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16-Civ-A4 Geotechnical Materials and Analysis · May 2017

Question 2 of 5: Effective Stress under a Dam Spillway (Flow Net)

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

Notes on this paper

Paper format: PEO/EGBC National Examination 16-Civ-A4 — Geotechnical Materials and Analysis, May 2017. Closed book, 3 hours, 100 marks. Answer ALL questions (5 × 20 marks). All required charts and equations were supplied at the back of the paper.

Reference texts: Das, B.M. & Sobhan, K., Principles of Geotechnical Engineering, 9th ed. (Cengage); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed.; Knappett & Craig, Craig’s Soil Mechanics, 8th ed.; Budhu, Soil Mechanics and Foundations, 3rd ed.

Check (source figures): Q2’s flow net is a hand-drawn sketch; the counts used (Nd = 10, one drop to B and nine to A) were taken from the printed net and carry the usual ±1-field tolerance. Void ratio (e0), Cc and the preconsolidation break in Q4’s Figure 4(a), and the OCR–Af read in Q3’s Figure 3, are read off hand-plotted curves. Conclusions are robust to these reading tolerances; each is flagged where it bites.

Question 2: Effective Stress under a Dam Spillway (Flow Net) (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. A concrete spillway with a central sheet-pile cut-off founded on a saturated soil, $\gamma_{sat}=20\ \text{kN/m}^3$. From Figure 2, taking the downstream ground/tailwater surface as datum (elevation $0$): upstream reservoir water level $= +5.00\ \text{m}$; points A (downstream side) and B (upstream side) both lie $5\ \text{m}$ below datum, i.e. at elevation $-5.0\ \text{m}$. A flow net is drawn beneath the structure.

QuantityValue
Upstream reservoir WL (above datum)+5.00 m
Downstream tailwater WL0 (datum)
Elevation of A and B−5.0 m
γsat / γw20 / 9.81 kN/m3

Find. The vertical effective stress $\sigma'$ at A and at B, accounting for the seepage-modified pore pressures read from the flow net.

[Figure not reproduced: Figure 2 from the exam: spillway flow net with points A and B. See the official exam paper or the cited reference text.]

Figure 2 (exam flow net): reservoir on the right, 5.00 m above ground; tailwater at ground level on the left (datum). Flow lines: the structure base and pile, two interior U-shaped lines, and the impermeable base, so Nf = 3. There are nine interior equipotentials, so Nd = 10. The line through B is the first equipotential in from the upstream entry; the line through A, which meets the downstream end of the apron, is the last one before the exit.

Approach. Effective stress $\sigma' = \sigma - u$. Compute the total stress $\sigma$ from the overburden (water + saturated soil) above each point, and the pore pressure $u$ from the flow net: $u=\gamma_w\,h_p$, where the pressure head $h_p = h_t - z$, the total head $h_t$ comes from the number of potential drops to the point, and $z$ is the elevation head.

  1. Total head loss. Head drives from the upstream reservoir to the downstream tailwater: $$H = 5.00 - 0 = 5.00\ \text{m}.$$
  2. Potential drop per field. Counting the printed net gives nine interior equipotentials, so $N_d = 10$ (and $N_f = 3$): $$\Delta h = \frac{H}{N_d}=\frac{5.00}{10}=0.50\ \text{m}.$$
  3. Total head at A and B. Referencing total head to the downstream water level, B lies on the first equipotential in from the upstream entry (1 drop lost) and A on the last equipotential before the downstream exit (9 drops lost, 1 drop still to go): $$h_B = H - 1\,\Delta h = 5.00-0.50 = 4.50\ \text{m}, \qquad h_A = H - 9\,\Delta h = 0.50\ \text{m}.$$
  4. Pore pressures. Both points are at $z=-5.0\ \text{m}$, so $h_p = h_t - z = h_t + 5.0$: $$u_A = \gamma_w(0.50+5.0)=9.81(5.50)=54.0\ \text{kPa},$$ $$u_B = \gamma_w(4.50+5.0)=9.81(9.50)=93.2\ \text{kPa}.$$
  5. Total vertical stresses. Above A (downstream) is $5\ \text{m}$ of saturated soil with the tailwater at ground level (no free-water surcharge); above B (upstream) is $5\ \text{m}$ of saturated soil plus the $5.00\ \text{m}$ reservoir column: $$\sigma_A = \gamma_{sat}(5)=100\ \text{kPa}, \qquad \sigma_B = \gamma_w(5.00)+\gamma_{sat}(5)=49.1+100=149.1\ \text{kPa}.$$
  6. Effective stresses. $$\boxed{\sigma'_A = 100 - 54.0 = 46.0\ \text{kPa}} \qquad \boxed{\sigma'_B = 149.1 - 93.2 = 55.9\ \text{kPa}}$$

The results are physically consistent: at B the seepage is directed downward (entering the ground), which reduces pore pressure below hydrostatic and raises $\sigma'$ above the no-flow value ($149.1-98.1=51.0$ kPa); at A the seepage is directed upward toward the exit, which raises pore pressure and lowers $\sigma'$ below the no-flow value. The downstream side is therefore the critical location for piping/heave.

Pointσ (kPa)u (kPa)σ′ (kPa)
A (downstream)100.054.046.0
B (upstream)149.193.255.9
Check (flow-net tolerance): $N_d$ and the drop counts to A and B are read from a hand-sketched net and carry ±1 field ($\pm 0.5$ m head, i.e. $\pm 5$ kPa in $u$). The method and the qualitative result — higher $\sigma'$ on the upstream (down-flow) side, lower on the downstream (up-flow) side — are unaffected; state the assumed net counts, as done here. The no-flow value is $20(5)-9.81(5)=51.0$ kPa at both points, and the answers bracket it as they should: 46.0 kPa on the up-flow side and 55.9 kPa on the down-flow side.