Seepage under a dam with a cut-off wall: 16-Civ-A4 Geotechnical Materials and Analysis, May 2014, Q5

Question 5: Seepage under a dam with 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. Head of water retained $H = 10\ \text{m}$ (upstream free surface 10 m above the ground; downstream water level at the ground surface F, i.e. zero head); silty sand stratum 20 m thick over impervious clay; dam base J–D 25 m long, founded 2.5 m below the ground (B–C); sheet pile at the upstream heel penetrating 7 m below the base (C–I), so its tip is 9.5 m below the ground and 10.5 m above the clay; $k = 2.0\times10^{-4}\ \text{cm/s} = 2.0\times10^{-6}\ \text{m/s}$; soil homogeneous and isotropic.

Figure 5 — flow net beneath the dam (flow lines blue, equipotentials red). Flow enters the upstream bed, passes down the embedded dam face and round the sheet-pile tip, runs under the 25 m base and rises to the downstream bed. The net has $N_f = 4$ flow channels and $N_d = 12.5$ equipotential drops (the last field is a half field).

Find. A flow net for the confined flow, and the seepage $q$ per metre run of wall.

Approach. The flow is confined (bounded above by the impervious dam base and below by the clay), so it is a Laplace-equation flow net problem; sketch curvilinear squares, count the flow channels $N_f$ and equipotential drops $N_d$, and apply the flow-net discharge formula.

Total head loss. The upstream reservoir surface is 10 m above the downstream ground surface, where the water head is zero, so the head driving the seepage is $\Delta H = H = 10\ \text{m}$, dissipated between the upstream and downstream beds.
Construct the flow net. The boundary conditions fix the net. The upstream bed A–B is an equipotential at $h = 10\ \text{m}$ and the downstream bed D–F an equipotential at $h = 0$. The embedded dam face B–C, both faces of the sheet pile C–I, the base J–D and the buried downstream face together form the uppermost flow line, and the top of the clay G–H is the lowest flow line. Sketching curvilinear squares between them with $N_f = 4$ flow channels needs $N_d \approx 12.5$ equipotential drops, so each drop is $\Delta h = \Delta H/N_d = 10/12.5 = 0.80\ \text{m}$.
Seepage per unit length. For a flow net, $q = k\,\Delta H\,\dfrac{N_f}{N_d}$. Substituting, $q = (2.0\times10^{-6})(10)\dfrac{4}{12.5}$, so $\boxed{q \approx 6.4\times10^{-6}\ \text{m}^3/\text{s per m} = 0.55\ \text{m}^3/\text{day per m}}$.

Seepage results (per metre of wall)
Quantity	Value
Head loss $\Delta H$	10 m
Flow net ($N_f\,/\,N_d$)	4 / 12.5
Drop per field $\Delta h$	0.80 m
Seepage $q$	$6.4\times10^{-6}\ \text{m}^3/\text{s} = 0.55\ \text{m}^3/\text{day}$

Check. A hand-sketched net carries roughly ±½ field of tolerance in $N_d$, so $q$ depends on the net. As a check on the sketch, a finite-difference solution of Laplace’s equation for this exact geometry gives a shape factor $N_f/N_d = 0.317$, i.e. $q = 6.3\times10^{-6}\ \text{m}^3/\text{s}$ per metre (0.55 m³/day), within 1% of the counted net. A shape factor near 0.5 applies only to a lone sheet pile reaching about half-way through the layer; the 25 m base lengthens every flow path and cuts the factor to about 0.32. Per the exam’s instruction, the flow-net counts used are stated explicitly.

Topic. Confined seepage and flow nets (Laplace flow beneath hydraulic structures).

Key relations. $q = k\,\Delta H\,\dfrac{N_f}{N_d}$ per unit width; $\Delta h = \Delta H/N_d$; head at a point $h_p = \dfrac{n_d}{N_d}\Delta H$.

Why this works. For homogeneous, isotropic soil the seepage obeys Laplace’s equation; a flow net of curvilinear squares makes each “square” carry the same flow $\Delta q = k\,\Delta h$, so the total discharge reduces to the shape factor $N_f/N_d$ times $k\,\Delta H$.

Common pitfall. Using the full 20 m stratum depth or the wall penetration as a “length” in the formula — the geometry enters only through the flow-net shape factor, not as an explicit dimension. The cut-off wall reduces seepage precisely by increasing $N_d$ (lengthening the flow paths).

Source. Das, Principles of Geotechnical Engineering, Ch. 8 (Seepage — flow nets, seepage calculation).

Question 5 of 6: Seepage under a dam with a cut-off wall

Question 5: Seepage under a dam with a cut-off wall (20 marks)