Question 1 of 6: True/False with Justification

Question 1: True/False with Justification (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.

Given. Five statements covering gradation and shear strength, stress history and permeability, the depth of influence of two very different loads, total- versus effective-stress friction angles, and plasticity as a predictor of compressibility.

Find. A True or False verdict on each, with the soil-mechanics reasoning that decides it.

(i) Well-graded versus uniform sand — TRUE

Shear resistance in a cohesionless soil comes from two sources: sliding friction between grains, which depends only on mineralogy and is essentially the same for both sands, and dilatancy plus interlocking, which depends entirely on how densely the grains are packed. A well-graded sand contains a continuous spread of sizes, so the smaller particles occupy the voids between the larger ones. It therefore reaches a much lower void ratio, and a much higher relative density, for the same compactive effort than a uniform sand whose single-size grains can pack no better than a stack of equal spheres.

The consequence is measured directly in the direct-shear or triaxial box. A dense, well-graded sand must ride its grains up and over one another before it can shear, and that expansion against the confining stress is extra work the applied load has to supply. The peak friction angle is written

$$\phi_{\text{peak}} = \phi_{cv} + \psi$$

where $\phi_{cv}$ is the constant-volume (critical-state) friction angle, typically 30° to 33° for quartz sand of either gradation, and $\psi$ is the dilatancy angle. For a dense well-graded sand $\psi$ commonly reaches 8° to 12°, giving $\phi_{\text{peak}} \approx 40^\circ$ to 45°, whereas a uniform sand at the same compactive effort rarely exceeds $\phi_{\text{peak}} \approx 32^\circ$ to 36°. The uniformity coefficient $C_u = D_{60}/D_{10}$ is the index that captures this: $C_u > 6$ with a coefficient of curvature between 1 and 3 defines “well graded” in the Unified Soil Classification System, and it is exactly the gradation an engineered granular fill is specified to.

The one qualification worth stating is that at very large strain both sands converge on the same critical-state angle, because dilatancy is a peak-strength phenomenon that disappears once the soil reaches constant volume. The statement as written concerns strength in the ordinary design sense, and in that sense it is true.

(ii) Permeability of normally consolidated versus over-consolidated clay — TRUE

The comparison must be made on the same clay at the same present effective stress, which is how the statement is meant. An over-consolidated clay reached that stress by being loaded to a higher pre-consolidation pressure $\sigma_c'$ and then unloaded along the flat swelling line; because $C_s \approx C_c/5$ to $C_c/10$, it recovers only a small fraction of the void ratio it lost. A normally consolidated clay arrived at the same stress along the virgin compression line and has never been denser. At any given $\sigma_0'$ the over-consolidated specimen therefore sits at the lower void ratio.

Permeability follows void ratio very steeply. For clays the standard correlation is a straight line on a $e$ versus $\log k$ plot,

$$e = e_{k} + C_k \log\!\left(\frac{k}{k_0}\right) \quad\text{with}\quad C_k \approx 0.5\,e_0$$

so a void-ratio difference of only 0.1 to 0.2 — entirely typical between the NC and OC states of the same clay — changes $k$ by roughly half an order of magnitude. Smaller voids mean smaller pore throats, and in a clay a larger share of the remaining pore water is immobilised in the adsorbed double layer, so it contributes nothing to flow. The normally consolidated clay is the more permeable of the two, and the statement is true.

(iii) Earth dam versus condominium strip footing — TRUE

Two things separate these loads: the magnitude of the contact stress and, far more importantly, the width of the loaded area, which fixes how far down the stress reaches. Taking a compacted embankment fill at $\gamma \approx 20\ \text{kN/m}^3$, the dam applies at its base

$$\Delta\sigma_{\text{dam}} = \gamma H = 20 \times 10 = 200\ \text{kPa}$$

and because an earth dam is tens of metres wide it behaves as a very wide loaded strip: the $2{:}1$ spread is negligible over the first several metres, so essentially the full 200 kPa is still felt at $z = 5$ m and beyond. The condominium footing is 1 m wide, and even a generous allowance of five storeys at roughly 30 kPa of tributary line load per storey gives a bearing pressure of order 150 kPa at the base. Spread through a $1$ m wide strip, however, that stress decays almost at once:

$$\Delta\sigma_z = \frac{qB}{B+z} = \frac{150 \times 1}{1 + 5} = 25\ \text{kPa at }z = 5\ \text{m below the footing}$$

So even where the two contact pressures are comparable, at any depth of engineering interest the dam is inducing roughly an order of magnitude more stress, and it does so over the entire footprint rather than in a narrow bulb. This is precisely why an embankment on soft ground is a consolidation-settlement problem reaching tens of metres down, while a narrow strip footing is a bearing-capacity and immediate-settlement problem confined to the first two or three metres. The statement is true.

(iv) Consolidated-undrained versus consolidated-drained friction angle — FALSE

The statement inverts the usual result. A consolidated-undrained test run without pore-pressure measurement can only be interpreted in total stresses, so the envelope fitted to the total-stress Mohr circles returns the total-stress parameters $c_{cu}$ and $\phi_{cu}$. In a normally consolidated or lightly over-consolidated clay the specimen wants to contract as it shears; prevented from draining, it generates a positive excess pore pressure $\Delta u_f$. Each effective-stress circle is therefore the same diameter as its total-stress twin but displaced to the left by $\Delta u_f$, and shifting circles left while holding the deviator stress fixed steepens the effective-stress envelope relative to the total-stress one:

$$\phi_{cu} \approx (0.5\ \text{to}\ 0.7)\,\phi' \quad<\quad \phi'$$

A consolidated-drained test holds the pore pressure at zero throughout, so it measures $\phi'$ directly — which is exactly what Question 6 of this paper does. Only a heavily over-consolidated clay, which dilates and develops negative excess pore pressure, can push $\phi_{cu}$ above $\phi'$, and even then the word “always” in the statement would make it false. The statement is false.

(v) Plasticity index and compressibility — FALSE

The statement has the ranking backwards. Compressibility rises with plasticity, because a high plasticity index means a large fraction of fine, high-specific-surface clay minerals whose plate-shaped particles reorient and squeeze water out of the double layer under load. The classical Skempton correlation, which is printed on the formula sheet of this very paper, makes the dependence explicit:

$$C_c = 0.009\,(LL - 10)$$

and the liquid limit tracks $I_p$ closely for soils near the A-line. Since all three soils are stated to be normally consolidated, the whole of the 200 kPa increment travels down the virgin compression line for each of them and the settlement is

$$s_c = \frac{C_c H}{1 + e_0}\,\log\!\left(\frac{\sigma_0' + \Delta\sigma'}{\sigma_0'}\right)$$

so with $H$, $e_0$ and the stress ratio held equal the settlements are in the ratio of the $C_c$ values. Soil A with $I_p = 0$ is non-plastic — a clean sand or a rock-flour silt — with $C_c$ of order 0.01 to 0.05; Soil C with $I_p = 50$ has $LL$ of order 75 and hence $C_c \approx 0.6$. Soil C therefore consolidates the most and Soil A the least, so the statement is false. Soil A does settle fastest, because its coefficient of consolidation is orders of magnitude higher, and confusing rate with magnitude is the usual reason this statement gets marked true.