24-Bld-A6 Geotechnical Materials and Analysis · May 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: (ii) The total stress is equal to pore-water pressure. A swimmer's body is surrounded by free water, not a soil skeleton — there are no particle-to-particle contacts to carry load. All of the stress at 5 m depth is transmitted through the water itself, so the total stress equals the hydrostatic pore-water pressure, $\sigma=u=\gamma_w z=9.81\times5\approx49.1\text{ kPa}$, and the effective stress (the inter-particle contact stress, $\sigma'=\sigma-u$) is zero — there is no soil skeleton to carry it. Statement (i) is wrong because hydrostatic pressure certainly exists in open water; statement (iii) inverts the correct relation (effective stress is zero, not equal to $u$).
(b) Answer: (i) sand. Liquefaction is the sudden loss of shear strength in a saturated, loose, cohesionless soil when cyclic/dynamic loading generates pore-water pressure faster than it can dissipate, driving the effective stress ($\sigma'=\sigma-u$) toward zero. Loose saturated sand has no cohesion to hold particles together once $\sigma'\to0$, so it behaves like a liquid. Clays (soft or stiff) are cohesive and much less permeable; even soft clay's plasticity and inter-particle bonding resist the sudden strength collapse that defines liquefaction (clays instead show cyclic softening, a related but distinct phenomenon).
(c) Answer: (iii) CD tests. A consolidated-drained (CD) test shears the specimen slowly enough that excess pore pressure never builds up ($u\approx0$ throughout shear), so the measured total stresses at failure are automatically the effective stresses — the test yields effective-stress parameters $c'$, $\phi'$ directly, with no pore-pressure measurement needed. A UU test only ever gives total-stress (undrained) parameters, and a CU test needs a separate pore-pressure transducer to back out $c'$, $\phi'$; only CD delivers them directly from the measured axial/confining stresses.
(d) Answer: (ii) approximately 300 kPa. For a footing in sand ($c'=0$), the hint formula reduces to $q_{ub}=\gamma D N_q+0.5\,\gamma B N_\gamma$. With the GWT at $D=5B$ (well below the zone of influence, roughly $1$–$2B$ below the footing, that governs $N_\gamma$), both terms use the moist/bulk unit weight $\gamma$, giving $q_{ub}=600\text{ kPa}$. When the GWT rises to the ground surface, the entire stressed zone becomes submerged and both terms must use the buoyant unit weight $\gamma'=\gamma_{sat}-\gamma_w$, which for typical sands ($\gamma_{sat}\approx18$–$20\text{ kN/m}^3$) is close to half of $\gamma$. Since $q_{ub}$ scales almost linearly with the governing unit weight, the bearing capacity falls to roughly half its original value, $\approx300\text{ kPa}$ — this is the classic result used to size footings above a rising water table.
(e) Answer: (i) increases. A dense sand is already packed tighter than its natural (critical-void-ratio) arrangement; as shear stress forces particles to ride up and over their neighbours (dilatancy), the sample must expand to keep shearing. At a low confining pressure of 10 kPa, particle crushing/rearrangement that would otherwise suppress dilation is minimal, so the classic dense-sand behaviour dominates and the volume increases (negative volumetric strain, positive dilation) right up to failure.
