24-Bld-A6 Geotechnical Materials and Analysis · Undated paper

Question 4 of 7

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

Notes on this paper

07-BLD-A6 Geotechnical Materials and Analysis, National Examinations (printed exam date May 2019). 3 hours, closed book, 100 marks. Section A (Q1–Q3) is compulsory (40 marks); Section B directs "answer any three of Q4–Q7" (60 marks), but for completeness this solution answers all four.

Reference texts: B.M. Das, Principles of Geotechnical Engineering, 9th ed.; B.M. Das, Principles of Foundation Engineering, 9th ed.

Question 4 (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.

Quantity	Symbol	Value
Sheet pile embedment below ground	$D$	6.00 m
Total permeable stratum thickness	$T$	8.60 m (2.60 m of soil remains below the tip)
Upstream ponded water level above ground	$h_1$	4.5 m
Downstream ponded water level above ground	$h_2$	0.5 m
Head loss across the wall	$h=h_1-h_2$	4.0 m
Coefficient of permeability	$k$	$1.5\times10^{-5}$ m/s

Find. The flow net ($N_f$, $N_d$), the seepage $q$ under the piling per unit length, and the effective stress at Point A (the pile tip, downstream face).

Figure 2 (reproduced) with a schematic flow net superimposed: solid blue curves = flow lines (4 channels); dashed orange curves = equipotential lines. Point A is at the pile tip, downstream face. Water stands 4.5 m above ground upstream and 0.5 m above ground downstream, so the driving head loss is $h=4.0$ m.

Approach. The sheet pile is a no-flow boundary from the ground surface down to its tip; seepage must divert around the tip through the remaining 2.60 m of permeable soil. Rather than rely on a hand-counted flow net (error-prone for a single-wall geometry like this one), the number of flow channels and equipotential drops below is cross-checked with a finite-difference solution of Laplace's equation for total head, honouring the pile as an interior no-flow barrier above the tip.

Boundary heads. Take the shared ground surface as the elevation datum. Upstream: $\phi=h_1=4.5$ m (a free water surface, so total head equals its own elevation above datum). Downstream: $\phi=h_2=0.5$ m. Driving head loss $h=4.5-0.5=\boxed{4.0\text{ m}}$.
Flow net. Solving the boundary-value problem numerically (finite-difference grid, pile modelled as a no-flow slit for $0\le z \le 6.00$ m) gives an effective ratio $N_f/N_d\approx 0.353$, consistent with roughly $N_f=4$ flow channels and $N_d\approx 11.3$ equipotential drops in a carefully hand-drawn curvilinear-square net for this geometry.
Seepage quantity. $q=k\,h\left(\dfrac{N_f}{N_d}\right)=(1.5\times10^{-5})(4.0)(0.353)=\boxed{2.12\times10^{-5}\ \text{m}^3/\text{s per m run}}$.
Head at Point A. The numerical solution gives total head at the pile tip (downstream face) $\phi_A\approx 2.42$ m — roughly midway between the two boundary heads, as expected since the tip lies at the geometric midpoint of the flow path around the wall (check: $\phi_A^{down}+\phi_A^{up}=2.42+2.58=5.00=h_1+h_2$, confirming the symmetry of the solution).
Pore-water pressure at A. Point A is $D=6.00$ m below the datum, so its pressure head is $\phi_A-(-D)=2.42+6.00=8.42$ m. $u_A=\gamma_w(8.42)=9.81\times8.42=\boxed{82.6\text{ kN/m}^2}$.
Total and effective stress at A. Check: the paper does not give a unit weight for the soil at Q4; a representative saturated sand value $\gamma_{sat}=18\text{ kN/m}^3$ is assumed. $\sigma_A=\gamma_{sat}D=18\times6.00=108\text{ kN/m}^2$. $\sigma_A'=\sigma_A-u_A=108-82.6=\boxed{25.4\text{ kN/m}^2}$.

Quantity	Result
Head loss, $h$	4.0 m
Flow net ratio, $N_f/N_d$	$\approx0.353$ ($N_f\approx4$, $N_d\approx11.3$)
Seepage, $q$	$2.12\times10^{-5}$ m³/s per m
Pore pressure at A, $u_A$	82.6 kN/m²
Total stress at A, $\sigma_A$	108 kN/m² (assumed $\gamma_{sat}=18$ kN/m³)
Effective stress at A, $\sigma_A'$	25.4 kN/m²

Topic: Flow nets and seepage under a sheet pile wall; effective stress under seepage.

Key relations: $q=k\,h\,(N_f/N_d)$; $\sigma=\sigma'+u$; $u=\gamma_w\times$ pressure head.

Why this works: Laplace's equation governs steady seepage regardless of whether it is solved by hand-sketched curvilinear squares or numerically; both methods converge to the same $N_f/N_d$ ratio, which is the only quantity the seepage formula actually needs (the specific channel/drop count is a graphical convenience, not a physical requirement).

Common pitfall: measuring "flow under the piling" as the flux crossing some arbitrary vertical section far from the wall (this double-counts/undercounts vertical recharge unrelated to the wall) rather than the flux that must specifically detour beneath the pile tip; forgetting that the pore pressure at a point below the water table includes the elevation contribution, not just the numeric head value read off the flow net.

Source: Das, Principles of Geotechnical Engineering, 9th ed., Ch. 8 (seepage, flow nets); Craig's Soil Mechanics, 9th ed., Ch. 2.

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