Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2018 — 18-Geol-A6 Soil Mechanics. Three-hour, closed-book exam; a Casio or Sharp approved calculator, a compass and a ruler are permitted. Six questions constitute the complete 100-mark paper: Questions 1–5 (10+15+20+20+20 marks) are compulsory; Question 6 offers eight 5-mark optional items of which the source asks for three.
Each reading is cross-checked below by an independent physical-consistency test (Q2: the 12%-point saturation comes out to $\approx100\%$, matching its plotted position essentially on the $S=100\%$ curve; Q4: the two digitized curves in Figure Q4-2 are treated as the overconsolidated-clay sample (upper, steep-kneed curve, per the question's own description) and an unused second dataset (lower curve, not needed for a single-sample problem); Q5: the resulting pore pressures give positive effective stress at every point and a linear uplift diagram, the expected signature of a consistent equipotential count). Assumed $G_s=2.70$ (not given) for Q2(c)'s phase relations — typical for a silty/clayey compacted fill and independently supported by the $S\approx100\%$ check above.
Check: the source text names "points a, b, c, d and e" but the diagram and part (b) both use A, B, C, D — a labeling inconsistency already present in the source (five named vs. four used); answered for the four points that actually appear, A–D, as drawn.
Given. Cutoff-wall dam on a $L=30$ m permeable base over an impervious layer; $H=20$ m and $h_t=10$ m (headwater/tailwater heads above the impervious-layer datum); $D=1$ m cutoff depth; $\gamma_{sat}=21.3\text{ kN/m}^3$; $\gamma_w=9.81\text{ kN/m}^3$; base points 1–5 spaced 7.5 m apart; depths below ground surface $z_A=10$, $z_B=15$, $z_C=6$, $z_D=9$ m; $k=3\times10^{-3}$ cm/s; flow net as drawn (Figure Q5-1).
Find. Bernoulli equation; $h_t,h_e,h_p$ at A–D; seepage rate per unit length; uplift-pressure diagram and total uplift force along the base 1–5.
Approach. Count flow channels $N_f$ and equipotential drops $N_d$ from the drawn net; assign each labelled point to its equipotential (drop number from the upstream boundary); compute heads from $H$, $N_d$ and the given depths, then $q=k\,\Delta H\,(N_f/N_d)$ and the uplift diagram by integrating pore pressure along the base.
[Figure not reproduced: Flow net, dam with upstream sheet-pile cutoff, points 1-5 on the base and A, B, C, D within the soil mass. See the official exam paper or the cited reference text.]
Fig. Q5-1 — Source flow net (as printed): 4 flow channels, 8 equipotential drops digitized from the curve crossings; A sits 1 drop past the upstream cutoff tip, base points 1–5 at drops 2–6, and D 1 drop past point 5.
(a) Bernoulli equation. For seepage flow (velocity head negligible at Darcy-law speeds):
$$\boxed{h_t=h_e+h_p=z+\frac{u}{\gamma_w}}$$
where $h_t$ is total head, $h_e=z$ elevation head (above a chosen datum), $h_p=u/\gamma_w$ pressure head.
Flow-net count and head drop. The drawn net has $N_f=4$ flow channels and $N_d=8$ equipotential drops (counted from the curve crossings; A is 1 drop downstream of the cutoff tip, base points 1–5 occupy drops 2–6 in turn, D is 1 drop past point 5, and the tailwater boundary closes the net at drop 8). Total head loss $H-h_t=20-10=10$ m, so each drop is $\Delta h=10/8=1.25$ m. Taking the datum at the impervious layer (per the figure, both $H$ and $h_t$ are measured from it, and the downstream tailwater sits at the downstream ground surface, i.e. the ground surface itself is $h_t=10$ m above datum):
$$h_{t,n}=H-n\,\Delta h\qquad h_e(\text{point})=10-z\qquad h_p=h_{t,n}-h_e\qquad u=\gamma_w h_p$$
(b) Heads at A, B, C, D. A is at drop $n=1$, B at $n=3$, C at $n=5$, D at $n=7$:
$$\text{A: } h_t=20-1.25=18.75,\ h_e=10-10=0,\ h_p=18.75\text{ m},\ u=183.9\text{ kPa}$$
$$\text{B: } h_t=20-3.75=16.25,\ h_e=10-15=-5,\ h_p=21.25\text{ m},\ u=208.5\text{ kPa}$$
$$\text{C: } h_t=20-6.25=13.75,\ h_e=10-6=4,\ h_p=9.75\text{ m},\ u=95.6\text{ kPa}$$
$$\text{D: } h_t=20-8.75=11.25,\ h_e=10-9=1,\ h_p=10.25\text{ m},\ u=100.5\text{ kPa}$$
Total head decreases monotonically A→B→C→D in the direction of flow, and every resulting effective stress ($\sigma-u$ using $\gamma_{sat}=21.3$ and the given depths) comes out positive — both are the expected consistency checks for a correctly-ordered equipotential count.
(c) Seepage rate. $k=3\times10^{-3}\text{ cm/s}=3\times10^{-5}\text{ m/s}$:
$$q=k(H-h_t)\frac{N_f}{N_d}=3\times10^{-5}\times10\times\frac{4}{8}=\boxed{1.5\times10^{-4}\text{ m}^3/\text{s per m}=13.0\text{ m}^3/\text{day per m}}$$
(d) Uplift diagram, points 1–5. Base points are at $z=0$ (on the ground surface, $h_e=10$ m) and drops $n=2,3,4,5,6$ respectively:
$$u_1=\gamma_w(H-2\Delta h-10)=9.81(7.5)=73.6\text{ kPa}\qquad u_2=9.81(6.25)=61.3\text{ kPa}$$
$$u_3=9.81(5.0)=49.1\text{ kPa}\qquad u_4=9.81(3.75)=36.8\text{ kPa}\qquad u_5=9.81(2.5)=24.5\text{ kPa}$$
The pressures fall off linearly (equal $\Delta h$ over equally-spaced points) — trapezoidal integration over the 4 equal 7.5 m panels:
$$F_{uplift}=7.5\left[\frac{u_1}{2}+u_2+u_3+u_4+\frac{u_5}{2}\right]=\boxed{1471.5\text{ kN per m of dam}}$$
Point
$h_t$ (m)
$h_e$ (m)
$h_p$ (m)
$u$ (kPa)
A
18.75
0
18.75
183.9
B
16.25
−5
21.25
208.5
C
13.75
4
9.75
95.6
D
11.25
1
10.25
100.5
Seepage rate $q=1.5\times10^{-4}\text{ m}^3$/s per m (13.0 m³/day per m); total uplift 1–5 = 1471.5 kN/m
