24-Bld-A6 Geotechnical Materials and Analysis · December 2017
Question 2 of 7
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
07-Bld-A6 Geotechnical Materials and Analysis — National Exam, December 2017. Closed book, 3 hours; drawing instruments and either a Casio or Sharp approved calculator required; the formula sheet and influence charts printed at the back of the exam are reproduced/used inline where needed. Section A (Questions 1–3, 40 marks, answer all) and Section B (Questions 4–7, 20 marks each, the paper asks for any three of four) — all seven questions are answered below.
Reference texts: B. M. Das, Principles of Geotechnical Engineering, 9th ed. (phase relations, seepage/flow nets, stress distribution, consolidation, shear strength); R. F. Craig / J. Knappett, Craig's Soil Mechanics, 9th ed. (flow nets, effective-stress strength parameters); Canadian Foundation Engineering Manual (CFEM), 4th ed.
Check — assumptions and source notes for this paper. (1) Question 4's flow net (Nf = 5 flow channels, Nd = 14 equipotential drops, point A read as three drops upstream of the downstream exit, exit-field length ≈ 2 m) is used with a soil total unit weight of 20 kN/m³. (2) Question 5(ii)'s "approximate" method is extended from the formula sheet's centre-of-rectangle formula to a corner point (point A here is not centred on the loaded area) using the standard mirror/quartering superposition trick, valid because the approximate formula is linear in load exactly like the exact one. (3) Question 6's least-squares fit through the three effective-stress points gives an intercept of −0.7 kPa — not distinguishable from zero with only three data points — so c′ is taken as 0, consistent with the expected behaviour of a normally consolidated clay. (4) Question 7's figure shows three distinct layer thicknesses — H1 = 1.5 m (sand above the water table), H2 = 1.5 m (sand below the water table), H3 = 2 m (the clay layer, void ratio e = 0.75) — and all three values are used below.
Given. ρ = 1.28 Mg/m³, e = 9.0, ρs = 2.75 Mg/m³, S = 95%, w = 311%.
Find. Which one of the five values is inconsistent with the other four, and its correct value.
Approach. Two independent phase relations connect these five quantities: the fundamental identity $Se = wG_s$ (which involves only e, S, w, Gs) and the bulk-density formula $\rho=(Se+G_s)\rho_w/(1+e)$ (which brings in ρ). Checking the first relation isolates whether e, S, w, Gs are mutually consistent before ρ is even considered; if they are, the fault must lie in ρ.
Check e, S, w, Gs against the fundamental identity. With $G_s=\rho_s/\rho_w=2.75$,
$$Se = (0.95)(9.0) = 8.55, \qquad wG_s = (3.11)(2.75) = 8.5525.$$
The two sides agree to within 0.03% — e, S, w and Gs are mutually consistent, so none of these four is the faulty value.
Recompute ρ from the consistent values.
$$\rho = \frac{(Se+G_s)\rho_w}{1+e} = \frac{(8.55+2.75)(1.0)}{1+9.0} = \frac{11.30}{10} \boxed{=\ 1.13\ \text{Mg/m}^3}.$$
Compare with the given value. The given $\rho=1.28$ Mg/m³ disagrees with the computed 1.13 Mg/m³ by 13% — far outside rounding — so the bulk density is the inconsistent value; the internally consistent figure is 1.13 Mg/m³.
Cross-check by elimination. If, instead, Gs were assumed wrong and ρ = 1.28 correct, solving $\rho(1+e)=Se+G_s=G_s(1+w)$ backward gives $G_s = \rho(1+e)/(1+w) = 12.8/4.11 = 3.11$ — far above any plausible mineral specific gravity for a clay (normally 2.65–2.80), which confirms ρ, not Gs, is the outlier.
