressure head is $h - z$ and not $h$ — below the datum the elevation head is negative and adds to the pressure head. A second classic slip is counting the boundary equipotentials as drops: eight drops means nine equipotential lines including both boundaries. A third is checking piping on the upstream face, where seepage is downward and stability is improved; the danger is always on the exit side.

Source. Das, Principles of Geotechnical Engineering, 9th ed., §8.4–8.7 (flow nets, seepage calculation) and §9.7–9.9 (uplift, heave, critical gradient); Knappett & Craig, 8th ed., Ch. 2; Canadian Geotechnical Society, Canadian Foundation Engineering Manual, 4th ed., Ch. 23 (excavation and dewatering, required factors of safety).

Question 5 of 6: Flow Net, Seepage and Stability of a Cutoff Wall

Question 5: Flow Net, Seepage and Stability of a Cutoff Wall (20 marks)