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18-Env-A3 Geotechnical and Hydrogeological Engineering · December 2018

Question 1 of 6: Soil Sample Phase Relationships

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2018 — 18-Env-A3 / Geotechnical & Hydrogeological Engineering. 3 hours duration; open book exam, any non-communicating calculator permitted. FIVE (5) questions constitute a complete exam paper (the first five as they appear in the answer book are marked, 20 marks each, 100 marks total); all six printed questions are solved below for completeness.

Reference texts. Braja M. Das, Principles of Geotechnical Engineering (9th ed.) — compaction, permeability and seepage, weight–volume relations; Craig & Knappett, Craig's Soil Mechanics (8th ed.) — flow nets and composite seepage barriers; Freeze & Cherry, Groundwater (1979) — Darcy's law, the Dupuit–Thiem equation and seepage velocity.

Question 1: Soil Sample Phase Relationships (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.

Given data
QuantitySymbolValue
Sample diameter$d$50 mm = 0.050 m
Sample length$L$0.5 m
Total (moist) mass$M_t$1.750 kg
Moisture content$w$20%
Specific gravity of solids (assumed)$G_s$2.70

Find. (a) degree of saturation $S$; (b) porosity $n$; (c) saturated unit weight $\gamma_{sat}$; (d) dry unit weight $\gamma_{d}$.

Approach. Get the total volume from the sample's cylindrical geometry, split the moist mass into solids and water via $w$, size the solids volume with an assumed reasonable $G_s$, then read every phase ratio off the resulting three-phase diagram.

  1. Sample volume. $V=\dfrac{\pi}{4}d^2L=\dfrac{\pi}{4}(0.050)^2(0.5)=9.817\times10^{-4}\ \text{m}^3$ (981.7 cm³).
  2. Assume $G_s$. No index test for solids density is given, so a reasonable value for an inorganic silt/clay (typical range 2.66–2.75, Das Table 3.1) is taken: $G_s=2.70$.
  3. Split the mass. $M_s=\dfrac{M_t}{1+w}=\dfrac{1.750}{1.20}=1.4583\ \text{kg}$; water mass $M_w=M_t-M_s=0.2917\ \text{kg}$.
  4. Phase volumes. $V_s=\dfrac{M_s}{G_s\rho_w}=\dfrac{1.4583}{2.70\times1000}=5.401\times10^{-4}\ \text{m}^3$; $V_w=\dfrac{M_w}{\rho_w}=2.917\times10^{-4}\ \text{m}^3$; void volume $V_v=V-V_s=4.416\times10^{-4}\ \text{m}^3$; void ratio $e=V_v/V_s=0.8177$.
  5. Part (a) — degree of saturation. $$S=\frac{V_w}{V_v}=\frac{2.917\times10^{-4}}{4.416\times10^{-4}}=\boxed{66.05\%}.$$
  6. Part (b) — porosity. $$n=\frac{V_v}{V}=\frac{4.416\times10^{-4}}{9.817\times10^{-4}}=\boxed{44.98\%}.$$
  7. Part (c) — saturated unit weight. Flooding the same void space to $S=100\%$: $$\gamma_{sat}=\frac{G_s+e}{1+e}\,\gamma_w=\frac{2.70+0.8177}{1+0.8177}\times9.81=\boxed{18.99\ \text{kN/m}^3}.$$
  8. Part (d) — dry unit weight. $$\gamma_{d}=\frac{G_s}{1+e}\,\gamma_w=\frac{2.70}{1.8177}\times9.81=\boxed{14.57\ \text{kN/m}^3}.$$
Check: $G_s=2.70$ is assumed (a "reasonable value" per the question's own wording, since no solids-density test is reported) — typical for inorganic silts/clays per Das Table 3.1. All four results are internally consistent: $\gamma_d$ recomputed directly from $M_s/V$ (14.57 kN/m³) and $\gamma_{sat}=\gamma_d+n\gamma_w$ (18.99 kN/m³) both match the $G_s$/$e$ formulas exactly.
QuantityValue
(a) Degree of saturation, $S$66.05%
(b) Porosity, $n$44.98% ($e=0.8177$)
(c) Saturated unit weight, $\gamma_{sat}$18.99 kN/m³
(d) Dry unit weight, $\gamma_d$14.57 kN/m³
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