07-Str-A3 · May 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. Engineers Canada / PEO National Examinations, May 2015 — 07-Str-A3 Geotechnical Materials and Analysis. Closed book, three hours, 100 marks, approved Casio or Sharp calculator, drawing instruments required. All six questions are compulsory and are weighted 20 / 10 / 10 / 20 / 20 / 20. The paper supplies its own appendix: a formula sheet, the rectangular-loading m–n influence chart and a Newmark influence chart with influence value $I_N = 0.005$ (200 elements). Values quoted from those sheets are used here.
Reference texts.
Check — two printing slips in the source, carried as printed. (1) Question 1 is headed “(4 x 5 = 20 marks)” but prints five lettered parts (i)–(v). All five are answered below and the header total of 20 marks is kept; read the header as five parts at four marks each. (2) In Question 5 the surcharge and the two cohesion intercepts are printed with the units $\text{kN/m}^{3}$ — dimensionally these are stresses and must be read as $\text{kPa}$ ($\text{kN/m}^{2}$); the unit weights carry the $\text{kN/m}^{3}$ correctly. The solution states each value in the unit its symbol requires.
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 statement joins a false claim to a true one, so as printed it is false. Effective stress has a very definite physical meaning: it is that part of the total normal stress which is carried by the soil skeleton through the inter-particle contacts, and Terzaghi’s principle states that it — and only it — governs the two things an engineer cares about, namely shear strength and volume change. Two elements of soil at the same total stress but different pore pressure behave completely differently; two elements at the same effective stress behave identically. That is as physical as a stress measure can be.
What is correct is the second half. There is no instrument that reads effective stress. It is obtained indirectly from two quantities that can be measured — the total stress and the pore-water pressure — through the relation on the formula sheet,
$$\sigma' = \sigma - u .$$The total stress follows from the weight of overburden and the applied loading, and the pore-water pressure is read with a piezometer or, in the triaxial cell, with a pore-pressure transducer at the specimen base. The correct answer is therefore False, with the qualification that only the “cannot be directly measured” clause is true.
[Figure not reproduced: Figure 1 (redrawn). Sand A is uniform and rounded, Sand B is gap graded, Sand C is well graded with a continuous range of sizes. See the official exam paper.]
Figure 1 is a gradation sketch, not a strength test, so the answer must be argued from packing. Sand A is a uniform (poorly graded) assembly of equal-sized rounded grains; Sand B is gap graded, a few coarse grains together with a small quantity of much finer material and nothing in between; Sand C shows a continuous, well-graded distribution running from coarse down to fine. In a well-graded sand the smaller particles occupy the voids between the larger ones, so at the same compactive effort the soil reaches a markedly lower void ratio and a higher relative density. That produces more inter-particle contacts per unit volume, greater mechanical interlocking, and a larger dilatancy component when the sample is sheared, all of which raise the peak angle of internal friction. Uniform rounded sand, by contrast, has the fewest contacts and the least interlocking and gives the lowest $\phi'$; the gap-graded Sand B sits between the two because its fines are too few to fill the coarse voids properly. Sand C therefore has the highest $\phi'$.
Saturated permeability is controlled by the size of the pore channels through which water must travel, and pore size scales with the finer particles present. For clean coarse soils Hazen’s relation $k \approx C\,D_{10}^{2}$ makes the dependence explicit: $k$ goes as the square of the effective size, so an order-of-magnitude change in $D_{10}$ moves $k$ by two orders of magnitude. A well-graded gravel (GW) has $D_{10}$ in the millimetre range and typically $k \approx 10^{-2}$ to $10^{-1}\ \text{m/s}$. A low-plasticity silt (ML) falls to roughly $10^{-7}$ to $10^{-5}\ \text{m/s}$, and a high-plasticity clay (CH), whose flow paths are the tortuous gaps between platelets bound to adsorbed water, is below $10^{-9}\ \text{m/s}$. The answer is (a) GW, by some eight to ten orders of magnitude over CH. Being well graded, GW is a little less pervious than a uniform gravel of the same $D_{50}$ would be, but that refinement does not change the ranking.
The compression index $C_c$ is the slope of the virgin compression line on the $e - \log \sigma'$ plot,
$$C_c = \frac{e_0 - e_1}{\log\left(\dfrac{\sigma_1'}{\sigma_0'}\right)}, \qquad C_c \approx 0.009\,(LL - 10),$$the second form being the Skempton correlation printed on the paper’s formula sheet. A normally consolidated clay is already sitting on that virgin line, so any load increment moves it down the steep branch and the full $C_c$ applies. An over-consolidated clay has previously carried a larger effective stress; a load increment first travels along the much flatter recompression (swelling) branch, whose slope $C_r$ is typically only $0.1$ to $0.2$ of $C_c$, and the steep branch is not reached until the effective stress exceeds the preconsolidation pressure $\sigma_c'$. For the same stress increment the normally consolidated clay is therefore both more compressible and more settlement-prone. The answer is Soil A.
[Figure not reproduced: Figure 2 (redrawn). Flocculated (random, edge-to-face) fabric on the left; dispersed, parallel-oriented fabric on the right. See the official exam paper.]
Compacting a fine-grained clay dry of optimum gives the flocculated structure, sketch A of Figure 2, for two reinforcing reasons. The first is electrochemical. At low water content there is not enough pore water to develop the diffuse double layer around each clay platelet, so the double layers stay thin, the face-to-face repulsion between particles is suppressed, and the net inter-particle force is the edge-to-face electrical attraction between the positively charged particle edges and the negatively charged faces. The platelets consequently meet edge-to-face in a random, open, card-house arrangement.
The second reason is mechanical. Dry of optimum the soil is stiff and the water films do not lubricate the contacts, so the compactive effort simply cannot generate the shear strain needed to slide the particles past one another into a parallel alignment. Add water and both effects reverse: wet of optimum the double layers expand, repulsion dominates, the soil is soft enough for the roller to shear it, and the platelets slide into the parallel dispersed-oriented fabric of sketch B.
The distinction matters in practice because the two fabrics behave differently at the same dry density. A dry-of-optimum, flocculated fill is stiffer and stronger at small strain, more permeable, more brittle, and has greater swell potential on wetting — which is why it is preferred for a structural fill but avoided for a clay liner or the core of an earth dam, where the low permeability and ductility of a wet-of-optimum, dispersed fabric are wanted instead.
