Free water in an aggregate sample, and HMA volumetrics: 16-Civ-B7 Transportation Planning and Engineering, December 2014, Q7

Question 7: Free water in an aggregate sample, and HMA volumetrics (5 + 15 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. Part (a): a wet aggregate sample weighing 300.0 N, the same sample oven-dry at 280.0 N, and an aggregate absorption capacity of 4.0 percent. Part (b): the mixture properties tabulated above, with $G_{se} = 2.724$, $G_b = 1.028$, $G_{mb} = 2.346$, $G_{mm} = 2.516$, binder content $P_b = 5.5$ percent by total mass and material passing the 75 micrometre sieve $P_{0.075} = 4.0$ percent.

Find. (a) The free (surface) water as a percentage; (b) the bulk density of the compacted mixture, the air void content, the voids in the mineral aggregate, the voids filled with asphalt and the dust proportion.

Approach. For part (a), split the total moisture into the part the aggregate has absorbed into its pores and the part carried on the particle surfaces; the surface fraction is the free water. For part (b), work in mass fractions of the total mixture and convert to volume fractions through the specific gravities, obtaining first the air voids from the two mixture gravities, then VMA from the aggregate content and its bulk gravity, then VFA and the dust proportion.

Part (a) — Free water

Part (a) — find the total water in the sample. The difference between the wet and the oven-dry weights is all of the water present, both absorbed and free: $$W_w = 300.0 - 280.0 = 20.0\ \text{N}$$ Expressed as a total moisture content on the oven-dry basis, which is the convention for aggregates, $$w_{\text{total}} = \frac{20.0}{280.0}\times100 = 7.143\ \text{percent}$$
Find the absorbed water. Absorption is a property of the aggregate itself and is defined on the oven-dry mass, so at full saturation the pores hold $$W_{\text{abs}} = 0.040 \times 280.0 = 11.20\ \text{N}$$ Since the sample is wet, the aggregate is certainly saturated, and this full absorption capacity is taken up.
Subtract to obtain the free water. Whatever is not in the pores is on the surfaces: $$W_{\text{free}} = 20.0 - 11.20 = \boxed{8.80\ \text{N}}$$ As a percentage on the oven-dry aggregate mass, the basis on which batching corrections are made, $$w_{\text{free}} = \frac{8.80}{280.0}\times100 = \boxed{3.14\ \text{percent}}$$ which is the same as the difference between the total moisture and the absorption, $7.143 - 4.0 = 3.14$ percent. Read literally as a fraction of the original wet sample, as the question words it, the free water is $8.80/300.0 = \boxed{2.93\ \text{percent}}$.

Check — which basis the question wants. The question asks for “the percent of free water in the original wet sample”. Both readings are given above because the phrase is genuinely ambiguous, and both are one line of arithmetic apart. The oven-dry basis, 3.14 percent, is the engineering answer: it is the number a batch plant uses to correct mixing water and aggregate weights, and it is what “free moisture” means in CSA A23.1 and ASTM C70. The wet basis, 2.93 percent, is the literal reading of the sentence. State the basis on the answer paper and either is defensible.

Part (b) — Volumetric properties of the compacted mixture

Figure 7.1 — The compacted specimen resolved into its volume fractions. VMA is the air plus the effective binder, that is everything between the aggregate particles; VFA is the share of that intergranular space occupied by binder rather than air.

Part (b) — compute the bulk density. The bulk specific gravity of the compacted mixture is its density relative to water, so $$\rho_{\text{mix}} = G_{mb}\,\rho_w = 2.346 \times 1000 = \boxed{2346\ \text{kg/m}^3}$$ that is 2.346 Mg/m$^3$, or 146.5 lb/ft$^3$. Using the AASHTO convention of water at 25 degrees Celsius, $\rho_w = 997.0$ kg/m$^3$, gives 2339 kg/m$^3$; the difference is immaterial at engineering precision.
Compute the air voids in the total mix. The maximum specific gravity $G_{mm}$ is that of the same mixture with no air at all, so the air void content is the proportional shortfall of the compacted gravity below it: $$V_a = 100\,\frac{G_{mm}-G_{mb}}{G_{mm}} = 100\,\frac{2.516-2.346}{2.516} = 100\,\frac{0.170}{2.516} = \boxed{6.76\ \text{percent}}$$ This is well above the 4 percent design target, so the specimen is under-compacted relative to a design condition — a realistic figure for a field core taken shortly after paving.
Compute the voids in the mineral aggregate. VMA is the volume of the intergranular void space, air plus effective binder, expressed as a percentage of the bulk volume of the compacted specimen: $$\mathrm{VMA} = 100 - \frac{G_{mb}\,P_s}{G_{sb}}$$ where $P_s = 100 - P_b = 94.5$ percent is the aggregate content by mass. The question supplies the effective aggregate gravity $G_{se}$ but not the bulk gravity $G_{sb}$, which is what this formula requires; taking $G_{sb} = G_{se} = 2.724$, as discussed in the callout below, $$\mathrm{VMA} = 100 - \frac{2.346 \times 94.5}{2.724} = 100 - \frac{221.697}{2.724} = 100 - 81.386 = \boxed{18.61\ \text{percent}}$$
Compute the voids filled with asphalt. VFA is the percentage of the VMA that the effective binder occupies, the remainder being air: $$\mathrm{VFA} = 100\,\frac{\mathrm{VMA}-V_a}{\mathrm{VMA}} = 100\,\frac{18.61-6.76}{18.61} = \boxed{63.70\ \text{percent}}$$ The effective binder therefore occupies $18.61 - 6.76 = 11.86$ percent of the specimen volume, which is the middle band of Figure 7.1.
Compute the dust proportion. The dust proportion is the ratio of the material passing the 75 micrometre sieve to the effective binder content, both by mass of total mixture: $$\begin{aligned} \mathrm{DP} &= \frac{P_{0.075}}{P_{be}} \\ P_{be} &= P_b - \frac{P_{ba}}{100}P_s \end{aligned}$$ With $G_{sb}$ taken equal to $G_{se}$ the absorbed binder content $P_{ba}$ is zero, so $P_{be} = P_b = 5.5$ percent and $$\mathrm{DP} = \frac{4.0}{5.5} = \boxed{0.73}$$ This sits comfortably inside the Superpave acceptance range of 0.6 to 1.2, so the mixture is neither dust-starved nor over-filled with fines.
Assess the results against the criteria. Setting the five results beside the usual acceptance limits for a 12.5 mm nominal maximum size surface mixture, the VMA of 18.6 percent is generously above the 14 percent minimum and the dust proportion is central in its range, but the air void content of 6.8 percent is well above the 3 to 5 percent design band and the VFA of 63.7 percent is correspondingly just below the 65 to 75 percent band. Both departures have the same cause and the same remedy: the specimen simply has too much air in it. Additional compaction that brought $V_a$ down to 4.0 percent would raise $G_{mb}$ to about 2.415, lower the VMA to about 16.2 percent and raise the VFA to roughly 75 percent, placing every property within its limits. The mixture design is sound; it is the density that is deficient.

Check — two data issues, both declared under Note 2. First, the question gives $G_{se}$ but the VMA, VFA and dust-proportion formulas all require the bulk aggregate gravity $G_{sb}$, which is not supplied. Taking $G_{sb} = G_{se}$ makes the absorbed binder $P_{ba}$ zero and therefore makes VMA an upper bound. The sensitivity is modest and can be quoted: if $G_{sb}$ were 1 percent below $G_{se}$, at 2.697, then $P_{ba} = 0.38$ percent, VMA falls to 17.79 percent, VFA to 62.0 percent and DP rises to 0.78; if $G_{sb}$ were 2 percent below, VMA falls to 16.95 percent, VFA to 60.1 percent and DP to 0.84. The conclusions of Step 6 are unchanged under either. Second, the two supplied mixture gravities are not quite self-consistent: from $G_{se} = 2.724$ the maximum specific gravity should be $G_{mm} = 100/(P_s/G_{se} + P_b/G_b) = 2.497$, against the 2.516 printed, a discrepancy of 0.75 percent; equivalently, the printed $G_{mm}$ implies $G_{se} = 2.748$. The measured $G_{mm}$ has been used for the air voids, which is standard practice because $G_{mm}$ is determined directly by the Rice test, while $G_{se}$ has been used where the formula calls for an aggregate gravity. Had $G_{se} = 2.748$ been used instead, VMA would rise to 19.31 percent and VFA to 65.0 percent, exactly on the lower limit of the acceptance band rather than just below it. The air void content is untouched by either choice, because it depends only on the two measured mixture gravities, so the finding of Step 6 — that the specimen is under-compacted and that this is what drags the VFA down — stands under every reading of the data.

Quantity	Value
(a) Total water in the sample	20.0 N (7.14 percent of oven-dry mass)
(a) Absorbed water	11.20 N (4.0 percent)
(a) Free water	8.80 N — 3.14 percent of oven-dry, 2.93 percent of the wet sample
(b) Bulk density of the compacted mixture	2346 kg/m³ (146.5 lb/ft³)
(b) Air voids, V_a	6.76 percent
(b) Voids in mineral aggregate, VMA	18.61 percent
(b) Voids filled with asphalt, VFA	63.70 percent
(b) Dust proportion, DP	0.73

Topic. Aggregate moisture states, and the volumetric analysis of a compacted hot-mix asphalt specimen.

Key relations. Free moisture $=$ total moisture $-$ absorption; $V_a = 100(G_{mm}-G_{mb})/G_{mm}$; $\mathrm{VMA} = 100 - G_{mb}P_s/G_{sb}$; $\mathrm{VFA} = 100(\mathrm{VMA}-V_a)/\mathrm{VMA}$; $P_{ba} = 100\,G_b(G_{se}-G_{sb})/(G_{sb}G_{se})$; $P_{be} = P_b - P_{ba}P_s/100$; and $\mathrm{DP} = P_{0.075}/P_{be}$.

Why this works. Aggregate exists in four moisture states — oven dry, air dry, saturated surface dry and wet — and absorption is the water content at the saturated-surface-dry condition, the boundary between water inside the particle and water on it. That distinction matters because absorbed water is chemically and mechanically part of the aggregate, taking no part in the mixture, while free water is available to change the water-cement ratio of a concrete or to cause foaming and stripping in a hot mix. The volumetric analysis in part (b) rests on the same inside-versus-outside distinction, applied to binder instead of water. Three specific gravities describe the aggregate: bulk, which includes all permeable pores; apparent, which excludes them; and effective, which lies between the two and counts only those pores the asphalt does not penetrate. VMA must be computed on the bulk gravity because it is a real, physical intergranular void, while the absorbed binder $P_{ba}$ is precisely the difference the effective gravity captures. When $G_{sb}$ and $G_{se}$ coincide, no binder has been absorbed and all of it is effective, which is the assumption made here and the reason the VMA obtained is an upper bound.

Common pitfall. In part (a) the standard error is to report the total moisture, 7.14 percent, as the free water, which double-counts the absorbed fraction and would lead a batch plant to under-add water by more than half. In part (b) the equivalent error is to substitute $G_{se}$ or $G_{mm}$ into the VMA formula without noticing that the formula demands $G_{sb}$, and then to report the result with no note that an assumption has been made. A third frequent error is computing the dust proportion on the total binder content $P_b$ rather than on the effective binder $P_{be}$; the two coincide only when absorption is zero, as here, and diverge sharply for an absorptive aggregate.

Source. Asphalt Institute, Asphalt Mix Design Methods, MS-2, 7th ed., Ch. 4 (volumetric relationships and the full set of gravity definitions); Mamlouk and Zaniewski, Materials for Civil and Construction Engineers, 4th ed., Chs. 5 and 9; ASTM C127 and C128 for aggregate specific gravity and absorption; CSA A23.1 for the Canadian free-moisture batching convention.

Question 7 of 7: Free water in an aggregate sample, and HMA volumetrics

Question 7: Free water in an aggregate sample, and HMA volumetrics (5 + 15 marks)

Part (a) — Free water

Part (b) — Volumetric properties of the compacted mixture