07-Str-A3 · May 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. Engineers Canada / PEO National Examinations, May 2016 — 07-Str-A3 Geotechnical Materials and Analysis. Closed book, three hours, 100 marks, one approved Casio or Sharp calculator, drawing instruments required. All six questions are compulsory and are weighted 20 / 10 / 10 / 20 / 20 / 20. The paper carries its own appendix (pages 8–11): a formula sheet, the rectangular-loading m–n influence chart and a Newmark influence chart whose influence value is printed on the formula sheet as $\sigma_z = 0.005\,N q$, i.e. $I_N = 0.005$ over 200 elements. Values quoted below are taken from those sheets.
Reference texts.
Check — three readings of the printed paper, carried as stated.
(1) Question 1 is headed “(4 x 5 = 20 marks)” but prints five lettered parts (i)–(v). All five are answered and the header total of 20 marks is kept as printed; the mismatch is a printing slip in the source, not a missing part.
(2) Figure 1 (part (ii)) prints no dimension for the depth of the permeable stratum. The dam base is dimensioned $L = 15$ m and the section is drawn to a single horizontal scale, so the stratum has been scaled off the drawing at $T \approx 5.6$ m ($T/L = 0.37$). The answer to (ii) does not depend on that number — it is read from the printed flow net — but the independent numerical check below does, and the sensitivity is stated where it is used.
(3) In Figure 4 the stem of Question 5(b) says point $A$ is at the “back of the piling”, while the dot is drawn on the upstream (headwater) face of the sheet pile, 1 m below ground. The drawn face is answered, and the value on the downstream face at the same elevation is reported alongside it.
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.
The marks on this question are carried by the justification, not by the letter. Each part is answered T or F and then defended from a definition or from a calculation.
The sentence describes a well-graded soil, not a uniformly graded one. The word “uniform” in the phrase refers to the uniformity of the particle sizes themselves — the grains are all much the same size — and not to the uniformity with which the size range is represented. A uniformly graded soil therefore plots as a steep, almost vertical grading curve spanning a narrow band of sizes, and is classified as poorly graded (SP or GP) in the Unified system.
The distinction is made quantitatively by the coefficients read off the grading curve:
$$C_u = \frac{D_{60}}{D_{10}}, \qquad C_c = \frac{(D_{30})^2}{D_{60}\,D_{10}}.$$A sand is well graded when $C_u > 6$ and $1 \le C_c \le 3$; a gravel when $C_u > 4$ over the same $C_c$ range. A uniformly graded soil has $C_u$ approaching unity. The engineering consequences follow directly from the shape: the uniform soil has no fines to fill the voids between its larger grains, so it reaches a low maximum dry density, is easily compacted to a loose state, and is comparatively permeable; the well-graded soil packs the smaller grains into the voids of the larger, giving a high maximum dry density and a lower permeability.
This part is arithmetic, not opinion, and the figure supplies everything needed. The lower panel of Figure 1 carries the flow net; the printed caption gives the equation to use.
Given.
| Quantity | Symbol | Value |
|---|---|---|
| Coefficient of permeability | $k$ | 1 m/day |
| Head across the dam | $h_w$ | 6 m |
| Base width of the dam | $L$ | 15 m |
| Flow channels counted on the printed net | $N_f$ | 3 |
| Potential drops counted on the printed net | $N_d$ | 11 |
| Depth of permeable stratum (read from the printed figure, see check note 2) | $T$ | ≈ 5.6 m |
Find. The seepage per metre width, and whether it equals 3 m3/day per m.
[Figure not reproduced: Figure 1 redrawn to the scale of the printed section: the flow net beneath the 15 m dam base. The lines are contours of a direct finite-difference solution of Laplace’s equation on the drawn geometry, at the eleven equal potential levels and the two interior stream-function levels that the pri. See the official exam paper.]
Approach. Count the net, substitute into the shape-factor form of Darcy’s law printed on the figure, and then confirm the count by solving Laplace’s equation on the drawn geometry independently of any counting.
| Result | Value |
|---|---|
| Flow net read from Figure 1 | $N_f = 3$, $N_d = 11$ |
| Seepage from the net | $q = 1.64$ m3/day per m |
| Seepage from a direct solution of Laplace’s equation | $q = 1.68$ m3/day per m |
| Value stated in the question | 3 m3/day per m |
| Answer | FALSE |
Below its preconsolidation pressure the clay is in the over-consolidated range: it is denser than it would be at that stress on the virgin line, and when it is sheared it tends to dilate. If drainage is prevented, the volume cannot increase, so the pore water is put into tension and the pore-water pressure falls — often below zero. Skempton’s expression makes the mechanism explicit:
$$\Delta u = B\big[\Delta\sigma_3 + A(\Delta\sigma_1 - \Delta\sigma_3)\big].$$For a saturated soil $B = 1$. The pore-pressure parameter $A$ at failure is about $0.5$ to $1.0$ for a normally consolidated clay, falls towards zero as the over-consolidation ratio rises, and is typically $-0.5$ to $0$ for a heavily over-consolidated clay. With $A$ negative the deviatoric term drives $\Delta u$ negative, and for a lightly loaded stiff clay the total pore pressure can go into suction. The same suction is what gives a stiff fissured clay its high short-term (undrained) strength and its slow, delayed loss of strength as the negative pore pressure equalises — the classic mechanism of delayed failure in cuts in London Clay and in Canadian glaciolacustrine clays.
Check — the caveat that keeps the statement true. The negative pore pressure comes from the deviatoric (shear) part of the loading. Under purely isotropic or one-dimensional compression, with $\Delta\sigma_1 = \Delta\sigma_3$, the equation gives $\Delta u = \Delta\sigma_3 > 0$ regardless of stress history. The statement is read here in the sense the paper intends — shear-induced pore pressure in the over-consolidated range — and is TRUE in that sense; a complete answer says so.
Compaction densifies a soil by expelling air, and no amount of mechanical energy expels the last of it: a fraction of the air becomes occluded in bubbles that the compactive effort cannot drive out. Full saturation is therefore an asymptote rather than an attainable state, and the compaction curve of any soil — sand or clay — lies wholly below the zero-air-voids line.
That line is the locus of dry unit weight at $S = 1$ and follows from the phase identity $wG_s = Se$ printed on the paper’s formula sheet:
$$\gamma_{d,\text{ZAV}} = \frac{G_s\gamma_w}{1 + wG_s}.$$At the optimum moisture content the degree of saturation is typically 80–90 %, with about 4–6 % air voids remaining; on the wet side of optimum the curve approaches the ZAV line and runs roughly parallel to it, but never touches it. The statement is therefore true, and it is true for both soil types — a clean sand compacted at its optimum retains air in the same way a clay does, even though its curve is flatter and its optimum less pronounced.
The requirement for an effective-stress (drained) test is that the pore-water pressure be zero, or known, throughout shearing, so that the applied total stresses are also the effective stresses. Both pieces of equipment can meet it.
In the consolidated drained triaxial test the specimen is back-pressure saturated, consolidated under the cell pressure with drainage open, then sheared with the drainage leads still open and slowly enough that no excess pore pressure builds. The cell pressure is $\sigma_3'$ and the deviator stress at failure gives $\sigma_1'$ directly; a set of tests at different cell pressures defines $c'$ and $\phi'$ from the tangent to the Mohr circles — which is exactly what Question 6 of this paper asks for.
In the direct shear (shear box) test the specimen is thin and sits between porous stones, so drainage is never restricted; the test is inherently drained provided the rate of displacement is slow enough. For a sand any ordinary rate is drained, and the shear box is in fact the standard way of obtaining $\phi'$ for a granular soil. For a clay the rate must be made very slow — the time to failure is commonly taken as $t_f \approx 12.7\,t_{100}$ after Gibson and Henkel — but the test is entirely capable of giving $c'$ and $\phi'$, and it is the usual test for residual strength when reversed repeatedly.
The shear box has real limitations — the failure plane is imposed rather than chosen by the soil, the stress distribution on it is not uniform, and pore pressures cannot be measured, so it cannot be used for undrained (CU) testing with pore-pressure measurement. None of these contradicts the statement, which is about capability under drained conditions.
