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07-Str-A3 · December 2015

Question 1 of 6: Five statements with justification

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

Notes on this paper

Paper format. Engineers Canada / PEO National Examinations, December 2015 — 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: a formula sheet, the rectangular-loading m–n influence chart and a Newmark influence chart whose influence value is printed as $I_N = 0.005$ (200 elements). Values quoted below are taken from those sheets.

Reference texts.

  • B. M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. — Ch. 6 (compaction), Ch. 7 (permeability), Ch. 9 (stresses in a soil mass), Ch. 10 (consolidation), Ch. 12 (shear strength).
  • R. F. Craig / J. A. Knappett, Craig’s Soil Mechanics, 8th ed. — Ch. 3 (effective stress, artesian profiles), Ch. 4 (pore pressure coefficients $A$ and $B$), Ch. 5 (shear strength and the choice of test), Ch. 6 (stress distribution).
  • Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — Ch. 4 (site investigation and sampling categories), Ch. 6 (excavations and base stability). The Canadian reference for practice, sampling classes and terminology.
  • K. Terzaghi, R. B. Peck & G. Mesri, Soil Mechanics in Engineering Practice, 3rd ed. — Art. 15–17 (consolidation and seepage), Art. 19–20 (shear strength of clays).

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. (2) The section figure for Question 5 is captioned “Figure 2” on page 4 although the stem of Question 5 calls it Figure 3 — the same drawing, mislabelled in the source. (3) On that figure the “3.0 m” dimension is drawn with its arrowhead at the top of the sand, but at the figure’s own vertical scale (calibrated on the 2 m of clay left beneath the cut) the dimensioned distance is $3.1$ m from the piezometric surface down to the base of the cut, and $5.1$ m down to the sand. The piezometric surface is therefore taken as $3.0$ m above the base of the cut, i.e. $5.0$ m above point $A$; this is also the only reading for which a positive depth of water $h$ exists, and it reproduces the depth of water drawn in the figure. The alternative reading is worked through and dismissed in Question 5.

Question 1: Five statements with justification (4 x 5 = 20 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.

Each part is worth four marks, and on this paper the marks are carried entirely by the justification. The figure below reproduces the two grain assemblies drawn in the source and adds the grading curves they correspond to, since part (ii) is answered from the shape of the gradation rather than from the size of the largest grain.

[Figure not reproduced: Figure 1 as printed: Sand A carries a coarse skeleton and a separate group of fines with a size gap between them; Sand B runs continuously from coarse to fine. The right-hand panel shows the grading curves the two sketches imply. See the official exam paper.]

(i) The correct statement is (a). The zero-air-voids line is the locus of dry density at complete saturation, $S = 1$, and follows directly from the phase relations on the paper’s formula sheet:

$$\gamma_{d,\text{ZAV}} = \frac{G_s\gamma_w}{1 + wG_s},$$

which follows from the phase identity $wG_s = Se$ with $S = 1$.

Only the specific gravity of solids and a chosen moisture content enter that expression, so the line can be plotted from a pycnometer test alone — no compaction test is needed. Because compaction never expels the last of the air, every real compaction point plots at $S < 1$ and therefore below the line; the curve approaches it asymptotically on the wet side of optimum but cannot touch or cross it. A test point that plots above the zero-air-voids line is not a soil property, it is an arithmetic or a specific-gravity error.

(ii) Sand A has the higher saturated coefficient of permeability. Both sketches contain grains of similar maximum size, so the answer cannot come from the coarse fraction. Sand A is gap graded: a coarse skeleton and a small, separate population of fines, with the intermediate sizes missing. Sand B is well graded, running continuously from coarse to fine, so its intermediate and fine particles pack into the voids between the coarse grains. The consequence is a lower void ratio and, more importantly, a smaller controlling pore throat and a smaller effective size $D_{10}$ in Sand B. Permeability scales with the square of that controlling dimension — Hazen’s relation $k \approx C\,D_{10}^{2}$ with $D_{10}$ in mm and $k$ in cm/s — and with the void ratio through the Kozeny–Carman form $k \propto e^{3}/(1+e)$. Sand A keeps large, open, continuous channels between the coarse grains because there is no intermediate fraction to plug them, so it is the more permeable of the two. This is exactly why a well-graded material is specified for a compacted core or a low-permeability liner and a uniform one for a filter or a drainage blanket.

(iii) The GW soil has the higher angle of internal friction. GW is a well-graded gravel: a coarse, hard, angular, granular material whose strength is generated by interparticle friction, interlocking and the dilation that has to occur before a dense assembly can shear. Typical values are $\phi' = 38^{\circ}$ to $45^{\circ}$, and the well-graded designation adds a few degrees over a uniform gravel because the finer sizes wedge the coarse grains apart. CH is a clay of high plasticity, dominated by platy minerals of the smectite family that align along the shear surface; its effective friction angle falls in the $15^{\circ}$ to $22^{\circ}$ range and its residual value can be under $10^{\circ}$. The ordering is a direct consequence of particle shape and mineralogy: rotund, angular, quartz-rich grains against aligned clay platelets separated by adsorbed water.

(iv) Soil B, the expansive clay, has the higher swelling index. The swelling index $C_s$ is the slope of the unloading (rebound) branch of the $e$ against $\log \sigma'$ plot, so it measures how much void ratio the soil recovers when the effective stress is reduced. An expansive clay is montmorillonitic; its 2:1 lattice has a weakly bonded interlayer that admits water, its specific surface is two orders of magnitude larger than that of a kaolinitic silty clay, and the double layer around each platelet expands as soon as the confining stress is relieved. A silty clay carries a significant non-plastic silt fraction whose grains simply do not swell, and its clay fraction is normally illitic or kaolinitic. Typical values are $C_s \approx 0.02$ to $0.05$ for the silty clay against $0.10$ to $0.25$ or more for the expansive clay, with the usual rule of thumb $C_s \approx C_c/5$ to $C_c/10$ in both cases. This is the parameter that governs heave in a shallow footing or a slab founded on Regina or Winnipeg clay.

(v) Positive shear-induced pore pressures are higher in (a), the normally consolidated clay. The sign and size of the pore pressure generated by shearing follow the volume change the soil would undergo if it were allowed to drain. A normally consolidated clay has never carried a higher effective stress, so its structure is loose relative to the critical state and it contracts on shearing; prevented from contracting by the undrained condition, the water carries the load and the pore pressure rises. A heavily over-consolidated clay is dense relative to critical state, tries to dilate, and generates a falling or negative pore pressure instead. In Skempton’s notation the pore pressure parameter at failure is $A_f \approx 0.5$ to $1.0$ for a normally consolidated clay and can fall to $-0.5$ for a heavily over-consolidated one. The practical corollary appears twice more in this paper: it is why the end of construction is critical for an embankment on soft clay (Question 2 and Question 6) and why the long-term case governs for a cut in a stiff over-consolidated clay.

PartAnswerGoverning reason
(i)Statement (a)$\gamma_d = G_s\gamma_w/(1+wG_s)$ needs only $G_s$; real points plot at $S < 1$
(ii)Sand AGap graded, so no intermediate sizes fill the voids: larger $D_{10}$ and larger $e$
(iii)GWAngular granular interlock and dilation, $\phi' \approx 38^{\circ}$–$45^{\circ}$ against $15^{\circ}$–$22^{\circ}$ for CH
(iv)Soil B (expansive clay)Montmorillonite interlayer and double-layer rebound, $C_s \approx 0.10$–$0.25$
(v)(a) Normally consolidatedContractive structure, $A_f \approx 0.5$–$1.0$; over-consolidated clay dilates
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