16-Civ-A4 Geotechnical Materials and Analysis · December 2013
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper: National Examinations — December 2013 · 98-Civ-A4 Geotechnical Materials and Analysis · 3 hours, closed book · 100 marks · answer all six questions. Charts (m–n influence, Newmark) and a formula sheet are supplied at the back of the paper.
Reference texts. R. F. Craig / Knappett & Craig, Craig’s Soil Mechanics (8th ed.); B. M. Das, Principles of Geotechnical Engineering; Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering; M. Budhu, Soil Mechanics and Foundations. Canadian practice: Canadian Geotechnical Society, Canadian Foundation Engineering Manual (CFEM, 4th ed.). Unit weight of water taken as $\gamma_w = 9.81\ \text{kN/m}^3$ throughout.
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.
(i) TRUE. A well-graded sand contains a wide range of particle sizes, so finer grains occupy the voids between coarser grains. This produces a denser packing (lower void ratio), a larger number of interparticle contacts and stronger mechanical interlocking. The peak friction angle $\phi'$ therefore rises, and since shear strength on the Mohr–Coulomb line is $\tau_f = c' + \sigma'\tan\phi'$ with $c'\approx 0$ for sand, a higher $\phi'$ means greater strength at the same effective stress. A uniform (poorly-graded) sand of the same mineralogy packs more loosely and mobilises a lower $\phi'$, hence lower strength.
(ii) TRUE. Permeability is governed chiefly by void ratio. For a given clay at a given current effective stress, the normally consolidated (NC) state is the loosest the soil can be — it has never carried a higher stress — so it retains the largest void ratio and the most continuous pore network. An over-consolidated (OC) sample of the same clay was previously loaded to a higher stress and then unloaded; at the same current stress it sits at a markedly lower void ratio. Because $k$ increases strongly with $e$ (Kozeny–Carman, and empirically $e \propto \log k$), the NC clay is the more permeable of the two.
(iii) TRUE (with the reading stated below). A 10 m earth dam applies a contact stress of order $\gamma H \approx 20 \times 10 = 200\ \text{kPa}$ over a very wide base. Because the loaded area is large relative to the depth of interest, the vertical stress it induces decays only slowly with depth — the pressure bulb extends tens of metres down. A 5-storey building on a 1 m-wide strip footing applies a comparable contact pressure, but over so narrow a width that the induced stress dissipates within a few metres (Boussinesq stress under a strip falls off as roughly $1/z$). At any depth more than a few metres the dam therefore induces the higher — and far more extensive — vertical stress, which is why dam foundations must be assessed to great depth.
Check: statement (iii) is worded ambiguously in the source (it trails off with “???”). The answer above assumes it compares the magnitude and reach of the induced stress at depth; taking $\gamma_{\text{soil}}\approx 20\ \text{kN/m}^3$ for the dam fill and a normal 5-storey strip-footing pressure. Under either reading the wide dam load governs at depth.
(iv) FALSE. For a saturated clay, the total-stress friction angle $\phi_{cu}$ obtained from a consolidated-undrained test without pore-pressure measurement is smaller than the effective-stress angle $\phi'$ from a consolidated-drained (or CU-with-$u$) test, not larger. During undrained shear of a normally consolidated clay the specimen develops positive excess pore pressure, so the effective stresses are lower than the totals; plotting the total-stress circles gives a flatter envelope, i.e. $\phi_{cu} < \phi'$ (typically $\phi_{cu}$ is roughly half of $\phi'$). Hence the statement is false.
(v) FALSE. Compressibility increases with plasticity: the compression index correlates with plasticity/liquid limit (e.g. $C_c \approx 0.009(LL-10)$, and $C_c$ rises with $I_p$). A soil with $I_p = 0$ is non-plastic (a silt or sand) and is the least compressible of the three, whereas Soil C ($I_p = 50$) is the most compressible. Under the same 200 kPa increment the $I_p = 0$ soil consolidates the least, so the claim that it consolidates the most is false.
