24-Bld-A6 Geotechnical Materials and Analysis · December 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
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.
(a) Answer: (i) Sand A. Sand B's grain-size curve lies to the left of Sand A's in Figure 1 — for a given "percentage smaller" ordinate (e.g., the 10% line used to read D10), Sand B always corresponds to a smaller particle size than Sand A. Sand A's D10 therefore exceeds Sand B's. By Hazen's approximation $k\approx C\,D_{10}^2$ (with $C\approx 100$ for $D_{10}$ in cm, $k$ in cm/s), the coefficient of permeability is governed by the smallest particles that control the narrowest pore throats — so the soil with the larger D10 (Sand A, the coarser of the pair at the fine end of its distribution) is the more permeable one.
(b) Answer: (iv) Both sands A and B will have no shear strength. Liquefaction is defined by excess pore pressure rising until it equals the total stress ($\Delta u\to\sigma$), driving the effective stress $\sigma'=\sigma-u$ to zero. Granular soils carry essentially all of their shear strength through the frictional term $\sigma'\tan\phi'$ (with $c'\approx0$), so once $\sigma'=0$ the available shear strength is also zero — for either sand, regardless of how their gradations differ. Grain-size distribution affects HOW READILY each sand builds and dissipates excess pore pressure (looser/finer sands liquefy more easily than denser/coarser ones), not the residual strength once liquefaction has actually occurred.
(c) Answer: (i) UU tests. "Total" shear-strength parameters — c and $\phi$ expressed directly in terms of total stress — come from a test that permits no drainage and no consolidation at any stage: the unconsolidated–undrained (UU) test, sheared rapidly with no volume change, yielding $c_u$ (and, for a saturated clay, $\phi_u\approx0$). A CU test instead needs a pore-pressure measurement to recover the corresponding effective-stress envelope $c',\phi'$, and a CD test is sheared slowly enough that $u\approx0$ throughout, so it too reports effective-stress parameters directly — neither is the route to a total-stress envelope on its own terms.
(d) Answer: (i) approximately 150 kPa. The problem is a surface footing ($D=0$) on cohesionless sand ($c'=0$), so the hint reduces to $q_{ub}=0.5\,\gamma B N_\gamma$. With the GWT at $5B$ — well below the roughly $1$–$2B$ depth that actually governs $N_\gamma$ beneath a shallow surface footing — the water table has no practical effect, and the given 300 kPa uses the full moist/bulk unit weight $\gamma$. When the GWT rises to the surface, the entire stressed zone beneath the footing becomes submerged, so $\gamma$ must be replaced by the buoyant unit weight $\gamma'=\gamma_{sat}-\gamma_w$. For typical sands ($\gamma_{sat}\approx18$–$20\text{ kN/m}^3$), $\gamma'\approx\gamma/2$, and since $q_{ub}$ scales almost linearly with the governing unit weight, the bearing capacity falls to roughly half its original value, $\approx150\text{ kPa}$.
(e) Answer: (a). The zero-air-voids (ZAV) line is a purely geometric bound — the locus of dry unit weight at $S=100\%$, $\gamma_{d,zav}=\dfrac{G_s\gamma_w}{1+wG_s}$ — and depends only on $G_s$ and the assumed water content, so it can be computed and plotted for any soil directly from $G_s$, without running a single compaction test. It always lies above the real compaction curve (compaction can never fully expel all air, so $S<100\%$ is always the actual state), unlike the compaction curve itself (option b), whose shape genuinely depends on the soil and compaction effort and can only be found by testing.
