07-Str-A3 · Undated paper
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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.
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 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. One mark each; the reasoning that earns the mark is given underneath.
| Item | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Answer | A | B | B | A | C | A | B | A and C | B and C | A |
1 — (A) $2 \times 10^{-6}$ m. The Unified / CSA grain-size boundary between silt and clay sizes is 2 µm; $75$ µm is the sand–silt boundary (the No. 200 sieve), so (B) is the distractor for the wrong boundary.
2 — (B) Plastic limit. For clays compacted with standard effort the optimum water content falls close to, and usually a little below, the plastic limit; the liquid limit is far wetter than any compactible state and the shrinkage limit far drier.
3 — (B) Permeability. Compaction closes voids and destroys the continuous flow paths, so $k$ falls — by one to three orders of magnitude for a clay compacted wet of optimum. Strength and bearing capacity both increase, which is the whole purpose of compacting.
4 — (A) Falling head test. With $k$ of the order of $10^{-7}$–$10^{-9}$ cm/s the discharge in a constant-head cell is too small to measure; the falling-head cell instead measures the time for a standpipe to drain, which is long and easily timed. (The piezometer or field permeability test is an in-situ method, not a laboratory determination of $k$.)
5 — (C) Expulsion of water from within the voids. This is the definition of primary consolidation. Both water and the mineral grains are effectively incompressible at engineering stress levels, so the volume change must come from the voids losing water.
6 — (A) True. Both are “dense” relative to their critical state: both show a stiff initial response, a distinct peak, dilation during shear and then softening toward a critical-state strength. A normally consolidated clay and a loose sand are the contractive, no-peak pair.
7 — (B) False. The word that makes the statement wrong is only. A CU test with pore-pressure measurement gives $c'$ and $\phi'$ from the effective-stress circles and $c_{cu}$ and $\phi_{cu}$ from the total-stress circles — both sets come out of the same test, as Question 6 of this very paper demonstrates.
8 — (A) and (C) are incorrect. (A) is wrong because an overconsolidated clay is precisely the case that does show a real effective cohesion intercept, $c' > 0$; the normally consolidated clay is the one whose envelope passes through the origin. (C) is wrong because a saturated specimen sheared undrained cannot change volume at all — there is no drainage path and water is incompressible; the tendency to contract shows up as positive pore pressure, not as volume reduction. (B) is correct (dilatant clays generate negative excess pore pressure) and (D) is correct (drained parameters govern long-term, fully-dissipated stability).
9 — (B) and (C) are NOT assumptions. Terzaghi’s one-dimensional theory assumes $m_v$ and $k$ are constant over the stress increment considered — that constancy is what makes $c_v = k/(m_v \gamma_w)$ a constant and the governing equation linear — so (B) states the opposite of the assumption. (C) is not an assumption either: the theory assumes flow and strain are one-dimensional and vertical, but it is indifferent to whether the layer is singly or doubly drained; that choice only sets the drainage path length $d$ in $T_v = c_v t / d^2$. (A), (D) and (E) are all genuine assumptions of the theory.
10 — (A) Consolidated drained triaxial test. It is the only drained test in the list: the vane shear test and the unconfined compression test are both rapid, undrained tests that measure $s_u$.
