24-Bld-A6 Geotechnical Materials and Analysis · Undated paper
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: (iii) Saturated density. For the same soil skeleton (fixed void ratio $e$ and grain specific gravity $G_s$), $\gamma_{sat}=\dfrac{(G_s+e)\gamma_w}{1+e}$ fills every void with water in addition to the solids, giving the largest possible weight per unit volume. Moist (partially saturated) density carries some air-filled voids and so is lighter; dry density removes the pore water entirely; submerged (buoyant) density, $\gamma'=\gamma_{sat}-\gamma_w$, subtracts the full weight of water and is the smallest of the four — often even less than the dry density. The ranking is therefore $\gamma_{sat}>\gamma_{moist}>\gamma_{dry}>\gamma_{submerged}$.
(b) Answer: (ii) The total stress is equal to the pore-water pressure. Free water has no solid skeleton, so it cannot carry any effective (grain-to-grain) stress: $\sigma'=0$ everywhere in the pool. By the effective-stress principle $\sigma=\sigma'+u$, the entire total stress at 4 m depth is therefore carried by the pore-water pressure, $u=\gamma_w\times 4\text{ m}\approx 39.2\text{ kN/m}^2$, and $\sigma=u$. Water absolutely does transmit pressure (option (i) is wrong — "no pores in water" confuses the concept of pore pressure in a soil skeleton with hydrostatic pressure in a fluid), and option (iii) reverses cause and effect ($\sigma'=\sigma-u=0$, not $\sigma'=u$).
(c) Answer: (iv) All of the above tests. Effective-stress parameters $c',\phi'$ require the pore pressure at failure to be known (or zero) so that $\sigma'=\sigma-u$ can be computed. A CD direct-shear test is sheared slowly enough that no excess pore pressure ever develops ($u\approx0$ throughout), so $\sigma\approx\sigma'$ directly. A CU triaxial test develops excess pore pressure but measures it, so $\sigma'=\sigma-u$ is recovered numerically. A CD triaxial test, like the direct-shear case, is sheared slowly enough that drainage keeps $u\approx0$, so no pore-pressure measurement is even needed to know $\sigma'\approx\sigma$. All three routes deliver a genuine effective-stress envelope.
(d) Answer: (iv) all of them have the same bearing capacity. Under undrained loading, a saturated clay is analysed in terms of total stress with $\phi_u=0$, so $N_q=1$ and $N_\gamma=0$ in the given hint — the bearing-capacity equation collapses to $q_{ult}=c_uN_c\,(\text{shape factor})+\gamma D$. Every term on the right depends only on the undrained shear strength $c_u$, the fixed embedment $D=1.5$ m, and the (fixed, square) footing shape — not on the footing's absolute width $B$, because the one term that would introduce $B$ ($0.5\gamma BN_\gamma$) vanishes when $N_\gamma=0$. All three footings therefore carry identical ultimate bearing pressure (in kPa), even though the 2 m × 2 m footing carries far more total load in kN.
(e) Answer: (iv) All sands have approximately the same coefficient of permeability. By Hazen's approximation, $k\approx C\,D_{10}^2$ — permeability is governed almost entirely by the effective (smallest-controlling) particle size $D_{10}$, which fixes the pore-throat size that bottlenecks flow. Whether a sand's overall grading is uniform, well-graded, or gap-graded describes the spread of particle sizes above $D_{10}$, not $D_{10}$ itself; a well-graded sand's larger particles simply pack around the same fine fraction. Without a stated difference in $D_{10}$, gradation shape alone gives no basis for ranking permeability — the intended trap is to assume "better graded" means "more permeable," when in practice a well-graded sand is often less permeable (denser packing, lower void ratio) at the same $D_{10}$.
