24-Bld-A6 Geotechnical Materials and Analysis · December 2017
Question 4 of 7
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
07-Bld-A6 Geotechnical Materials and Analysis — National Exam, December 2017. Closed book, 3 hours; drawing instruments and either a Casio or Sharp approved calculator required; the formula sheet and influence charts printed at the back of the exam are reproduced/used inline where needed. Section A (Questions 1–3, 40 marks, answer all) and Section B (Questions 4–7, 20 marks each, the paper asks for any three of four) — all seven questions are answered below.
Reference texts: B. M. Das, Principles of Geotechnical Engineering, 9th ed. (phase relations, seepage/flow nets, stress distribution, consolidation, shear strength); R. F. Craig / J. Knappett, Craig's Soil Mechanics, 9th ed. (flow nets, effective-stress strength parameters); Canadian Foundation Engineering Manual (CFEM), 4th ed.
Check — assumptions and source notes for this paper. (1) Question 4's flow net (Nf = 5 flow channels, Nd = 14 equipotential drops, point A read as three drops upstream of the downstream exit, exit-field length ≈ 2 m) is used with a soil total unit weight of 20 kN/m³. (2) Question 5(ii)'s "approximate" method is extended from the formula sheet's centre-of-rectangle formula to a corner point (point A here is not centred on the loaded area) using the standard mirror/quartering superposition trick, valid because the approximate formula is linear in load exactly like the exact one. (3) Question 6's least-squares fit through the three effective-stress points gives an intercept of −0.7 kPa — not distinguishable from zero with only three data points — so c′ is taken as 0, consistent with the expected behaviour of a normally consolidated clay. (4) Question 7's figure shows three distinct layer thicknesses — H1 = 1.5 m (sand above the water table), H2 = 1.5 m (sand below the water table), H3 = 2 m (the clay layer, void ratio e = 0.75) — and all three values are used below.
Given. A concrete dam on a permeable foundation over an impermeable stratum, with a flow net of Nf = 5 flow channels and Nd = 14 equipotential drops; the upstream water level is 6.3 m and the downstream (tailwater) level 1.6 m above the ground surface. Point A lies 9.4 m below the downstream ground surface (17.2 m is the full water-surface-to-impermeable-base dimension).
Given data — Question 4
Quantity
Symbol
Value
Permeability
k
25×10-6 m/s
Upstream water level above ground
—
6.3 m
Downstream (tailwater) level above ground
—
1.6 m
Flow channels / equipotential drops
Nf / Nd
5 / 14
Depth of A below downstream ground
—
9.4 m
Total unit weight of soil
γ
20 kN/m³
Find. (i) seepage q per metre of dam; (ii) effective stress at A; (iii) maximum exit gradient and its implication.
Figure 1 — seepage under the concrete dam. Head loss H = 6.3 − 1.6 = 4.7 m drives flow through Nf = 5 channels and Nd = 14 drops; A is 9.4 m below the downstream ground surface, about three drops upstream of the exit.
Approach. Use the flow-net discharge formula for q; find the total head at A from the number of potential drops counted from the downstream exit, convert to pore pressure via the pressure head, and subtract from the total vertical stress; then compare the exit gradient (drop over the shortest exit field) with the critical gradient $\gamma'/\gamma_w$.
(i) Seepage. Net head loss $H=6.3-1.6=4.7$ m, head loss per drop $\Delta h=H/N_d=4.7/14=0.336$ m,
$$q=kH\frac{N_f}{N_d}=(25\times10^{-6})(4.7)\left(\frac{5}{14}\right)=4.20\times10^{-5}\ \text{m}^3/\text{s per m}.$$
Per day: $q=4.20\times10^{-5}\times86400=\boxed{3.63\ \text{m}^3/\text{day per m of dam}}.$
Total head at A. Taking the downstream tailwater surface as datum, A sits about three equipotential drops upstream of the exit, so
$$h_A = 3\,\Delta h = 3(0.336) = 1.01\ \text{m above datum}.$$
Pore pressure at A (ii). A lies $1.6+9.4=11.0$ m below the datum, so the pressure head is $h_p=h_A+11.0=12.0$ m and
$$u_A = \gamma_w h_p = 9.81(12.0) = 117.8\ \text{kPa}.$$
Total and effective stress at A. With 1.6 m of standing tailwater and 9.4 m of saturated soil ($\gamma=20$ kN/m³) above A,
$$\sigma_A = 1.6\gamma_w + 9.4\gamma = 1.6(9.81)+9.4(20) = 15.7+188.0 = 203.7\ \text{kPa},$$
$$\sigma_A' = \sigma_A-u_A = 203.7-117.8 \boxed{=\ 85.9\ \text{kPa}}.$$
Critical and exit gradients (iii).
$$i_{cr}=\frac{\gamma-\gamma_w}{\gamma_w}=\frac{20-9.81}{9.81}=1.04.$$
Scaling the shortest exit-flow field from the net (length ≈ 2 m at the downstream toe),
$$i_{exit}=\frac{\Delta h}{l}\approx\frac{0.336}{2}\approx0.17,\qquad FS=\frac{i_{cr}}{i_{exit}}\approx6.2.$$
The exit gradient is well below critical, so the dam as drawn is safe against piping. If the exit gradient instead exceeded icr, the effective stress at the downstream ground surface would fall to zero, the soil there would boil (quick condition), and progressive backward erosion (piping) beneath the dam could enlarge into a channel and ultimately undermine and fail the structure. Standard remedies are a downstream loaded/graded filter, a longer or deeper cutoff, and relief wells.
Question 4 — final results
Quantity
Value
(i) Seepage, q
3.63 m³/day per m of dam
(ii) Pore pressure at A, uA
117.8 kPa
(ii) Effective stress at A, σ′A
85.9 kPa
(iii) Critical gradient, icr
1.04
(iii) Exit gradient, iexit (FS)
≈ 0.17 (FS ≈ 6.2)
Check (hand-drawn flow net, ±1 field): q depends only on H, Nf, Nd and is firm. The effective stress at A depends on reading A as ≈3 drops from the downstream exit; at ±1 drop the answer ranges roughly 80–92 kPa. The exit-field length l ≈ 2 m is likewise scaled from the net.
