Seepage Flow Net — Heads and Pore Pressures: 16-Civ-A4 Geotechnical Materials and Analysis, December 2018, Q5

Question 5: Seepage Flow Net — Heads and Pore Pressures (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. From Figure 3: the datum is the downstream (tailwater) water surface. The tailwater stands 1.00 m above the downstream ground and the reservoir 5.00 m above the upstream ground (the two ground surfaces are at the same level). The base of the dam’s downstream apron (points 1, 2, 3) is 0.80 m below ground, and the main base (points 4 to 7½, at the cutoff wall) is 1.40 m below ground. Equipotentials are numbered from the downstream end: 1–7 meet the base, and 7½ marks the base/cutoff corner half-way between lines 7 and 8.

Find. $N_d$ and $\Delta h$, then the total head $h$ and pore pressure $u$ at points 2, 4 and 6.

[Figure not reproduced: Figure 3 from the December 2018 exam: flow net under a concrete dam with a downstream apron and an upstream cutoff wall. See the official exam paper or the cited reference text.]

Figure 3 (reproduced from the exam paper) — datum at the tailwater surface; tailwater 1.00 m and reservoir 5.00 m above the ground; apron base (points 1–3) 0.80 m and main base (points 4–7½) 1.40 m below ground. Equipotentials are numbered from the downstream end.

Approach. The total head loss equals the difference between the two water levels; dividing it by the number of potential drops counted on the net gives $\Delta h$. With the datum at tailwater level the downstream ground has total head 0, so the head on equipotential $k$ is $h_k = k\,\Delta h$; the elevation head $z$ of each point comes from the base depths, and $u=\gamma_w(h-z)$.

Total head loss. The seepage is driven by the difference of the two free-water surfaces: $$H = 5.00 - 1.00 = 4.00\ \text{m}.$$
Potential drops and head per drop. Follow any flow line from the downstream ground to the upstream ground and count the equipotentials it crosses: lines 1–7 under the base, line 8 on the downstream face of the cutoff, lines 9–11 fanning round the cutoff tip, lines 12–13 upstream of it, and line 14 on the upstream face of the cutoff — 14 internal equipotentials between the two boundary equipotentials (the downstream and upstream ground surfaces), so $$N_d = 15,\qquad \Delta h = \frac{H}{N_d}=\frac{4.00}{15}= 0.267\ \text{m per drop}.$$ (The label 7½ is not the last line. It marks the base corner at the cutoff, half-way between lines 7 and 8; about half of the head is lost around the cutoff.)
Total head at each point. Counting $k$ drops up from the downstream ground ($h=0$ at the datum): $$h_2 = 2(0.267)=0.533\ \text{m},\quad h_4 = 4(0.267)=1.067\ \text{m},\quad h_6 = 6(0.267)=1.600\ \text{m}.$$ (Check: $k=15$ gives 4.00 m, the reservoir level above the datum.)
Elevation heads. The datum is 1.00 m above the ground, so point 2 on the apron base is at $z_2 = -(1.00+0.80) = -1.80$ m, and points 4 and 6 on the main base are at $z_4 = z_6 = -(1.00+1.40) = -2.40$ m.
Pore-water pressure. $u=\gamma_w(h-z)$: $$u_2 = 9.81\,(0.533+1.80) = 9.81\times2.333 = \boxed{22.9\ \text{kPa}}$$ $$u_4 = 9.81\,(1.067+2.40) = 9.81\times3.467 = \boxed{34.0\ \text{kPa}}$$ $$u_6 = 9.81\,(1.600+2.40) = 9.81\times4.000 = \boxed{39.2\ \text{kPa}}$$

Check: $N_d = 15$ comes from counting the printed net, and it is the value in the textbook example this figure is taken from (Craig). A $\pm1$ miscount changes $\Delta h$ by about 7% and $u$ by about 1–2 kPa. The datum and base depths are dimensioned on the figure, so they are exact.

Final results — Question 5
Quantity	Point 2	Point 4	Point 6
Total head $h$ above datum (m)	0.533	1.067	1.600
Elevation head $z$ (m)	−1.80	−2.40	−2.40
Pressure head $h-z$ (m)	2.33	3.47	4.00
Pore pressure $u$ (kPa)	22.9	34.0	39.2

($H=4.00$ m, $N_d=15$, $\Delta h=0.267$ m/drop; datum at tailwater level.)

Topic: Confined seepage under a dam by the flow-net method.

Key relations: $\Delta h=H/N_d$; total head $h=k\,\Delta h$ (datum at tailwater); pressure head $h_p=h-z$; $u=\gamma_w h_p$.

Why this works: In a valid flow net every equipotential represents an equal drop $\Delta h$ of total head, so with the datum at tailwater level the head at a point is just the number of drops between it and the downstream ground times $\Delta h$. Splitting total head into elevation and pressure head then delivers the pore pressure — here the uplift acting on the base of the structure, which increases toward the upstream heel.

Common pitfall: Taking the highest label under the base (7½) as $N_d$ and ignoring the head lost round the cutoff, or putting the points at $z=0$ when the figure’s datum is the tailwater surface — the base is 1.80–2.40 m below it, and that elevation head is part of the uplift pressure.

Source: Craig & Knappett, Craig’s Soil Mechanics, ch. 2 (seepage, flow nets); Das, ch. 8.

Question 5 of 6: Seepage Flow Net — Heads and Pore Pressures

Question 5: Seepage Flow Net — Heads and Pore Pressures (20 marks)