07-Str-A3 · Undated paper

Question 5 of 6: Seepage through a flow net around a sheet pile cut-off

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

Notes on this paper

Paper format. Engineers Canada / PEO National Examinations, May 2019 — 07-Str-A3 Geotechnical Materials and Analysis. Closed book, three hours, 100 marks, one approved Casio or Sharp calculator. All six questions are compulsory and are weighted 10 / 10 / 15 / 20 / 15 / 30. The paper carries its own appendix: two sheets of blank semi-logarithmic and squared graph paper (pages 7–9) for the plotted answers, the rectangular-loading $m$–$n$ influence chart (page 10), a Newmark influence chart (page 11) and a two-page formula sheet (pages 12–13) on which the Newmark influence value is printed as $\sigma_z = 0.005\,N q$, i.e. $I_N = 0.005$ over 200 elements. Every chart value quoted below is taken from those appendix sheets.

Sitting. The printed cover reads NATIONAL EXAMINATIONS – May 2019, 07-Str-A3 Geotechnical Materials and Analysis, 3 hours duration, and every interior page repeats it. The sitting is May 2019.

Reference texts.

B. M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. — Ch. 2–3 (grain size and phase relations), Ch. 6 (compaction), Ch. 7–8 (permeability and seepage, flow nets), Ch. 9 (stresses in a soil mass, the $m$–$n$ chart and Newmark’s chart), Ch. 11 (consolidation, Casagrande’s construction), Ch. 12 (shear strength and the triaxial test).
R. F. Craig / J. A. Knappett, Craig’s Soil Mechanics, 8th ed. — Ch. 2 (seepage and flow nets), Ch. 3 (effective stress), Ch. 4 (consolidation), Ch. 5 (shear strength, the stress-path or “modified” failure envelope and the choice of test), Ch. 6 (stress distribution).
M. E. Harr, Groundwater and Seepage — Ch. 4 (the conformal solution for a single sheet pile in a stratum of finite depth, used here to audit the drawn flow net).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — Ch. 4 (site investigation and sampling), Ch. 8 (settlement), Ch. 11 (laboratory strength testing). The Canadian reference for practice and terminology.
K. Terzaghi, R. B. Peck & G. Mesri, Soil Mechanics in Engineering Practice, 3rd ed. — Art. 14 (one-dimensional consolidation theory and its assumptions), Art. 16–17 (seepage and piping), Art. 19–20 (shear strength).

Check — two readings of the printed paper, carried as stated.

(1) The printed paper has exactly six questions worth 10 + 10 + 15 + 20 + 15 + 30 = 100 marks, and that numbering is used here.

(2) Question 4 lists the void ratio at $\sigma' = 500$ kPa as $e = 0.925$. Taken with the $400$ kPa point that implies $C_c = 0.57$ over the last increment, against $0.27$–$0.30$ over every earlier virgin increment — the last row is inconsistent with the rest of the test. It is carried exactly as printed, plotted, and excluded from the straight-line fit for $C_c$, with the effect of including it stated where it matters.

Question 5: Seepage through a flow net around a sheet pile cut-off (15 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.

Quantity	Symbol	Value
Upstream water surface	—	elevation 100.0 m
Ground surface, both sides (downstream tailwater)	—	elevation 93.0 m
Total head loss across the wall	$H$	$100.0 - 93.0 = 7$ m
Thickness of the permeable stratum	$T$	$93.0 - 80.0 = 13$ m
Penetration of the sheet piling (tip at elev. 86.0)	$D$	7 m
Number of flow channels counted on the figure	$N_f$	4
Number of equipotential drops counted on the figure	$N_d$	10
Coefficient of permeability	$k$	$22 \times 10^{-4}$ cm/s
Point 5 — upstream face of the pile	—	elevation 87.4 m

Find. The seepage loss per metre run of the wall, and the pressure head at point 5.

[Figure not reproduced: Figure 2 redrawn to the elevations printed on the original. The lines are contours of a direct finite-difference solution of Laplace’s equation on the drawn geometry, at the nine interior potential levels and the three interior stream-function levels that the printed net shows — that is,. See the official exam paper.]

Approach. Count the net, convert $k$ to SI, apply the shape-factor form of Darcy’s law from the formula sheet, then get the total head at point 5 by counting equipotential drops and subtract its elevation head.

Read the net off the figure. The numbered points are the intersections of the equipotential lines with the sheet pile: points 2 and 12 sit on the upstream and downstream ground surfaces (the two boundary equipotentials), points 3, 4, 5 and 6 mark interior equipotentials down the upstream face, point 7 is the equipotential through the tip, and points 8 to 11 mirror them on the downstream face. Eleven equipotential lines means $$N_d = 11 - 1 = 10 \ \text{drops}.$$ The count is self-checking: by symmetry the equipotential through the tip must be the middle one, and point 7 is indeed the fifth of ten drops from the upstream side. Three interior flow lines are drawn between the pile and the impervious base on each side, so with the pile and the base as the two bounding flow lines, $N_f = 4$ channels.
Head loss per drop. The total head loss is the difference between the two water surfaces, $H = 100.0 - 93.0 = 7$ m, shared equally between the drops: $$\Delta h = \frac{H}{N_d} = \frac{7}{10} = 0.700\ \text{m per drop}.$$
Convert the permeability. Units matter more than anything else in this question: $$k = 22 \times 10^{-4}\ \text{cm/s} = 22 \times 10^{-6}\ \text{m/s} = 2.20 \times 10^{-5}\ \text{m/s}.$$
Seepage loss per metre of wall. The formula sheet gives the shape-factor form of Darcy’s law for a flow net, $q = k\,h\,(N_f/N_d)\times(\text{width})$. Per metre run, $$q = k H \frac{N_f}{N_d} = (2.20\times10^{-5})(7.00)\left(\frac{4}{10}\right) = 6.16 \times 10^{-5}\ \text{m}^3/\text{s per m}$$ and converting to the units a report would use, $$q = 6.16\times10^{-5} \times 86\,400 = \boxed{5.32\ \text{m}^3/\text{day per metre of wall}}$$ that is 0.0616 litres per second for every metre run of wall.
Total head at point 5. Point 5 lies on the third equipotential counted from the upstream ground surface, so three of the ten drops have been used up in reaching it. Taking elevations above the same datum as the printed elevation scale, the total head there is $$h_5 = 100.0 - 3\,\Delta h = 100.0 - 3(0.700) = 100.0 - 2.10 = 97.9\ \text{m}.$$
Pressure head at point 5. Total head is the sum of elevation head and pressure head, so subtracting the elevation of point 5 (87.4 m from the printed scale), $$h_p = h_5 - z_5 = 97.9 - 87.4 = \boxed{10.5\ \text{m of water}}$$ equivalent to a pore-water pressure $u_5 = \gamma_w h_p = 9.81 \times 10.5 = 103.0$ kPa. As a check on the sign of the counting, the mirror point 9 on the downstream face sits at the same elevation but seven drops down, giving $h_9 = 100.0 - 7(0.7) = 95.1$ m and a pressure head of 7.7 m — lower on the discharge side, as it must be.

Quantity	Value
Total head loss across the wall, $H$	7.00 m
Flow net as drawn	$N_f = 4$, $N_d = 10$, shape factor 0.400
Head loss per drop, $\Delta h$	0.700 m
Permeability, $k$	$2.20 \times 10^{-5}$ m/s
Seepage loss per metre of wall, $q$	$6.16 \times 10^{-5}$ m$^3$/s/m = 5.32 m$^3$/day/m
Total head at point 5	97.9 m
Pressure head at point 5	10.5 m
Pore-water pressure at point 5	103.0 kPa

Check — the printed net is about 15 % coarse. The drawn net gives a shape factor $N_f/N_d = 4/10 = 0.400$. Two independent checks of the same geometry disagree with it slightly. Harr’s conformal solution for a single sheet pile of penetration $D$ in a stratum of thickness $T$, $q/(kH) = K(m')/[2K(m)]$ with $m = \sin(\pi D / 2T)$, gives 0.473 at $D/T = 7/13$; a direct finite-difference solution of Laplace’s equation on the same section converges to 0.470. The drawn net therefore understates the true seepage by about 15 %, giving 6.3 m$^3$/day/m rather than 5.32. The question says to use the flow net shown, so 5.32 m$^3$/day/m is the answer to the question as set; a design estimate should use the larger figure or, better, a net drawn with a fractional outermost channel. The pressure head at point 5 is not affected — it depends only on the drop count, which the symmetry check confirms.

Topic. Two-dimensional confined seepage — reading a flow net, the shape factor, and recovering pore-water pressures from equipotential counts.

Key relations. $q = kH\,N_f/N_d$ per unit width; $\Delta h = H/N_d$; $h_p = h_{tot} - z$; $u = \gamma_w h_p$; and for the audit, $q/(kH) = K(m')/[2K(m)]$ with $m = \sin(\pi D/2T)$ (Harr).

Why this works. A flow net is a hand-drawn solution of Laplace’s equation in which every “square” carries the same discharge $\Delta q = k\,\Delta h$ and drops the same head $\Delta h$. Multiplying the per-channel discharge by the number of channels gives the total, which is why only two integers and the total head are needed. The same net gives pressures because total head is a linear function of the drop number: count the drops from the upstream boundary to the point, subtract from the upstream head to get the total head, then subtract the elevation. Note that the pressure head is not simply the depth of the point below the upstream water surface — that would be the hydrostatic value, and the whole point of the seepage is that head is being lost along the way.

Common pitfall. Two, and both are fatal to the arithmetic. The first is leaving $k$ in cm/s, which makes $q$ a hundred times too large. The second is counting the drops from the wrong end, or counting equipotential lines instead of the spaces between them: there are $N_d + 1$ lines and $N_d$ drops, and the symmetry of a single sheet pile — the tip equipotential must be the middle one — is the cheapest way to confirm the count before using it.

Source. Das & Sobhan, Principles of Geotechnical Engineering, 9th ed., §8.4–8.6 (flow nets, seepage calculation, uplift pressure under a structure); Craig / Knappett, 8th ed., §2.5–2.6; Harr, Groundwater and Seepage, Ch. 4.

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