Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, December 2019 — 18-Geol-A6, Soil Mechanics (3 hours, closed book). Six questions constitute a complete exam: Questions 1–5 are compulsory; for Question 6, candidates choose 3 of 8 optional sub-questions (5 marks each) — all 8 are solved below as a complete study resource.
graphically-read values (compaction curve extrapolation, oedometer Casagrande construction, flow-net equipotential positions) are flagged with individual check callouts below where the read/assumed value materially affects the answer.
Standard and modified Proctor compaction curves, with the $S=100/95/90\%$ zero-air-voids family and the $w=4\%$ read-off point used in part (c).
(a)–(b) Curve labelling and interpretation
(a) Vertical axis: dry density, $\rho_d$ (kg/m$^3$), range $1700$–$2200$; horizontal axis: moulding water content, $w$ (%), range $0$–$20\%$.
(b)(i) The Standard Proctor curve (12,400 J/m$^3$ compactive effort) rises from a low dry density at low $w$, peaks near $w\approx 9.2\%$ at $\rho_{d,\max}\approx 2050\text{ kg/m}^3$, then falls on the "wet side" as extra pore water begins to occupy space the solids would otherwise fill. (ii) The Modified Proctor curve (56,000 J/m$^3$, roughly $4.5\times$ the standard effort) is shifted up and left of the standard curve: higher compactive energy squeezes air out more effectively at any given $w$, so the peak dry density is higher ($\approx 2130\text{ kg/m}^3$) and occurs at a lower optimum water content ($\approx 8\%$), because less water is needed to lubricate particle rearrangement when more mechanical energy is available. (iii) Standard Proctor: $OMC\approx 9.2\%$, $MDD\approx 2050\text{ kg/m}^3$ ($\gamma_{d,\max}\approx 20.1\text{ kN/m}^3$). Modified Proctor: $OMC\approx 8.0\%$, $MDD\approx 2130\text{ kg/m}^3$ ($\gamma_{d,\max}\approx 20.9\text{ kN/m}^3$). (iv) The line of optimums is the locus joining the peak point of every compaction curve for the same soil at increasing compactive effort (standard → modified → …); it runs roughly parallel to, and just below, the $S=90\%$ zero-air-voids curve, since achievable peak density is bounded by (but never reaches) $100\%$ saturation.
(c) Phase relations at $w=4\%$
Given. Standard Proctor data points $(w\%,\rho_d\text{ kg/m}^3)$: $(6,1975),(7.5,2025),(9.2,2050),(10.8,2040),(12.2,2010)$; test water content $w=4\%$; assume $G_s=2.70$ (not printed on this exam page — a typical value for this soil's mineralogy, flagged below).
Find. Void ratio $e$, degree of saturation $S_r$, total unit weight $\gamma_t$, volumetric water content $\theta$, porosity $n$, dry density $\rho_d$ at $w=4\%$.
Check: $w=4\%$ sits below the lowest tested Standard Proctor point ($w=6\%$); $\rho_d$ at $4\%$ is read from a least-squares parabola fit through the five printed Standard Proctor points, extrapolated slightly left of the data — the dry-side data are sparse here so this is a short, defensible extrapolation, not an interpolation. $G_s=2.70$ is also assumed (a typical value for a silty/sandy soil) since no specific gravity is printed anywhere on this exam.
Approach. Fit the five Standard Proctor points to get $\rho_d(4\%)$, then work through the standard phase-relation identities from $\rho_d$, $w$, and $G_s$.
Extrapolate $\rho_d$ at $w=4\%$. Least-squares quadratic through the five Standard Proctor points gives $\rho_d = -5.87w^2+112.4w+1512$ (kg/m$^3$, $w$ in %); at $w=4\%$:
$$\boxed{\rho_d(4\%) = 1868\text{ kg/m}^3}$$
(ii) Degree of saturation. $S_r=\dfrac{wG_s}{e}=\dfrac{0.04\times 2.70}{0.446}=\boxed{24.2\%}$
(iii) Total unit weight. $\rho_t=\rho_d(1+w)=1868(1.04)=1943\text{ kg/m}^3$, so $\gamma_t=\rho_t g = 1943\times 9.81/1000=\boxed{19.06\text{ kN/m}^3}$ (and $\gamma_d=1868\times9.81/1000=18.32\text{ kN/m}^3$).
(iv) Volumetric water content. $\theta=\dfrac{w\rho_d}{\rho_w}=\dfrac{0.04\times 1868}{1000}=\boxed{7.47\%}$
(vi) Dry density. Already found in step 1 above: $\rho_d=\boxed{1868\text{ kg/m}^3}$.
Phase relations at $w=4\%$ ($G_s=2.70$ assumed)
$\rho_d$
$e$
$S_r$
$\gamma_t$
$\theta$
$n$
1868 kg/m$^3$
0.446
24.2%
19.06 kN/m$^3$
7.47%
30.8%
(d) Truck count for the 50 m$^3$ excavation
Given. Specification: compact to $102\%$ of Standard Proctor MDD ($\rho_{d,\max}=2050\text{ kg/m}^3$) at $OMC=9.2\%$; excavation volume $V=50\text{ m}^3$; truck capacity $10\text{ Mg}$ each, soil delivered at OMC.
Find. Number of truckloads required.
Approach. Convert the target field dry density to a target moist (in-place) mass for the full 50 m$^3$, then divide by truck capacity.
Target field dry density. $\rho_{d,\text{field}}=1.02\times 2050=2091\text{ kg/m}^3$.
Target field moist density at OMC. $\rho_{t,\text{field}}=\rho_{d,\text{field}}(1+w)=2091(1.092)=2283\text{ kg/m}^3$.
Moist mass needed to fill 50 m$^3$ compacted. $m=\rho_{t,\text{field}}\times V=2283\times 50=114{,}170\text{ kg}=114.17\text{ Mg}$.
Check: assumes the soil is delivered and placed with no net bulking/shrinkage volume change beyond compaction itself (no swell factor given in the question) — i.e., truck-delivered moist mass equals the moist mass that ends up compacted in place.
Trucks required. $n=114.17/10=11.42$ → round UP to a whole truckload:
$$\boxed{n = 12\text{ truckloads}}$$
