Question 5 of 6: Seepage

Question 5: Seepage (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 symmetric double sheet-pile cofferdam (FE seepage output), Figure Q5-1: standing water/riverbed on both sides at the same level, ground surface (top of soil) at elevation $+12$ m with $6$ m of standing water above it to elevation $+18$ m; sheet piles driven to tip elevation $\approx+4$–$4.5$ m; the excavation inside the piles is dewatered to a water level at $B\approx+9$ m. $\gamma_{sat}=18$ kN/m³. Points: $A(30,14)$, $B(22,8.7)$, $C(11,12)$, $D(8,0)$, $X(21,7)$, $Y(9,3)$ (all in metres, using the printed grid as coordinates).

Find. Axis/datum labelling; boundary conditions at A–D; identity and head values of lines 1–3; flow direction; the Bernoulli head equation; head components and vertical effective stress at X and Y.

Approach. Recognize the standing water above the ground surface as a constant-head (Dirichlet) boundary equal to its own free-surface elevation, the dewatered excavation floor as the low-head exit boundary, and the base of the FE mesh as a no-flow (impervious) boundary; read the equipotential/flow-line pattern of the net to get boundary values, then apply $h_t=h_e+h_p$ and $\sigma'=\sigma-u$ at X and Y.

1. (a)/(b) Axes and datum. Both axes are drawn to the same metre scale printed on the figure: horizontal = lateral distance (m); vertical = elevation (m). The datum for elevation head is taken at $z=0$ (the bottom edge of the plotted domain), so elevation head at any point equals its plotted $y$-coordinate directly.
2. (c) Boundary conditions. A (downstream ground surface, under standing water): constant total head $h=18$ m (Dirichlet). C (upstream ground surface, symmetric): constant total head $h=18$ m (Dirichlet) — the standing water is static above the soil, so at any submerged point on the ground surface the total head simply equals the free-water-surface elevation. B (dewatered excavation floor, inside the cofferdam): constant total head $h\approx9$ m (Dirichlet) — the low-head boundary that drives seepage into the excavation. D (base of the modelled soil domain): no-flow (Neumann, $\partial h/\partial n=0$) boundary, representing either an impervious stratum or the practical limit of the FE mesh.
3. (d) i) Lines 1, 2, 3. They are equipotential lines — contours of constant total head through the flow domain. Together with the (unlabelled) flow lines they form the curvilinear-square flow net used to read head, pore pressure, and seepage quantity at any point; each equipotential must cross the impervious base boundary (D) at right angles, and they bunch together near the pile tips where the flow converges most sharply.
4. (d) ii) Values of lines 1, 2, 3. Total head loss $=18-9=9$ m. Reading the net as $N_d=8$ roughly-equal drops, $\Delta h=9/8=1.125$ m/drop, and lines 1–3 (the first three equipotentials in from the $h=18$ boundary) carry $h_1=18-1.125=\boxed{16.9\text{ m}}$, $h_2=18-2.25=\boxed{15.75\text{ m}}$, $h_3=18-3.375=\boxed{14.6\text{ m}}$.
5. (e) Direction of flow. Water enters at the high-head boundary (the standing water over the ground surface, both sides), flows down and around the sheet-pile tips, and rises up into the excavation toward the low-head boundary at B — i.e. inward and downward outside the piles, curving upward beneath and inside them.
6. (f) Bernoulli equation. $$h_t=\frac{v^2}{2g}+\frac{u}{\gamma_w}+z=h_p+h_e$$ For the slow (laminar, Darcy-regime) seepage velocities in soil the velocity head $v^2/2g$ is negligible, leaving total head as simply the sum of pressure head $h_p=u/\gamma_w$ and elevation head $h_e=z$; total head is what drives flow (from high to low $h_t$), not pressure alone.
7. (g) Head components at X and Y. Each point has three head components: elevation head $h_e=z$ (position relative to datum), pressure head $h_p=u/\gamma_w$ (pore pressure expressed as an equivalent water column), and total head $h_t=h_e+h_p$ (read from the flow net). At X: $h_e=7$ m; reading the net near X (just below the excavation floor B, innermost flow channel) $h_t\approx9.5$ m, so $h_p=h_t-h_e=2.5$ m. At Y: $h_e=3$ m; reading the net near Y (upstream side, just inside the outer equipotentials) $h_t\approx16.5$ m, so $h_p=h_t-h_e=13.5$ m.
8. (h) Point X — i) vertical total stress. Saturated soil column between the excavation floor (elev. 9 m) and X (elev. 7 m): $\sigma_v=\gamma_{sat}(9-7)=18(2)=\boxed{36.0\text{ kPa}}$. ii) pore pressure: from the flow-net head, $u_X=\gamma_w h_p=9.81(2.5)=\boxed{24.5\text{ kPa}}$. iii) effective vertical stress: $\sigma_v'=36.0-24.5=\boxed{11.5\text{ kPa}}$.
9. (h) Point Y — i) vertical total stress. The 6 m standing-water surcharge plus the saturated soil column down to Y: $\sigma_v=\gamma_w(18-12)+\gamma_{sat}(12-3)=9.81(6)+18(9)=58.9+162.0=\boxed{220.9\text{ kPa}}$. ii) pore pressure: $u_Y=\gamma_w h_p=9.81(13.5)=\boxed{132.4\text{ kPa}}$. iii) effective vertical stress: $\sigma_v'=220.9-132.4=\boxed{88.4\text{ kPa}}$.

The comparatively low $\sigma_v'$ at X (inside the excavation, on the seepage exit side) versus the much higher $\sigma_v'$ at Y (on the recharge/entry side) is the physically expected pattern for a dewatered excavation: upward seepage on the exit side reduces effective stress toward the base — the classic precursor to base heave or piping if it approaches zero — while downward-directed seepage on the entry side increases it.