Question 1 of 5

Question 1 (4 × 5 = 20 marks)

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) Answer: (a). The zero-air-voids (ZAV, or saturation) line is not derived from a compaction test at all — it is the locus of dry unit weight a soil would have if it were fully saturated (S = 100%) at each water content, computed purely from the phase relation $$\gamma_{zav} = \frac{G_s\,\gamma_w}{1+wG_s},$$ which needs only the specific gravity $G_s$ and the water content $w$, no laboratory compaction data. Because a compacted soil always traps some occluded air (100% saturation is a theoretical limit, never actually reached by mechanical compaction), the real compaction curve for any given compactive effort lies everywhere below and to the left of the ZAV line, approaching it most closely on the wet side of optimum. Statement (b) is wrong on both counts: the ZAV line cannot sit below the compaction curve, and it needs no compaction testing to plot.

(ii) Compacting fine-grained clay wet of optimum produces a dispersed (oriented) structure. At water contents above optimum the double-layer repulsion between clay platelets is large relative to the net inter-particle attraction, so the platelets are free to slide past one another under the compactive shear and rotate into a face-to-face, roughly parallel arrangement (Lambe, 1958) — type B in Figure 1. Dry of optimum, the thinner double layers let edge-to-face attraction dominate, and the compactive energy is not enough to overcome it, leaving the random, "house-of-cards" flocculated structure (type A). For permeability: a soil compacted dry of optimum (flocculated) has a markedly higher coefficient of permeability than the same soil compacted wet of optimum. The flocculated fabric leaves large, well-connected, randomly-oriented pore channels, whereas the dispersed fabric produced wet of optimum aligns the platy particles broadside to the flow path, both shrinking the average pore size and lengthening the tortuosity that water must follow — the classical result is that k measured dry of optimum can be one to two orders of magnitude larger than k measured wet of optimum for the same soil and the same dry density.

(iii) Answer: GW. A well-graded gravel has particle and pore sizes several orders of magnitude larger than a high-plasticity clay, and permeability in granular soil is governed almost entirely by the square of an effective particle/pore diameter (Hazen-type relations, $k\propto D_{10}^{2}$) with no double-layer or adsorbed-water effects. CH clay, by contrast, has an enormous specific surface area, a thick adsorbed water layer that further constricts the already microscopic pore throats, and flow paths that must wind between platy particles. Typical values illustrate the gap directly: $k_{GW}\approx 10^{-1}$ to $10^{-3}$ cm/s versus $k_{CH}\approx 10^{-7}$ cm/s or lower — four to six orders of magnitude apart.

(iv) Sand B (C_u = 1, uniformly graded) has the greater coefficient of permeability: with all particles nearly the same size there are no smaller grains available to migrate into and clog the voids between the larger ones, so the pore network stays large and well-connected. Sand A (C_u = 4, well-graded) has the greater effective friction angle φ′: the wider particle-size range lets smaller grains pack into the voids between larger ones, giving a denser fabric with more particle-to-particle contacts and better interlocking for the same relative density, both of which raise the mobilised friction angle. The two properties are governed by opposite aspects of gradation — permeability by the size of the smallest interconnected pore throat, friction angle by packing density and interlock — so it is entirely consistent that no single soil is best at both.

(v) Answer: none of them — all four have $\phi_u = 0$, regardless of their drained φ′. The stem's closing clause ("all soils will have the same value of φ_u") is the answer stated in the question; every mark here is for the justification, which is this: for a fully saturated soil sheared under undrained conditions, any increase in the applied confining (cell) pressure is carried entirely by an equal increase in pore pressure ($\Delta u = \Delta\sigma_3$ for saturated, incompressible pore fluid with no volume change permitted), so the mobilised EFFECTIVE confining stress — and therefore the effective-stress state at failure — is unchanged. The Mohr circles of total stress from a series of undrained tests on the same soil at different confining pressures are therefore all the same size, merely shifted along the σ-axis, and their common tangent (the total-stress failure envelope) is horizontal: $\tau_f = c_u$, $\phi_u = 0$. This holds for the expansive clay, the glacial till, and the silt alike, PROVIDED each is tested fully saturated; it is a total-stress artefact of the pore-pressure response, not a true friction angle. Soil D (sand) is the outlier only because it is rarely tested undrained in practice — its high permeability means load applied at any ordinary rate drains as it is applied, so an "undrained" strength envelope is not physically meaningful for it unless the loading is essentially instantaneous (e.g. seismic), in which case the same $\phi_u = 0$ argument still applies if it is saturated.