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18-Geol-A6 Soil Mechanics · December 2019

Question 5 of 6: Seepage

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

Notes on this paper

National Exams, December 2019 — 18-Geol-A6, Soil Mechanics (3 hours, closed book). Six questions constitute a complete exam: Questions 1–5 are compulsory; for Question 6, candidates choose 3 of 8 optional sub-questions (5 marks each) — all 8 are solved below as a complete study resource.

Reference texts: Das, Principles of Geotechnical Engineering; Craig's Soil Mechanics (Craig & Knappett); Freeze & Cherry, Groundwater (seepage/flow-net topics).

graphically-read values (compaction curve extrapolation, oedometer Casagrande construction, flow-net equipotential positions) are flagged with individual check callouts below where the read/assumed value materially affects the answer.

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.

Check: Figures Q5-1a/1b carry no printed numeric headwater/tailwater elevations. Boundary heads are read directly off the drawn geometry — $H_A=15\text{ m}$ at the top of the embankment's saturated (wetted) zone (reservoir level) and $H_B=10\text{ m}$ at the downstream ground surface (seepage exit, $u=0$). Equipotential-drop positions for points 1 and 2 are likewise engineering estimates read from the figure's flow-net grid, not exact values; both figures are re-drawn schematically below.

Given. $k=4\times10^{-6}\text{ m/s}$ (homogeneous clay foundation); boundary heads $H_A=15\text{ m}$ (upstream/embankment), $H_B=10\text{ m}$ (downstream ground surface); point elevations $z_1=3\text{ m}$, $z_2=4\text{ m}$; Figure Q5-1a has no cutoff wall; Figure Q5-1b has a partial cutoff wall extending from the base of the embankment down into the foundation (not reaching the impermeable bedrock).

Find. (a) boundary conditions at A, B. (b) total/elevation/pressure head at points 1, 2, both figures. (c) a representative gradient. (d) base pore-pressure-head distribution. (e) which configuration has higher uplift, and why.

[Figure not reproduced: Figure Q5-1a — no cutoff wall. Illustrative flow lines shown; boundary heads and points 1/2 as read from the source figure. See the official exam paper.]

Figure Q5-1b — with partial cutoff wall 0 10 20 30 40 0 4 8 12 16 Distance (m) Elevation (m) A (H=15m) B (H=10m) 1 (z=3m) 2 (z=4m) cutoff wall
Figure Q5-1b — with a partial cutoff wall below the upstream heel, forcing seepage to detour down and around its tip before reaching the downstream side.

(a) Boundary conditions at A and B

Point A (upstream, at the top of the embankment's saturated/wetted zone) is a constant-head boundary: total head there equals the reservoir/headwater elevation, $H_A=15\text{ m}$, because it is in direct contact with standing water and there is no velocity head in this slow (Darcy) seepage regime. Point B (downstream ground surface, where the phreatic surface daylights) is also a constant-head boundary, but at atmospheric pressure ($u=0$): total head there equals the local ground-surface elevation, $H_B=10\text{ m}$, since pressure head is zero at a free seepage-exit point.

(b) Total, elevation, and pressure head at points 1 and 2

Approach. Total head drops linearly with each equipotential-line crossing between the two boundaries. With $\Delta H_{\text{total}}=H_A-H_B=5\text{ m}$ split over $N_d=10$ equal drops ($\Delta h=0.5\text{ m}$ per drop), interpolate each point's total head by how many drops it sits past boundary A, then split into elevation head ($z$) and pressure head ($h_t-z$).

  1. Figure Q5-1a (no cutoff). Point 1 sits 2 drops in from A: $h_{t,1}=15-2(0.5)=14.0\text{ m}$. Point 2 sits 8 drops in from A (2 short of B): $h_{t,2}=15-8(0.5)=11.0\text{ m}$. $$\text{Pt 1: } h_p=14.0-3.0=11.0\text{ m}\ (u=107.9\text{ kPa}) \qquad \text{Pt 2: } h_p=11.0-4.0=7.0\text{ m}\ (u=68.7\text{ kPa})$$
  2. Figure Q5-1b (with cutoff). The cutoff sits between A and point 1's position, so point 1 (still relatively close to A, upstream of most of the cutoff-induced squeezing) reads about the same as in (a): $h_{t,1}=14.0\text{ m}$. Point 2, downstream of the cutoff, has now had one MORE drop's worth of head dissipated getting around the cutoff tip by the time seepage reaches it — 9 drops in from A: $h_{t,2}=15-9(0.5)=10.5\text{ m}$. $$\text{Pt 1: } h_p=11.0\text{ m}\ (u=107.9\text{ kPa}) \qquad \text{Pt 2: } h_p=10.5-4.0=6.5\text{ m}\ (u=63.8\text{ kPa})$$
Total / elevation / pressure head at points 1 and 2 (both configurations)
Figure Q5-1aFigure Q5-1b
Point$h_t$$z$$h_p$ ($u$)$h_t$$z$$h_p$ ($u$)
114.0 m3.0 m11.0 m (107.9 kPa)14.0 m3.0 m11.0 m (107.9 kPa)
211.0 m4.0 m7.0 m (68.7 kPa)10.5 m4.0 m6.5 m (63.8 kPa)

(c) Gradient

Taking one representative flow-net square adjacent to point 1 in Figure Q5-1a, with head drop $\Delta h=0.5\text{ m}$ over a square side $L\approx3\text{ m}$: $$i=\frac{\Delta h}{L}=\frac{0.5}{3}=\boxed{0.167}$$ In Figure Q5-1b, the squares bunch up tightly around the cutoff tip (down to roughly $L\approx1.2\text{ m}$ there), giving a locally much steeper gradient: $$i_{\text{near cutoff}}=\frac{0.5}{1.2}=\boxed{0.417}$$ — nearly $2.5\times$ Figure Q5-1a's gradient, which is exactly the mechanism by which a cutoff wall forces extra head loss into a short stretch of the flow path.

(d) Pore-pressure head along the dam base

Pore-pressure head distribution along the dam base 10 14 18 22 26 30 0 20 40 60 80 100 120 Distance along dam base, x (m) Pressure head, hp (m) — (a) no cutoff — (b) with cutoff
Pore-pressure head along the dam base, Figure Q5-1a (smooth, roughly linear decline) vs. Figure Q5-1b (a sharp early drop concentrated near the cutoff, then a flatter tail) — both curves start near the same upstream value and both remain everywhere below Figure Q5-1a's.

(e) Which dam has the highest uplift force?

Figure Q5-1a (no cutoff) is subject to the higher uplift force. Uplift is the integral of pore-pressure head acting upward on the dam base; with no cutoff, water has a shorter, less-obstructed path from the high-head upstream boundary to underneath the dam, so less head is dissipated by the time seepage reaches the base — leaving MORE residual pressure head (and hence more uplift) under the full footprint. The partial cutoff in Figure Q5-1b forces seepage to travel down and around its tip, a materially longer path that dissipates a larger share of the total head before the flow ever gets under the downstream portion of the dam, which is exactly why point 2's computed pressure head is lower in (b) than in (a) above. This is the standard engineering rationale for adding a cutoff or sheet-pile curtain beneath a dam: it does not change the total head drop across the site, but it redistributes where that drop happens, trading it away from the high-uplift zone directly under the structure.