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07-Str-A3 · May 2014

Question 1 of 6: Multiple Choice with Justification

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

Notes on this paper

Paper format: PEO / Engineers Canada National Examinations, May 2014, 07-Str-A3 Geotechnical Materials and Analysis. Three hours, closed book, drawing instruments required, one approved Casio or Sharp calculator. All charts and equations are supplied at the back of the paper (the m–n influence chart, a Newmark chart with $I_N = 0.005$, and a two-page formula sheet). Total value 100 marks over six compulsory questions (20 + 10 + 10 + 20 + 20 + 20). Every question and every sub-part is solved in full below.

Reference texts: Das, Principles of Geotechnical Engineering, 9th ed. (Ch. 2 grain size and gradation, Ch. 6 compaction, Ch. 7 permeability, Ch. 8 seepage and flow nets, Ch. 10 stresses in a soil mass, Ch. 11 consolidation, Ch. 12 shear strength); Knappett & Craig, Craig’s Soil Mechanics, 8th ed. (Ch. 2 seepage and flow-net construction, Ch. 5 shear strength and stress-path plots, Ch. 11 lateral earth pressure); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed. (compacted-fill fabric, Proctor behaviour by USCS group); Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for earth pressures, seepage control and settlement; ASTM D698 / D1557 (Proctor compaction), ASTM D2435 (one-dimensional consolidation), ASTM D4767 (consolidated-undrained triaxial with pore-pressure measurement).

Check — one printing slip in the source, carried as stated. The Question 1 header reads “(4 × 5 = 20 marks)” but five statements (i)–(v) are printed. The solution treats the question as 5 × 4 = 20 marks and answers all five parts; the total of 100 marks over the six questions is unaffected.


Question 1: Multiple Choice with Justification (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.

Given. Five short statements covering gradation and shear strength, gradation and permeability, the Proctor behaviour of three USCS groups, the effective friction angles of four soil types, and the fabric produced by compaction on the wet side of optimum.

Find. The correct selection in each case together with the soil-mechanics reasoning that decides it — the marks here are carried entirely by the justification, not by the choice.

Summary of answers
PartQuestion (abbreviated)Answer
(i)Greater shear strength: Sand A ($C_u = 4$) or Sand B ($C_u = 1$)Sand A
(ii)Lower coefficient of permeability: Sand A, B or CSand C
(iii)Higher dry density and higher optimum moisture content: GW, ML or CHNo single soil — GW gives the higher $\gamma_{d,\max}$, CH the higher OMC
(iv)Higher $\phi'$ from a CD triaxial test: expansive clay, glacial till, silt or sandSoil B — glacial till
(v)Fabric of a clay compacted wet of optimumDispersed (oriented)

(i) Sand A, with $C_u = 4$

The coefficient of uniformity $C_u = D_{60}/D_{10}$ measures the spread of grain sizes. Sand B with $C_u = 1$ is the limiting case of a perfectly uniform soil: every grain is the same size, so the packing is that of equal spheres and no smaller particle is available to sit in the void between three larger ones. Sand A at $C_u = 4$ carries a range of sizes, the finer fraction occupies part of the void space, and at the same compactive effort the soil reaches a markedly lower void ratio.

Shear resistance in a cohesionless soil has two components. The first is sliding friction between mineral surfaces, which depends on mineralogy and surface roughness and is essentially identical for two quartz sands. The second is the work that must be done to lift grains over one another before they can slide — interlocking and dilatancy — and that component grows rapidly as the void ratio falls. Writing the peak angle as $\phi_{\text{peak}} = \phi_{cv} + \psi$, where $\phi_{cv}$ is the constant-volume angle and $\psi$ the dilatancy angle, the two sands share $\phi_{cv}$ but the well-graded Sand A develops the larger $\psi$. In practice a dense well-graded sand reaches $\phi_{\text{peak}} \approx 38^\circ$ to $45^\circ$, while a uniform sand at comparable effort rarely exceeds $32^\circ$ to $36^\circ$. Sand A is therefore the stronger soil in shear.

(ii) Sand C

Permeability in a granular soil is controlled not by the average grain size but by the size of the narrowest pore throats along the flow path, and those throats are set by the finest fraction present. Hazen’s relation $k \approx C\,D_{10}^{2}$ and the more general Kozeny–Carman expression, in which $k$ varies as $e^{3}/(1+e)$ divided by the square of the specific surface, both say the same thing: fines and a low void ratio each drive $k$ down.

Sand A is a uniform coarse sand — large grains, large voids, large $D_{10}$ — and is by far the most permeable of the three. Sand B is gap-graded: it has coarse grains and fine grains but nothing in between, so the fines are too few to fill the coarse skeleton completely and a network of open voids survives. Sand C is continuously (well) graded from coarse down to very fine, so each size fraction occupies the voids left by the fraction above it. That gives Sand C both the smallest $D_{10}$ and the lowest void ratio, and therefore the lowest coefficient of permeability. This is exactly why a well-graded material is specified for a compacted core or a low-permeability liner, and why filter and drainage layers are deliberately specified as uniform.

(iii) GW gives the higher dry density, CH the higher optimum moisture content — no one soil gives both

Maximum dry density and optimum moisture content are inversely related, so the question as posed has no single-soil answer, and saying so is the answer. Water in a Proctor test acts as a lubricant that lets particles slide into a denser arrangement; the finer and more plastic the soil, the more water its specific surface must first satisfy before any is left to lubricate, and the more water is held in the compacted mass at peak density. Consequently the soils that need the most water reach the lowest dry density.

Taking standard-Proctor values typical of Canadian practice: a well-graded gravel GW reaches $\gamma_{d,\max} \approx 20$ to $22\ \text{kN/m}^3$ at an optimum moisture content of roughly 7 % to 11 %; a low-plasticity silt ML reaches about $17$ to $18\ \text{kN/m}^3$ at 14 % to 18 %; and a high-plasticity clay CH reaches only $13$ to $16\ \text{kN/m}^3$ at 22 % to 30 %. GW therefore has the highest dry density and the lowest optimum moisture content, CH the reverse, and ML sits between them. The correct examination answer is to name GW for dry density, CH for optimum moisture content, and to state that no soil can be highest in both because the two quantities move in opposite directions along the USCS sequence.

moulding water content, w (%) dry unit weight, γd OMC dry of optimum wet of optimum flocculated dispersed Flocculated (edge-to-face) Dispersed (face-to-face) Compacting wet of optimum leaves enough pore water to let the double layers expand, so the platelets slide parallel.
Figure 1.1 — A standard-Proctor curve, showing why the fabric question in part (v) and the density question in part (iii) are two views of the same curve. Compacting dry of optimum leaves the clay platelets flocculated; compacting wet of optimum leaves them dispersed.

(iv) Soil B — glacial till

In a consolidated-drained triaxial test the pore pressure stays at its back-pressure value throughout, so the measured strength is a genuine effective-stress strength and $\phi'$ reflects gradation, angularity, density and mineralogy alone. Ranking the four soils on those grounds:

The general rule the examiner is testing is that $\phi'$ falls as particles become smaller, rounder, more uniform and more platy, and glacial till is the only one of the four candidates that scores well on every count.

(v) A dispersed (oriented) structure

The fabric of a compacted clay is decided by the electrical forces between platelets at the moment of compaction, and those forces depend on how thick the diffuse double layer is. Compacting dry of optimum leaves the pore water in short supply, the double layers are suppressed, inter-particle attraction at the platelet edges dominates, and the compactive effort cannot overcome it — the particles lock into the random edge-to-face “card house” of Figure 2(A), the flocculated structure.

Compacting wet of optimum supplies enough water for the double layers to expand fully. Repulsion between the negatively charged platelet faces then dominates, inter-particle contact is lubricated, and the shear strains imposed by the roller or rammer are free to rotate the platelets into parallel alignment — the face-to-face arrangement of Figure 2(B), the dispersed or oriented structure. That is the answer to part (v).

The distinction matters in design as well as in theory. A clay compacted wet of optimum is less permeable (the parallel platelets close off the flow paths), more ductile, less prone to swelling, and less brittle on shearing, which is why liners and dam cores are specified 1 % to 3 % wet of optimum. A clay compacted dry of optimum is stiffer and stronger at small strain but is more permeable, more brittle and much more swell-prone on later wetting.

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