07-Str-A3 · May 2013
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
EGBC / Engineers Canada National Examination — Structural Engineering (legacy), 07-Str-A3 Geotechnical Materials and Analysis, May 2013. Three hours; closed book; one Casio or Sharp approved calculator; drawing instruments required; all charts and equations are supplied at the back of the paper. 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 distribution and classification, Ch. 6 compaction, Ch. 7 permeability, Ch. 8 seepage and flow nets, Ch. 9 in-situ stresses and capillary rise, Ch. 10 stresses in a soil mass, Ch. 11 consolidation, Ch. 12 shear strength); Knappett & Craig, Craig’s Soil Mechanics, 8th ed. (flow nets, effective stress, heave and piping); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed.; Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM), 4th ed. — the governing Canadian practice document for settlement, excavation dewatering and factors of safety; ASTM D2487 (USCS classification) and ASTM D6913 (grain-size analysis).
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 statements about compaction, plasticity, consolidation, shear-strength testing and permeability, each to be marked True or False and defended.
Find. The correct verdict on each statement, with the soil-mechanics reason that decides it.
| Part | Statement (abbreviated) | Answer |
|---|---|---|
| (i) | Sands: higher γd,max and lower OMC than clays | TRUE |
| (ii) | Ip(bentonite) > Ip(sandy clay) | TRUE |
| (iii) | Dam settles more on NC clay than on OC clay | TRUE |
| (iv) | φcu is always greater than φ′ | FALSE |
| (v) | The non-plastic soil (Ip = 0) has the highest k | TRUE |
Both halves of the statement are correct, and they are two faces of the same mechanism. A sand is coarse-grained, so its specific surface is small — of the order of $10^{-2}\ \text{m}^2/\text{g}$ against $10$ to $800\ \text{m}^2/\text{g}$ for a clay. Water in a compacting sand acts only as a short-lived lubricant that lets grains slide past one another into a denser packing, so very little of it is needed and the surplus simply drains or occupies void space that solids could have filled. A standard Proctor test on a clean to silty sand therefore peaks at about $\gamma_{d,\max} = 18$ to $20\ \text{kN/m}^3$ at an OMC of roughly 8 to 12 %, while a plastic clay peaks at about $14$ to $17\ \text{kN/m}^3$ at an OMC of 18 to 28 %. The clay is penalised twice: the adsorbed double-layer water is held so tightly that it behaves as part of the particle rather than as free pore fluid, and the plate-shaped particles cannot pack as efficiently as bulky rotund sand grains.
The two effects move the Proctor curve together, up and to the left for the sand, down and to the right for the clay, so a statement that reverses only one of them would be false. Note the qualifier “typically”: a very uniform fine sand can show a flat, almost bulking-dominated curve, and a well-graded gravelly clay can out-densify a poorly graded sand. The general trend quoted in the statement is nevertheless the one every earthworks specification is written around.
Bentonite is a rock whose clay fraction is dominated by sodium montmorillonite, a 2:1 expanding lattice mineral. Its unit layers are held together only by exchangeable cations and adsorbed water, so the mineral has a specific surface approaching $800\ \text{m}^2/\text{g}$ and a cation-exchange capacity of 80 to $150\ \text{meq}/100\ \text{g}$. Enormous quantities of water can be held in the diffuse double layer before the material begins to flow, which is exactly what the liquid limit measures. Typical index values are $LL = 300$ to $700\ \%$ and $PL = 50$ to $100\ \%$, so
$$I_p = LL - PL \approx 250\ \text{to}\ 600$$
A sandy clay, by contrast, is dominated by kaolinite or illite (specific surface 10 to $100\ \text{m}^2/\text{g}$, non-expanding) and carries a large inert sand fraction that contributes no plasticity at all but does contribute mass to the sample on which the limits are reported. Typical values are $LL \approx 30$ to $45\ \%$ and $PL \approx 15$ to $22\ \%$, so $I_p \approx 10$ to $25$. The two materials differ by more than an order of magnitude, which is why bentonite is the material of choice for slurry walls, drilling mud and landfill liners.
The two clays are the same thickness and carry the same stress increment from the same 5 m dam, so the comparison is purely one of compressibility. A normally-consolidated clay is already at its maximum past pressure, so the whole of the increment is taken along the virgin compression line and the settlement is governed by the compression index:
$$S_{c,\text{NC}} = \frac{C_c H}{1+e_0}\,\log\!\left(\frac{\sigma_0' + \Delta\sigma'}{\sigma_0'}\right)$$
An over-consolidated clay first has to travel back up the much flatter recompression (swelling) line, whose slope $C_s$ is characteristically only one fifth to one tenth of $C_c$. Provided the final stress stays below the pre-consolidation pressure the settlement is
$$S_{c,\text{OC}} = \frac{C_s H}{1+e_0}\,\log\!\left(\frac{\sigma_0' + \Delta\sigma'}{\sigma_0'}\right) \quad\Longrightarrow\quad \boxed{\;\frac{S_{c,\text{NC}}}{S_{c,\text{OC}}} = \frac{C_c}{C_s} \approx 5\ \text{to}\ 10\;}$$
If the dam load is large enough to push the over-consolidated clay past its pre-consolidation pressure the calculation becomes a two-part one — recompression up to $\sigma_c'$ then virgin compression beyond it — and the ratio narrows, but the normally-consolidated deposit still settles more. The statement is therefore true as written, and it is the reason a site investigation over soft ground always reports the over-consolidation ratio rather than just the strength.
The statement inverts the usual result. When a consolidated-undrained (CU) test is run without pore-pressure measurement the only stresses known are total stresses, so the envelope drawn through the total-stress Mohr circles yields the total-stress parameters $c_{cu}$ and $\phi_{cu}$. In a normally-consolidated or lightly over-consolidated clay the specimen contracts on shearing, which under undrained conditions generates a positive excess pore pressure. Every effective-stress circle is therefore displaced to the left of its total-stress counterpart by $\Delta u_f$, while the two circles have the same diameter. Shifting a circle to the left while holding the deviator stress constant lowers the slope of the envelope that touches it, so
$$\phi_{cu} \approx (0.5\ \text{to}\ 0.7)\,\phi' \quad<\quad \phi'$$
A consolidated-drained (CD) test allows the pore pressure to stay at zero throughout, so it measures $\phi'$ directly. The only clays that can drive $\phi_{cu}$ above $\phi'$ are heavily over-consolidated ones which dilate and develop negative excess pore pressure; even then the word “always” in the statement makes it false. The correct general statement is that $\phi_{cu}$ is normally appreciably smaller than $\phi'$, and that only a CU test with pore-pressure measurement, or a CD test, recovers the true effective-stress friction angle.
A plasticity index of zero means the soil is non-plastic: it is a clean sand, a gravel or a rock-flour silt with no active clay minerals. Indices of 100 and 200 place soils B and C in the extremely plastic range of the Casagrande chart, well above the A-line, which requires a large proportion of fine, high-surface-area clay minerals. Permeability is controlled by the size of the pore channels, and pore-channel size scales with the effective grain size; Hazen’s empirical rule captures the dependence:
$$k\ (\text{cm/s}) \approx C\,D_{10}^2 \quad\text{with } D_{10}\ \text{in mm and } C \approx 1.0$$
Typical values are $k \approx 10^{-3}$ to $10^{-5}\ \text{m/s}$ for a non-plastic sand and $k \approx 10^{-9}$ to $10^{-11}\ \text{m/s}$ for a high-plasticity clay — a difference of five to eight orders of magnitude. Two mechanisms act together: the pore throats are far smaller in the clay, and a large fraction of the clay’s pore water is immobilised in the adsorbed double layer and does not participate in flow at all. Soil A therefore has by far the highest coefficient of permeability, and the statement is true.
