07-Str-A3 · Undated paper

Question 2 of 6: Five statements, each to be justified

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

Notes on this paper

Paper format. Engineers Canada / PEO National Examinations, May 2019 — 07-Str-A3 Geotechnical Materials and Analysis. Closed book, three hours, 100 marks, one approved Casio or Sharp calculator. All six questions are compulsory and are weighted 10 / 10 / 15 / 20 / 15 / 30. The paper carries its own appendix: two sheets of blank semi-logarithmic and squared graph paper (pages 7–9) for the plotted answers, the rectangular-loading $m$–$n$ influence chart (page 10), a Newmark influence chart (page 11) and a two-page formula sheet (pages 12–13) on which the Newmark influence value is printed as $\sigma_z = 0.005\,N q$, i.e. $I_N = 0.005$ over 200 elements. Every chart value quoted below is taken from those appendix sheets.

Sitting. The printed cover reads NATIONAL EXAMINATIONS – May 2019, 07-Str-A3 Geotechnical Materials and Analysis, 3 hours duration, and every interior page repeats it. The sitting is May 2019.

Reference texts.

B. M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. — Ch. 2–3 (grain size and phase relations), Ch. 6 (compaction), Ch. 7–8 (permeability and seepage, flow nets), Ch. 9 (stresses in a soil mass, the $m$–$n$ chart and Newmark’s chart), Ch. 11 (consolidation, Casagrande’s construction), Ch. 12 (shear strength and the triaxial test).
R. F. Craig / J. A. Knappett, Craig’s Soil Mechanics, 8th ed. — Ch. 2 (seepage and flow nets), Ch. 3 (effective stress), Ch. 4 (consolidation), Ch. 5 (shear strength, the stress-path or “modified” failure envelope and the choice of test), Ch. 6 (stress distribution).
M. E. Harr, Groundwater and Seepage — Ch. 4 (the conformal solution for a single sheet pile in a stratum of finite depth, used here to audit the drawn flow net).
Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — Ch. 4 (site investigation and sampling), Ch. 8 (settlement), Ch. 11 (laboratory strength testing). The Canadian reference for practice and terminology.
K. Terzaghi, R. B. Peck & G. Mesri, Soil Mechanics in Engineering Practice, 3rd ed. — Art. 14 (one-dimensional consolidation theory and its assumptions), Art. 16–17 (seepage and piping), Art. 19–20 (shear strength).

Check — two readings of the printed paper, carried as stated.

(1) The printed paper has exactly six questions worth 10 + 10 + 15 + 20 + 15 + 30 = 100 marks, and that numbering is used here.

(2) Question 4 lists the void ratio at $\sigma' = 500$ kPa as $e = 0.925$. Taken with the $400$ kPa point that implies $C_c = 0.57$ over the last increment, against $0.27$–$0.30$ over every earlier virgin increment — the last row is inconsistent with the rest of the test. It is carried exactly as printed, plotted, and excluded from the straight-line fit for $C_c$, with the effect of including it stated where it matters.

Question 2: Five statements, each to be justified (10 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.

Answers. Two marks each — one for the choice, one for the reason.

1 — (ii) Sample B, the dense sand. The stem and the options were printed from two different versions of this item: the stem names an overconsolidated and a normally consolidated clay, while the two options offer a loose and a dense sand. Answered on the options as printed, the dense sand carries the greater peak strength. The physics is the same under either reading and so is the mark: in a drained test, the material that sits denser than its critical state must dilate in order to shear, the dilation absorbs work, and the measured peak strength rises above the critical-state value. Dense sand and overconsolidated clay are the two members of that family, loose sand and normally consolidated clay of the other. So under the option wording B (dense) wins, and under the stem wording A (overconsolidated) wins — in both cases for the same reason.

2 — (iii) Both of them will be the same. An overconsolidated clay with an OCR of 1 is a contradiction in terms: $OCR = \sigma'_c / \sigma'_0 = 1$ means the present effective stress is the largest the soil has ever carried, which is the definition of normally consolidated. Samples C and D are therefore the same material in the same state and give the same peak strength.

3 — Sample F, the one tested CD, has the higher shear strength. Both specimens are normally consolidated, so both want to contract when sheared. The drained specimen is allowed to do so and its effective stress stays equal to the applied cell pressure throughout; the undrained specimen cannot, so the contractive tendency appears instead as a positive excess pore pressure that reduces the effective confining stress at failure. Since strength is controlled by effective stress ($\tau_f = c' + \sigma' \tan \phi'$), the CD specimen fails at the higher deviator stress. Question 6 of this paper shows the same effect numerically: specimen A reaches failure with 82 kPa of positive pore pressure, so its effective confining stress at failure is 68 kPa rather than the applied 150 kPa.

4 — (i) Increases. A dense sand sheared drained must dilate: the grains are interlocked and cannot move past one another without riding up over their neighbours, so the specimen expands. Being a drained test, that volume change is free to occur and water is drawn in.

5 — (ii) Atmospheric pressure, i.e. the pore-water pressure at C is zero. Point C is the point at which the phreatic surface (the uppermost flow line, BC) meets the horizontal toe drain. The phreatic surface is by definition the locus on which $u = 0$, and the drain surface CD is a free-discharge boundary printed on the figure as $h_w = 0$. C lies on both, so $u_C = 0$ from either boundary condition. Below C the pore pressure in the dam is positive; above the phreatic surface, in the capillary zone, it is negative.

[Figure not reproduced: Figure 1 redrawn: the seepage regime in a homogeneous earth dam with a horizontal toe drain. C is where the phreatic surface BC lands on the drain CD, so it lies on the $u = 0$ line and on the $h_w = 0$ boundary at the same time. See the official exam paper.]

Topic. Dilatancy and the critical state; what OCR actually measures; why a drained test on a normally consolidated clay is stronger than an undrained one; the phreatic surface as a zero-pressure boundary.

Key relations. $OCR = \sigma'_c / \sigma'_0$; $\tau_f = c' + \sigma' \tan \phi'$; $\sigma' = \sigma - u$; on a phreatic surface $u = 0$ so the total head equals the elevation head.

Why this works. Four of these five items are one idea seen from different angles: strength is a function of effective stress and of density relative to the critical state, and the drainage condition of the test decides whether a volumetric tendency shows up as a volume change or as a pore pressure. Item 5 is the seepage counterpart of the same bookkeeping — the free surface is where the water pressure has fallen to atmospheric, which is exactly why its position is unknown in advance and has to be found as part of the solution.

Common pitfall. Reading “overconsolidated clay with an OCR of 1” in item 2 as a genuinely overconsolidated soil and answering (ii); the phrase is the trap. In item 5 the usual error is to reason that C is at the base of a body of saturated soil and therefore must carry a positive pressure — it is at the base of the flow region, but it is also on the discharge face, and the discharge face is open to the atmosphere.

Source. Craig / Knappett, Craig’s Soil Mechanics, 8th ed., §5.4–5.6 (drained and undrained triaxial behaviour, dilatancy, critical state) and §2.6 (the phreatic surface and seepage through an earth dam); Das & Sobhan, 9th ed., §12.11–12.13 and §8.9.

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