18-Env-A3 Geotechnical and Hydrogeological Engineering · December 2017
Question 1 of 6: Soil Phase Relationships
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2017 — 04-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.) — weight–volume relations, compaction, seepage/flow nets, and consolidation chapters; Craig & Knappett, Craig's Soil Mechanics (8th ed.) — cross-reference for flow-net theory and finite-difference seepage; Freeze & Cherry, Groundwater (1979) — Darcy's law, the Thiem confined-flow equation, and radial travel time.
Find. (a) void ratio $e$; (b) porosity $n$; (c) moisture content $w$; (d) dry unit weight $\gamma_d$.
Approach. Split the total mass into solids and water using $w$, convert each mass to a volume with $G_s$ and $\rho_w$, and take the volumes left over as voids.
Solids and water masses. With $M_t=M_s(1+w)$:
$$\begin{aligned} M_s&=\frac{M_t}{1+w}=\frac{100}{1.30}=76.92\ \text{kg} \\ M_w&=M_t-M_s=23.08\ \text{kg} \end{aligned}$$
Solids, water and void volumes. $V_s=\dfrac{M_s}{G_s\rho_w}=\dfrac{76.92}{2.65\times1000}=0.02903\ \text{m}^3$; $V_w=\dfrac{M_w}{\rho_w}=0.02308\ \text{m}^3$; $V_v=V-V_s=0.125-0.02903=0.09597\ \text{m}^3$ (of which $V_a=V_v-V_w=0.07290\ \text{m}^3$ is air — the block is not saturated).
Part (a) — void ratio.
$$e=\frac{V_v}{V_s}=\frac{0.09597}{0.02903}=\boxed{3.31}.$$
Part (b) — porosity.
$$n=\frac{V_v}{V}=\frac{0.09597}{0.125}=\boxed{76.8\%}\quad(\text{check: } e/(1+e)=3.31/4.31=76.8\%\ \checkmark).$$
Part (c) — moisture content. This is the value stated in the problem itself, $w=M_w/M_s=23.08/76.92=\boxed{30.0\%}$ — sub-part (c) simply asks the candidate to restate/confirm it from the phase masses found in Step 1.
Part (d) — dry unit weight.
$$\gamma_d=\frac{M_s\,g}{V}=\frac{76.92\times9.81/1000}{0.125}=\boxed{6.04\ \text{kN/m}^3}.$$
Check: with $e=3.31$ the sample is an unusually soft/loose clay (degree of saturation $S=V_w/V_v=24.0\%$, and bulk unit weight $\gamma=M_t g/V=7.85\ \text{kN/m}^3$, both far below a typical intact clay) — the numbers are solved literally and consistently from the stated mass and dimensions; no correction is applied.
