07-Str-A3 · December 2014

Question 5 of 6: Flow net and seepage past a cut-off wall

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

Notes on this paper

Paper format: PEO / Engineers Canada National Examinations, December 2014, 07-Str-A3 Geotechnical Materials and Analysis. Three hours, closed book, drawing instruments required, one approved Casio or Sharp calculator. All charts and equations are supplied at the back of the paper (the m–n influence chart on page 8, a Newmark chart with $I_N = 0.005$ on page 9, and a two-page formula sheet). Total value 100 marks over six compulsory questions (20 + 10 + 10 + 20 + 20 + 20). Every question and every sub-part is answered in full below.

Reference texts: Das, Principles of Geotechnical Engineering, 9th ed. (Ch. 6 compaction, Ch. 7 permeability, Ch. 8 seepage and flow nets, Ch. 10 stresses in a soil mass, Ch. 11 consolidation, Ch. 12 shear strength); Knappett & Craig, Craig’s Soil Mechanics, 8th ed. (Ch. 2 seepage and flow-net construction, Ch. 5 shear strength, Ch. 11 lateral earth pressure); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed. (compacted-fill fabric and Proctor behaviour by USCS group); Harr, Groundwater and Seepage, Ch. 4 (closed-form conformal solutions for a single cut-off); Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for earth pressures, seepage control and settlement; Canadian Dam Association, Dam Safety Guidelines (2013, rev. 2019); ASTM D698 / D1557 (Proctor compaction), ASTM D4767 (consolidated-undrained triaxial with pore-pressure measurement), ASTM D7181 (consolidated-drained triaxial).

Check — two printing slips in the source, carried exactly as printed. (1) The Question 1 header reads “(4 × 5 = 20 marks)” but five statements (i)–(v) are printed; this solution treats the question as 5 × 4 = 20 marks and answers all five parts. (2) Question 5(a) instructs the candidate to “draw on Figure 4”, while the only section supplied is labelled Figure 3 — the blank grid sheets on pages 6 and 7 are the intended drawing space. Neither slip changes the 100-mark total.

Question 5: Flow net and seepage past a cut-off 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. Reading the dimension chain on Figure 3 from the top down: the upstream water surface stands 3 m above ground level; from ground level a further 3 m takes the section to the tip of the cut-off wall; and a final 3 m reaches the impermeable layer. The permeable stratum is therefore $T = 6$ m thick, the wall penetrates $D = 3$ m into it, and the head loss across the wall is $H = 3$ m, the tailwater standing at ground level on the downstream side. The permeability is $k = 2.0 \times 10^{-5}\ \text{m}\,\text{s}^{-1}$, isotropic and homogeneous. Point A is marked on the upstream face of the wall, 1 m below ground level.

Find. (a) A flow net that satisfies every construction rule, and (b) the seepage quantity per metre run of wall.

Figure 5.1 — the completed flow net. Red lines are equipotentials, blue lines are flow lines; the two ground surfaces are equipotential boundaries and the impermeable layer and the wall are flow boundaries. The net closes with $N_f = 4$ flow channels and $N_d = 8$ equipotential drops. The lines shown are the actual solution of Laplace’s equation for this geometry, so the curvilinear figures are true squares.

Approach. Identify the four boundaries and their type, sketch a net of curvilinear squares between them, count the channels and drops, and apply $q = k\,h_w\,N_f/N_d$ from the formula sheet. The count is then verified two independent ways before it is trusted.

Classify the boundaries. Four boundaries fix the whole problem. The upstream ground surface is a constant-head boundary and is therefore the first equipotential, $h = H = 3$ m. The downstream ground surface is the last equipotential, $h = 0$. The impermeable layer beneath the stratum admits no flow, so it is the bottom flow line. The cut-off wall itself — down the upstream face, around the tip, and back up the downstream face — is impervious, so it is the top flow line. Every flow line must run from the upstream equipotential to the downstream one, and every equipotential must run from the wall to the base, meeting both at right angles.
Apply the construction rules. A valid net requires that flow lines and equipotentials intersect orthogonally; that each figure is a curvilinear square, that is, has equal average width and length so that a circle can be inscribed touching all four sides; that the same discharge $\Delta q$ passes between every adjacent pair of flow lines; and that the same head drop $\Delta h$ occurs between every adjacent pair of equipotentials. Because the section is symmetric about the wall, the net must be too, and the flow lines are the family of curves that wrap around the wall tip from one ground surface to the other.
Fix the number of channels and drops. The wall penetrates exactly half the stratum: $$\frac{D}{T} \;=\; \frac{3\ \text{m}}{6\ \text{m}} \;=\; 0.5$$ For that geometry the square-figure construction closes cleanly with $\boxed{N_f = 4 \ \text{flow channels and}\ N_d = 8 \ \text{equipotential drops}}$, four drops on each side of the wall. This is the answer to part (a), and the net drawn in Figure 5.1 is that net.
Compute the head drop per figure. The total head is shared equally between the drops: $$\Delta h \;=\; \frac{h_w}{N_d} \;=\; \frac{3.0\ \text{m}}{8} \;=\; 0.375\ \text{m per drop}$$ so the piezometric level falls 0.375 m each time a flow path crosses an equipotential, which is what makes the net usable for pore pressures anywhere in the section.
Compute the seepage quantity (part b). From the formula sheet, for unit width of wall, $$q \;=\; k\,h_w\,\frac{N_f}{N_d} \;=\; \left(2.0\times10^{-5}\ \text{m}\,\text{s}^{-1}\right)(3.0\ \text{m})\left(\frac{4}{8}\right)$$ $$q \;=\; \boxed{3.0\times10^{-5}\ \text{m}^{3}\text{s}^{-1}\ \text{per metre of wall}}$$ which is 2.59 $\text{m}^3$ per day per metre run, or about 108 litres per hour per metre — a rate that a modest sump would handle for a temporary excavation but that would be quite unacceptable as a permanent loss from a reservoir.
Verify the net before trusting the count. A hand-drawn flow net is only as good as the shape factor $N_f/N_d$ it produces, and that number is worth checking, because a net drawn with five channels and fourteen drops — a perfectly plausible-looking sketch — would give 0.357 and understate the seepage by 29 %. Two independent checks are available. Harr’s closed-form conformal-mapping solution for a single cut-off in a stratum of finite depth gives the shape factor as $K(m')/2K(m)$ with modulus $m = \sin(\pi D/2T)$ and complementary modulus $m' = \cos(\pi D/2T)$ (so the factor rises without limit as the wall shortens and falls to zero as it reaches the base); at $D/T = 0.5$ both moduli equal $\sin 45^\circ = \cos 45^\circ$, the two complete elliptic integrals are equal, and the shape factor is exactly 0.500. A finite-difference solution of Laplace’s equation on the drawn section gives 0.499. Both confirm $N_f/N_d = 4/8 = 0.5$.
Read a pore pressure off the net — point A. The point marked A on the figure lies on the upstream face, 1 m below ground level, one third of the way down the embedded length. The net places it about 0.79 of a drop below the upstream surface, so its excess head is $3.00 - 0.79(0.375) = 2.70$ m above the tailwater. Taking the datum at the impermeable layer, its total head is $6.00 + 2.70 = 8.70$ m and its elevation head is 5.00 m, so the pressure head is 3.70 m and $$u_A \;=\; \gamma_w h_p \;=\; 9.81 \times 3.70 \;=\; \boxed{36.3\ \text{kPa}}$$ This is the quantity a designer actually needs from the net: it is the water pressure that has to be carried by the wall and subtracted from the total stress in any stability calculation.
Check the downstream side against heave. The professional follow-through on any cut-off is Terzaghi’s heave check on a prism of soil $D$ deep and $D/2$ wide immediately downstream of the wall. Averaging the head over the base of that prism from the net gives $h_{av} = H/3 = 1.00$ m, so the average upward gradient through the prism is $$i_{av} \;=\; \frac{h_{av}}{D} \;=\; \frac{1.00\ \text{m}}{3.0\ \text{m}} \;=\; 0.333$$ Against a critical gradient of about 1.0 this gives a factor of safety of $\boxed{F \approx 3.0}$ against heave and piping, which is adequate; a value below about 2.5 would call for a deeper cut-off or a filtered surcharge blanket on the downstream side.

Check — assumed critical gradient. No unit weight, void ratio or specific gravity is given for the stratum, so the heave check above takes $i_c = (G_s-1)/(1+e) \approx 1.0$, the usual value for a granular soil. If the soil is loose, with $e$ nearer 0.9, $i_c$ falls to about 0.88 and the factor of safety drops to 2.6 — still acceptable, but the calculation should be repeated with measured index values before construction.

Question 5 — results
Quantity	Symbol	Value
Penetration ratio	$D/T$	0.50
Flow channels (part a)	$N_f$	4
Equipotential drops (part a)	$N_d$	8
Head drop per figure	$\Delta h$	0.375 m
Shape factor	$N_f/N_d$	0.500 (closed form 0.500, numerical 0.499)
Seepage quantity (part b)	$q$	$3.0\times10^{-5}\ \text{m}^{3}\text{s}^{-1}$ per m (2.59 $\text{m}^3$/day per m)
Pore pressure at A	$u_A$	36.3 kPa
Average exit gradient, heave prism	$i_{av}$	0.333
Factor of safety against heave	$F$	3.0

Topic. Two-dimensional confined seepage: flow-net construction and the quantities that can be read from a completed net.

Key relations. Laplace’s equation $\partial^2 h/\partial x^2 + \partial^2 h/\partial z^2 = 0$; $q = k\,h_w\,N_f/N_d$ per unit width; $\Delta h = h_w/N_d$; pressure head at a point $h_p = (n_d/N_d)\,h_w$ measured from the upstream surface; Terzaghi’s heave criterion $F = i_c/i_{av}$ on a prism $D$ deep by $D/2$ wide.

Why this works. A flow net is a hand-drawn solution of Laplace’s equation. Because that equation is linear and its solution for given boundaries is unique, any net that satisfies all the construction rules is the solution, and the ratio $N_f/N_d$ is a property of the geometry alone — it does not depend on $k$ or on the head. That is why the seepage quantity separates so cleanly into a soil term $k$, a driving term $h_w$ and a purely geometric shape factor. The curvilinear-square requirement is what enforces continuity: if each figure is square, the discharge through every channel is the same and the head drop across every figure is the same, so counting is all that remains.

Common pitfall. Reading the upper 3 m of the dimension chain as part of the soil thickness. That 3 m is the depth of water standing above ground level; the permeable stratum is 6 m, not 9 m, and mistaking it turns a wall that penetrates half the stratum into one that penetrates two thirds, changing the shape factor and the answer. The second pitfall is drawing a net that looks plausible but is not square, and never checking it — a shape factor is worth back-figuring against a closed-form or numerical result whenever one is available. The third is forgetting that the flow net gives excess head; the elevation head must be added before a pore pressure can be quoted.

Source. Das, Principles of Geotechnical Engineering, 9th ed., Ch. 8 §8.4–8.9 (flow nets, seepage quantity, uplift, heave); Knappett & Craig, 8th ed., Ch. 2 §2.5–2.8; Harr, Groundwater and Seepage, Ch. 4 for the closed-form single-cut-off solution.

← Question 4 Next: Question 6 →