Question 1 of 6: Justified true/false statements

Question 1: Justified true/false statements (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) FALSE. A soil with a good and equal representation of all sizes from largest to smallest is a well-graded soil — its grain-size curve is smooth and spread over a wide range, with a high uniformity coefficient $C_u = D_{60}/D_{10}$ and a coefficient of curvature $C_c = D_{30}^2/(D_{10}D_{60})$ between 1 and 3. A uniformly (poorly) graded soil is the opposite: its particles fall in a narrow band of sizes, so the curve is steep and nearly vertical and $C_u \to 1$. The statement therefore describes a well-graded, not a uniformly graded, soil.

(ii) FALSE. Given. $k=1\ \text{m/day}$, driving head $h_w=6\ \text{m}$, and the flow net printed beneath the dam in Figure 1. Find. Whether $q=3\ \text{m}^3/\text{day}$ per metre width. The flow-net discharge is

Count the net as printed. The flow lines are the dam base, the two curved interior flow lines and the impermeable floor, so there are $N_f = 3$ flow channels. Eight interior equipotential lines run from the underside of the dam down to the impermeable floor, between the upstream-bed and downstream-bed boundary equipotentials, so there are $N_d = 9$ potential drops. Hence

not 3 m³/day, so the statement is FALSE. A discharge of 3 m³/day would need $N_f/N_d = 0.5$ (for example a 3-by-6 net), which this long, flat dam base on a shallow permeable layer cannot produce. As an independent check, a finite-difference Laplace solve for a 15 m flat base on a layer of the drawn proportions gives $N_f/N_d \approx 0.28\text{–}0.33$, in line with the hand count of $3/9$.

(iii) TRUE. Below its preconsolidation pressure a saturated clay is over-consolidated (on the “dry”, heavily-over-consolidated side). When such a clay is sheared it tends to dilate; because undrained volume change is suppressed, this dilatant tendency is expressed as a drop in pore-water pressure, i.e. a negative excess pore-water pressure (Skempton's $A_f\lt 0$). Only once the load exceeds $\sigma_p'$ does the response become contractive and the excess pore pressure positive. Strictly, the statement holds for a heavily over-consolidated clay sheared undrained. A lightly over-consolidated clay (OCR below about 2–4) still develops a small positive shear-induced $\Delta u$. Any undrained increase in all-round stress also first raises $u$ by $B\,\Delta\sigma_3$. The TRUE verdict therefore refers to the dilatant shear response that the statement is testing.

(iv) TRUE. Compaction expels air but never all of it. On the compaction curve the optimum moisture content lies to the left of the zero-air-voids ($S=100\%$) line, at $S \approx 80\text{–}90\%$ for both sands and clays. Reaching full saturation by compaction alone would require the OMC point to sit on the ZAV line, which it never does, so the degree of saturation at OMC is always less than 100%.

(v) TRUE. In a fully drained (CD) test the excess pore pressure is held at zero, so total stress equals effective stress and the measured failure envelope is the effective-stress envelope, giving $c'$ and $\phi'$ directly. Both the triaxial cell and the direct-shear box can be run drained. This is routine for free-draining sands; for clays it is still valid provided the shearing rate is slow enough to keep $u\approx 0$.