Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2016 — 04-Geol-A6 Soil Mechanics. Three-hour, closed-book exam; a Casio or Sharp approved calculator, compass and ruler are permitted. Six questions constitute the complete 100-mark paper: Questions 1–5 are compulsory; Question 6 offers five 5-mark optional items of which the source asks for three.
Given. $k=4\times10^{-6}$ m/s. Read from Figures Q5-1/Q5-2 (elevation axis): clay from the impermeable base (elevation 0) to the ground surface (elevation 10 m) on both sides; dam base $a$–$d$ from $x=10$ to $30$ m at elevation 8 m (embedded 2 m); dam crest at elevation 16 m. Configuration B adds a thin cutoff at the heel ($x=10$ m) from the base down to elevation $\approx2$ m. Point 1 at $x\approx4$ m, elevation $\approx3.1$ m; point 2 at $x\approx34.6$ m, elevation $\approx5.0$ m (same positions in both figures).
Find. Boundary-condition labels; total/elevation/pressure head at points 1 and 2; pore-pressure-head distribution along the base; flow rate $q$ under each dam; which configuration has the higher uplift.
Approach. The printed nets are numerical contour plots: solid lines are equipotentials at equal head intervals, dashed lines are flow paths. Count the equipotential drops $N_d$, take the effective number of flow channels $N_f$ from the shape of the elements under the dam, then use $h=H_A-n\,\Delta H/N_d$, $h_p=h-z$ and $q=k\Delta H\,N_f/N_d$. Datum: the impermeable base of the clay (elevation 0), so the elevation head equals the elevation read off the figure.
Check: the source figure prints no water-surface elevations. The reservoir is taken as full to the dam crest (headwater $H_A=16$ m) and the downstream side as dry (tailwater at the ground surface, $H_B=10$ m), so $\Delta H=6$ m. Every head below equals $10+6\,\phi$, where $\phi$ is the fraction of $\Delta H$ still left at that point, so the answers can be rescaled for any other assumed headwater. The equipotential counts ($N_d=10$ and $20$) and the positions of points 1 and 2 were read from the printed figure. The resulting shape factors are $q/(k\Delta H)=0.250$ (A) and $0.178$ (B).
1. Boundary conditions. $A$ (upstream ground surface, under the reservoir) is a constant-total-head (equipotential) boundary: $h_A=16$ m (elevation head 10 m + pressure head 6 m). $B$ (downstream ground surface, dry) is also a constant-total-head boundary: $h_B=10$ m (pressure head 0). The dam base and its 2 m embedded faces, the cutoff wall (Configuration B), the impermeable base of the clay and the side boundaries at $x=0$ and $x=40$ m are no-flow boundaries (flow lines).
Flow-net reading. $\Delta H=16-10=6$ m for both configurations. Configuration A: 9 solid equipotentials between $A$ and $B$, so $N_d=10$ and each drop is $6/10=0.60$ m. The net is symmetric about $x=20$ m, and the middle equipotential (the vertical line at $x=20$ m) carries $\phi=0.5$. Configuration B: 19 equipotentials, so $N_d=20$ and each drop is $6/20=0.30$ m. Most of these are crowded around the cutoff tip.
2. Heads at points 1 and 2.
Config A: point 1 lies on equipotential 1, so $h_1=16-1(0.60)=\boxed{15.4\text{ m}}$, $z_1=\boxed{3.1\text{ m}}$, $h_{p1}=15.4-3.1=\boxed{12.3\text{ m}}$. Point 2 lies on equipotential 9, so $h_2=16-9(0.60)=\boxed{10.6\text{ m}}$, $z_2=\boxed{5.0\text{ m}}$, $h_{p2}=\boxed{5.6\text{ m}}$.
Config B: point 1 lies on equipotential 2 of 19 (below the first contour, which ends on the upstream face of the cutoff), so $h_1=16-2(0.30)=\boxed{15.4\text{ m}}$, $z_1=3.1$ m, $h_{p1}=\boxed{12.3\text{ m}}$. Point 1 is far upstream, so the cutoff barely changes its head. Point 2 lies between equipotentials 18 and 19, about 0.6 of the way across ($n\approx18.6$), so $h_2=16-18.6(0.30)=\boxed{10.4\text{ m}}$, $z_2=5.0$ m, $h_{p2}=\boxed{5.4\text{ m}}$.
3. Pore-pressure head along the base (elevation 8 m). Read $\phi$ where the equipotentials meet the base, then $h_p=(10+6\phi)-8$:
Config A: $a$ ($x=10$): $\phi\approx0.87$ → $h_p\approx7.2$ m; $x=15$: $0.66$ → $6.0$ m; $x=20$: $0.50$ → $5.0$ m; $x=25$: $0.34$ → $4.0$ m; $d$ ($x=30$): $0.13$ → $2.8$ m.
Config B: $a$ (just downstream of the cutoff): $\phi\approx0.51$ → $h_p\approx5.1$ m; $x=15$: $0.45$ → $4.7$ m; $x=20$: $0.35$ → $4.1$ m; $x=25$: $0.24$ → $3.4$ m; $d$: $0.10$ → $2.6$ m.
Both distributions fall from heel to toe (plotted below). About half of $\Delta H$ is lost around the cutoff before the water reaches the base in B, while in A only about 13% is lost before the heel.
4. Flow under the dam. Under the middle of the dam in A, the three drawn channels have width-to-length ratios of about 0.8, 1.0 and 0.8, so the effective $N_f\approx2.5$. In B the equipotentials are closer together (about 2.25 m apart instead of 3.1 m), so the same three channels are each about 1.2 squares wide and $N_f\approx3.6$.
Config A: $q_A=k\Delta H\dfrac{N_f}{N_d}=4\times10^{-6}\times6\times\dfrac{2.5}{10}=\boxed{6.0\times10^{-6}\text{ m}^3/\text{s per m}}$ ($\approx0.52$ m³/day per metre).
Config B: $q_B=4\times10^{-6}\times6\times\dfrac{3.6}{20}=\boxed{4.3\times10^{-6}\text{ m}^3/\text{s per m}}$ ($\approx0.37$ m³/day per metre).
The numerical shape factors (0.250 and 0.178) confirm both values. The cutoff reduces seepage by about 28%.
5. Uplift comparison (no calculation). Uplift on the dam equals $\gamma_w$ times the area under the pressure-head diagram along $a$–$d$. The cutoff dissipates about half of the available head before the water reaches the base, so the pressure head in B is lower at every point along the base. $\Rightarrow\boxed{\text{Configuration A (no cutoff) has the higher uplift}}$. As a check only, integrating the step 3 values gives about $981$ kN/m (A) against about $787$ kN/m (B). A cutoff does not stop seepage. It lengthens the flow path near the heel, so less head is left to act under the structure.
[Figure not reproduced: Equipotential contours for Configurations A and B. See the official exam paper or the cited reference text.]
Fig. Q5-A/B — the dam foundation redrawn with the printed equipotential spacing ($\phi$ steps of 0.1 in A and 0.05 in B). Points 1 and 2 are marked.
Fig. Q5.3 — pore-pressure head along the dam base (elevation 8 m) for $H_A=16$ m and $H_B=10$ m. Config A's curve lies above Config B's everywhere, so A has the larger uplift area.
