18-Env-A3 Geotechnical and Hydrogeological Engineering · December 2014
Question 1 of 6: Compaction — Weight–Volume Relations from Mold Dimensions
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2014 — 04-Env-A3 / Geotechnical & Hydrogeological Engineering. 3 hours duration; open book exam, any non-communicating calculator permitted. The first five questions as they appear in the answer book are marked (20 marks each, 100 marks total); all six are solved below for completeness.
Reference texts. Braja M. Das, Principles of Geotechnical Engineering (9th ed.) — weight–volume relations, compaction, seepage/flow nets, effective stress and shear strength chapters; Craig & Knappett, Craig's Soil Mechanics (8th ed.) — cross-reference for flow nets, method of fragments and shear strength; Freeze & Cherry, Groundwater (1979) — Darcy's law and the Dupuit–Thiem equation for unconfined radial flow to a well.
Find. (a) porosity $n$; (b) moisture content $w$; (c) bulk (moist) density $\rho$; (d) dry unit weight $\gamma_d$; (e) saturated unit weight $\gamma_{sat}$.
Approach. Compute the mold volume from its cylindrical dimensions, then solve the mass–volume identities simultaneously for the void volume — since here $S$, not $w$, is the given saturation-linked datum — and read every other ratio off the resulting phase diagram.
Mold volume. The mold is a right cylinder:
$$V=\frac{\pi}{4}d^2L=\frac{\pi}{4}(0.038\ \text{m})^2(0.088\ \text{m})=9.980\times10^{-5}\ \text{m}^3=99.80\ \text{cm}^3.$$
Solve for the void volume. With $M=M_s+M_w$, $M_s=G_s\rho_wV_s$, $M_w=S\,\rho_wV_v$ and $V_s=V-V_v$,
$$M=G_s\rho_w(V-V_v)+S\rho_wV_v=G_s\rho_wV+\rho_wV_v(S-G_s)\ \ \Rightarrow\ \ V_v=\frac{M-G_s\rho_wV}{\rho_w(S-G_s)}.$$
Substituting $M=0.185\ \text{kg}$, $G_s\rho_wV=2.5\times1000\times9.980\times10^{-5}=0.24951\ \text{kg}$, and $\rho_w(S-G_s)=1000(0.33-2.5)=-2170$,
$$V_v=\frac{0.185-0.24951}{-2170}=2.973\times10^{-5}\ \text{m}^3=29.73\ \text{cm}^3,\qquad n=\frac{V_v}{V}=\frac{29.73}{99.80}=\boxed{29.78\%}.$$
Part (b) — moisture content. $V_s=V-V_v=70.08\ \text{cm}^3$ and $e=V_v/V_s=0.4242$. The solids mass is $M_s=G_s\rho_wV_s=2.5\times70.08=175.19\ \text{g}$, so $M_w=M-M_s=185-175.19=9.81\ \text{g}$ and
$$w=\frac{M_w}{M_s}=\frac{9.81}{175.19}=\boxed{5.60\%}.$$
Part (c) — bulk density.
$$\rho=\frac{M}{V}=\frac{0.185\ \text{kg}}{9.980\times10^{-5}\ \text{m}^3}=\boxed{1854\ \text{kg/m}^3}.$$
Part (d) — dry unit weight.
$$\gamma_d=\frac{G_s}{1+e}\gamma_w=\frac{2.5}{1.4242}\times9.81=\boxed{17.22\ \text{kN/m}^3}.$$
Part (e) — saturated unit weight. Flooding the SAME compacted void ratio ($e=0.4242$) to $S=100\%$,
$$\gamma_{sat}=\frac{G_s+e}{1+e}\gamma_w=\frac{2.5+0.4242}{1.4242}\times9.81=\boxed{20.14\ \text{kN/m}^3}.$$
Check: this problem gives degree of saturation $S$ rather than moisture content $w$, so the void volume must be solved for algebraically from the total mass/volume/$S$/$G_s$ identity before any other ratio can be read off — part (e) still assumes the SAME void ratio is simply flooded to full saturation, not a re-compaction at a different density.
