NivaarExam PrepOfficial exam papers ↗

16-Civ-B7 Transportation Planning and Engineering · May 2013

Question 5 of 7: Standard Proctor Compaction and the Zero-Air-Voids Line

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

Notes on this paper

Paper format. 98-Civ-B7 Highway Engineering, National Examinations May 2013 — a three-hour open-book examination; any non-communicating calculator is permitted. The cover page states that a total of five solutions is required, that only the first five as they appear in the answer book will be marked, and that all questions are of equal value. The grading scheme on the last page confirms 20 marks per question: Q1 (a) and (b) 10 marks each; Q2 (a) through (e) 4 marks each; Q3 (a) to (j) 2 marks each; Q4 (a) and (b) 10 marks each; Q5 (a) and (b) 10 marks each; Q6 (a) through (e) 4 marks each; Q7 20 marks. All seven printed questions are worked here, because this set is a study resource rather than a timed attempt; on exam day a candidate submits only the first five, in order. The paper also states that any data not given but required may be assumed, and that assumptions should be recorded with the answer — several questions below need that licence, and each assumption is flagged where it is made.

Reference texts. N.J. Garber and L.A. Hoel, Traffic and Highway Engineering, 5th ed. (geometric design, sight distance, vertical curves, earthwork, pavement design); AASHTO, Guide for Design of Pavement Structures (1993) (rigid and flexible thickness design, reliability, drainage and load-transfer coefficients); Transportation Association of Canada, Geometric Design Guide for Canadian Roads (Canadian design-domain values for sight distance and vertical curvature); Asphalt Institute, Mix Design Methods MS-2, 7th ed. (mixture volumetrics, VMA, VFA, absorbed binder); B.M. Das, Principles of Geotechnical Engineering, 9th ed. (compaction, Proctor testing, zero-air-voids line, CBR); M.S. Mamlouk and J.P. Zaniewski, Materials for Civil and Construction Engineers, 4th ed. (concrete and asphalt materials); A.M. Neville, Properties of Concrete, 5th ed., and CSA A23.1 (air entrainment, curing, joints in concrete pavement).

Check — assumptions carried through this paper. Three items are not supplied by the exam and are assumed under the paper’s own Note 2 (“any data, not given but required, can be assumed”), each stated again at the point of use: (i) Question 5 gives the mass of the Proctor mould but not its volume, so the ASTM D698 / AASHTO T99 standard 101.6 mm mould volume of 944 cm3 is used; (ii) Question 6 does not name a design speed, so the available stopping sight distance is computed from the Canadian/AASHTO eye and object heights of 1.08 m and 0.60 m; (iii) Question 7 lists the modulus of subgrade reaction as “1.0 MPa”, which is dimensionally incomplete — it is read as 1.0 MPa/m and the sensitivity of the answer to that reading is reported with the result.

Question 5: Standard Proctor Compaction and the Zero-Air-Voids Line (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. Six Standard Proctor trials on one soil, each reported as the gross mass of the compacted specimen plus its mould together with the moulding moisture content, plus a specific gravity of solids for part (b).

Given data — Standard Proctor trials (mass of mould = 2 450 g)
TrialSoil + mould (g)Moisture content w (%)
14 15010.0
24 21011.0
34 32012.0
44 42013.0
54 41014.0
64 36015.0
Specific gravity of solids, Gs2.80

Find. The dry density at every trial, the moisture-density curve and from it the optimum moisture content and maximum dry density, and the zero-air-voids line plotted on the same axes.

Check — assumed mould volume. The question gives the mass of the mould but not its volume, which is indispensable. Note 2 of the exam permits any required datum to be assumed, so the Standard Proctor mould of ASTM D698 / AASHTO T99 is adopted: 101.6 mm internal diameter by 116.4 mm high, V = 944 cm3 (1/30 ft3). Every density below scales inversely with this figure; the optimum moisture content, which is what part (a) is really testing, does not depend on it at all.

Approach. Subtract the mould mass to get the wet mass, divide by the mould volume for bulk density, divide by (1 + w) for dry density, plot dry density against moisture content and read the peak; then compute the theoretical dry density at full saturation for the same moisture contents to draw the zero-air-voids line.

  1. (a) Wet mass and bulk density. For each trial the compacted soil mass is the gross mass less the mould, and the bulk (wet) density follows from the assumed mould volume. Taking trial 1 as the worked example, $$M=4\,150-2\,450=1\,700\ \text{g},\qquad \rho=\frac{M}{V}=\frac{1\,700}{944}=1.801\ \text{g/cm}^3$$
  2. (a) Dry density. Removing the water mass, $$\rho_d=\frac{\rho}{1+w}=\frac{1.801}{1+0.100}=1.637\ \text{Mg/m}^3$$ and repeating for all six trials gives the table below. Note that 1 g/cm3 = 1 Mg/m3, and the corresponding dry unit weight is $\gamma_d=9.81\rho_d$ in kN/m3.
Computed densities
Trialw (%)Wet soil (g)ρ (Mg/m3)ρd (Mg/m3)ρd,ZAV (Mg/m3)
110.01 7001.8011.6372.188
211.01 7601.8641.6802.141
312.01 8701.9811.7692.096
413.01 9702.0871.8472.053
514.01 9602.0761.8212.011
615.01 9102.0231.7591.972
1011121314151.601.802.002.20Moulding moisture content w (%)Dry density (Mg/m³)zero air voids (S = 100 %)OMC 13.25 %, 1.850 Mg/m³Standard Proctor moisture-density curve and the zero-air-voids lineThe compaction curve never crosses the ZAV line; the gap is the residual air
Figure 5.1 — Moisture-density curve from the six trials with the zero-air-voids line for Gs = 2.80. The peak gives the optimum moisture content and the maximum dry density.
  1. (a) Locate the peak of the curve. The dry density rises to trial 4 and falls thereafter, so the optimum lies between 12 % and 14 %. Fitting a parabola through the three points that bracket the peak — (12, 1.769), (13, 1.847) and (14, 1.821) — and differentiating gives $$w_{opt}=13+\frac{(1.821-1.769)/2}{(2\times 1.847)-1.769-1.821} =\boxed{13.3\ \%}$$ $$\rho_{d,max}=\boxed{1.850\ \text{Mg/m}^3}\quad (\gamma_{d,max}=9.81\times 1.850=18.15\ \text{kN/m}^3)$$ Reading the peak off a carefully drawn curve by eye gives the same result to the precision the test deserves; the parabola is used here only so the answer is reproducible.
  2. (b) The zero-air-voids relation. At full saturation there is no air, so the total volume is just solids plus water. Writing the dry density in terms of $G_s$ and $w$, $$\rho_{d,ZAV}=\frac{G_s\,\rho_w}{1+w\,G_s}$$ which for $G_s=2.80$ at $w=10\ \%$ gives $$\rho_{d,ZAV}=\frac{2.80\times 1.000}{1+0.100\times 2.80}=\frac{2.80}{1.28} =2.188\ \text{Mg/m}^3$$ The same evaluation at 11, 12, 13, 14 and 15 % gives 2.141, 2.096, 2.053, 2.011 and 1.972 Mg/m3, the last column of the table and the dashed line on the plot.
  3. (b) Check the two curves against each other. The zero-air-voids line lies above the compaction curve everywhere and the gap narrows as the soil gets wetter, which is the behaviour the test must show. At the optimum the void ratio implied by the measured dry density is $$e=\frac{G_s\rho_w}{\rho_{d,max}}-1=\frac{2.80}{1.850}-1=0.5133$$ so the degree of saturation and the air content there are $$S=\frac{w\,G_s}{e}=\frac{0.1325\times 2.80}{0.5133}=\boxed{72.3\ \%},\qquad A_v=\frac{e(1-S)}{1+e}=\boxed{9.4\ \%}$$ About 9 % of the compacted volume is still air at maximum density, which is exactly why the compaction curve can never touch the zero-air-voids line.

The shape of the result is as informative as the numbers. On the dry side of optimum the soil is too stiff for the rammer to rearrange the particles, and on the wet side the water occupies volume that the rammer cannot expel in the time available, so the dry density passes through a maximum in between. Because the curve here is fairly flat near the peak — the dry density changes by less than 0.03 Mg/m3 between 12.5 % and 13.8 % — a field specification of 95 % of Standard Proctor could be met over a usefully wide moisture window, which is a favourable characteristic for a subgrade soil placed in variable Canadian weather.

QuantityValue
Assumed Standard Proctor mould volume944 cm3
(a) Optimum moisture content13.3 % (13.25 % from the fitted parabola)
(a) Maximum dry density1.850 Mg/m3 (γd = 18.15 kN/m3)
(b) Zero-air-voids densities, w = 10 to 15 %2.188, 2.141, 2.096, 2.053, 2.011, 1.972 Mg/m3
Void ratio at optimume = 0.513
Degree of saturation at optimumS = 72.3 %
Air content at optimum9.4 % by volume