07-Str-A3 · December 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format: PEO / Engineers Canada National Examinations, December 2014, 07-Str-A3 Geotechnical Materials and Analysis. Three hours, closed book, drawing instruments required, one approved Casio or Sharp calculator. All charts and equations are supplied at the back of the paper (the m–n influence chart on page 8, a Newmark chart with $I_N = 0.005$ on page 9, and a two-page formula sheet). Total value 100 marks over six compulsory questions (20 + 10 + 10 + 20 + 20 + 20). Every question and every sub-part is answered in full below.
Reference texts: Das, Principles of Geotechnical Engineering, 9th ed. (Ch. 6 compaction, Ch. 7 permeability, Ch. 8 seepage and flow nets, Ch. 10 stresses in a soil mass, Ch. 11 consolidation, Ch. 12 shear strength); Knappett & Craig, Craig’s Soil Mechanics, 8th ed. (Ch. 2 seepage and flow-net construction, Ch. 5 shear strength, Ch. 11 lateral earth pressure); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed. (compacted-fill fabric and Proctor behaviour by USCS group); Harr, Groundwater and Seepage, Ch. 4 (closed-form conformal solutions for a single cut-off); Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for earth pressures, seepage control and settlement; Canadian Dam Association, Dam Safety Guidelines (2013, rev. 2019); ASTM D698 / D1557 (Proctor compaction), ASTM D4767 (consolidated-undrained triaxial with pore-pressure measurement), ASTM D7181 (consolidated-drained triaxial).
Check — two printing slips in the source, carried exactly as printed. (1) The Question 1 header reads “(4 × 5 = 20 marks)” but five statements (i)–(v) are printed; this solution treats the question as 5 × 4 = 20 marks and answers all five parts. (2) Question 5(a) instructs the candidate to “draw on Figure 4”, while the only section supplied is labelled Figure 3 — the blank grid sheets on pages 6 and 7 are the intended drawing space. Neither slip changes the 100-mark total.
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.
Each part is a one-line answer supported by the physical mechanism; the mechanism is what carries the marks.
Optimum moisture content tracks specific surface. Water in a compacted soil has to do two jobs: satisfy the adsorbed double layer around each particle, and then lubricate the fabric so that grains can slide into a denser packing. Clay minerals are plate-shaped and enormously fine — kaolinite has a specific surface near $15\ \text{m}^2\text{g}^{-1}$ and montmorillonite can exceed $700\ \text{m}^2\text{g}^{-1}$, against well under $0.1\ \text{m}^2\text{g}^{-1}$ for sand. A far greater mass of water is therefore locked into double layers before any is free to lubricate, so the peak of the compaction curve is pushed to the right. Typical standard-Proctor values are 6–10 % for a clean sand, 12–18 % for a silt and 18–30 % for a plastic clay, and the maximum dry density falls in the same order because the clay fabric can never pack as tightly as a well-graded granular skeleton.
Permeability is controlled by the size of the pore throats, not by how much void space there is. Both the Hazen relation $k \approx C\,D_{10}^{2}$ and the Kozeny–Carman relation
$$k \;\propto\; \frac{D_s^{2}\,e^{3}}{1+e}$$scale with the square of a characteristic particle (hence pore) dimension, and that dimension changes by three or four orders of magnitude from sand to clay. Representative values are $10^{-2}$ to $10^{-5}\ \text{m}\,\text{s}^{-1}$ for sand, $10^{-5}$ to $10^{-9}\ \text{m}\,\text{s}^{-1}$ for silt and below $10^{-9}\ \text{m}\,\text{s}^{-1}$ for clay. The classic trap here is to argue from void ratio: a soft clay may have $e = 1.2$ against $e = 0.6$ for a dense sand, yet it is still a million times less permeable, because its pores are sub-micron and much of the pore water is held immobile in the double layer.
At a given compactive effort, a clay compacted dry of optimum develops a flocculated, randomly oriented fabric and retains a high matric suction. Negative pore-water pressure is an effective stress, so through $\tau_f = c' + (\sigma - u_w)\tan\phi'$ the as-compacted specimen is stiffer and stronger, and the unconfined compressive strength $q_u$ is highest. Compacted wet of optimum the platelets are dispersed into a parallel, oriented fabric, suction is largely destroyed, and $q_u$ falls. Two qualifications belong in the answer: the comparison holds only in the as-compacted, unsoaked condition — on soaking, the dry-of-optimum specimen loses much of that strength and may collapse or swell; and the dry-side material is brittle and prone to cracking. That is exactly why a clay core or a landfill liner is deliberately placed 1–3 % wet of optimum, trading strength for a ductile, low-permeability fabric.
On the $e$–$\log \sigma'$ plot a normally consolidated clay sits on the virgin compression line, whose slope is $C_c$. An over-consolidated clay loaded within its past maximum stress $\sigma'_p$ follows the much flatter recompression line of slope $C_r$, typically
$$C_r \;\approx\; \frac{C_c}{5} \ \ \text{to}\ \ \frac{C_c}{10}$$so for the same stress increment the NC clay compresses several times more. Strictly, $C_c$ is a property of the virgin line and is the same for a given clay whatever its current stress history; what the question is testing is the settlement response, and it is the NC clay that mobilises $C_c$ rather than $C_r$. Terzaghi and Peck’s correlation $C_c = 0.009\,(LL - 10)$ applies to the virgin line of a normally consolidated clay of moderate sensitivity.
For a normally consolidated clay the effective-stress failure envelope passes essentially through the origin, so $c' \approx 0$ and the strength is purely frictional. An over-consolidated clay tested at effective stresses below $\sigma'_p$ gives an envelope that is curved and, when a straight line is fitted over the working stress range, has a positive intercept — typically 5–25 kPa. The intercept is a consequence of the denser fabric, interparticle bonding and dilatancy inherited from unloading; it is a fitting parameter for a stress range, not a true bond strength, which is why the CFEM cautions against extrapolating it to low stresses. Note also that $c'$ is not the undrained strength $c_u$: an OC clay has both a higher $c'$ and a higher $c_u$, but for entirely different reasons.
| Part | Property compared | Answer | Controlling mechanism |
|---|---|---|---|
| (i) | Optimum moisture content | (C) clay | Specific surface and double-layer water |
| (ii) | Coefficient of permeability | (A) sand | Pore-throat size, $k \propto D^{2}$ |
| (iii) | Unconfined compressive strength | (C) dry of optimum | Flocculated fabric, high matric suction |
| (iv) | Compression index $C_c$ | (A) NC clay | Virgin line ($C_c$) versus recompression line ($C_r$) |
| (v) | Effective cohesion $c'$ | (B) OC clay | Envelope intercept from stress history and dilatancy |
