Question 4 of 7: Asphalt Mixture Volumetrics and Aggregate Relative Density

Question 4: Asphalt Mixture Volumetrics and Aggregate Relative Density (10 + 10 = 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.

Find. (a) the air voids V_a, the voids in the mineral aggregate VMA and the voids filled with binder VFB of the compacted sample; (b) the saturated surface-dry relative density and the absorption of the coarse aggregate.

Approach. Part (a): recover the effective aggregate relative density from the stated binder absorption, use it to compute the theoretical maximum relative density, and then take the three void quantities in the order V_a → VMA → VFB, closing the calculation against an independent effective-binder volume. Part (b): apply the buoyancy relations directly, the displaced volume being B − C.

1. Part (a) — recover the effective aggregate relative density. Absorbed binder is defined by $$P_{ba} = 100\,G_{b}\,\frac{G_{se} - G_{sb}}{G_{sb}\,G_{se}} \quad\Longrightarrow\quad \frac{1}{G_{sb}} - \frac{1}{G_{se}} = \frac{P_{ba}}{100\,G_{b}}$$ Substituting P_ba = 0.60 and G_b = 1.03, $$\frac{1}{G_{se}} = \frac{1}{2.67} - \frac{0.60}{103} = 0.374532 - 0.005825 = 0.368707$$ $$\boxed{G_{se} = 2.712}$$ The effective value exceeds the bulk value, as it must: sealing the surface pores with binder removes volume that water would otherwise enter.
2. Compute the theoretical maximum relative density. With the aggregate content P_s = 100 − 5.5 = 94.5 percent, $$G_{mm} = \frac{100}{\dfrac{P_{s}}{G_{se}} + \dfrac{P_{b}}{G_{b}}} = \frac{100}{\dfrac{94.5}{2.712} + \dfrac{5.5}{1.03}} = \frac{100}{34.843 + 5.340}$$ $$\boxed{G_{mm} = 2.489}$$
3. Compute the air voids. The air voids are the fraction of the compacted volume that the void-free mixture does not occupy: $$V_{a} = 100\,\frac{G_{mm} - G_{mb}}{G_{mm}} = 100 \times \frac{2.489 - 2.400}{2.489}$$ $$\boxed{V_{a} = 3.56\ \text{percent}}$$ This sits at the low end of the 3–5 percent design window for a dense-graded surface course.
4. Compute the voids in the mineral aggregate. VMA is the volume outside the aggregate particles themselves, and it uses the bulk aggregate relative density because the volume of the aggregate includes its own permeable pores: $$VMA = 100 - \frac{G_{mb}\,P_{s}}{G_{sb}} = 100 - \frac{2.400 \times 94.5}{2.67} = 100 - 84.94$$ $$\boxed{VMA = 15.06\ \text{percent}}$$ For a 12.5 mm nominal maximum aggregate size the Asphalt Institute minimum is 14 percent, so the mixture has adequate room for binder.
5. Compute the voids filled with binder. $$VFB = 100\,\frac{VMA - V_{a}}{VMA} = 100 \times \frac{15.06 - 3.56}{15.06}$$ $$\boxed{VFB = 76.3\ \text{percent}}$$
6. Close the calculation independently. The effective binder content and its volume are $$P_{be} = P_{b} - \frac{P_{ba}\,P_{s}}{100} = 5.5 - \frac{0.60 \times 94.5}{100} = 4.933\ \text{percent}$$ $$V_{be} = P_{be}\,\frac{G_{mb}}{G_{b}} = 4.933 \times \frac{2.400}{1.03} = 11.49\ \text{percent by volume}$$ and indeed VMA − V_a = 15.06 − 3.56 = 11.49 percent, so the three void quantities are mutually consistent. Interpreting the result: with V_a at 3.6 percent and VFB at 76 percent the mixture is slightly rich for heavy traffic (the Asphalt Institute heavy-traffic VFB band is 65–75 percent) and would be expected to be comfortable for medium traffic (65–78 percent).
7. Part (b) — compute the displaced volume. By Archimedes, the volume of the saturated surface-dry sample in millilitres is numerically $$B - C = 2029 - 1272 = 757\ \text{mL}$$
8. Compute the saturated surface-dry relative density. $$G_{SSD} = \frac{B}{B - C} = \frac{2029}{757}$$ $$\boxed{G_{SSD} = 2.680}$$
9. Compute the absorption. Absorption is the mass of water held in the permeable pores expressed against the oven-dry mass: $$\text{Absorption} = 100\,\frac{B - A}{A} = 100 \times \frac{2029 - 2016.1}{2016.1} = 100 \times \frac{12.9}{2016.1}$$ $$\boxed{\text{Absorption} = 0.64\ \text{percent}}$$
10. Quote the companion densities as a check. The same three masses give $$G_{bulk} = \frac{A}{B - C} = \frac{2016.1}{757} = 2.663, \qquad G_{app} = \frac{A}{A - C} = \frac{2016.1}{744.1} = 2.710$$ and the identity G_SSD = G_bulk(1 + absorption/100) = 2.663 × 1.0064 = 2.680 closes exactly. The bulk value 2.663 is also reassuringly close to the 2.67 used as G_sb in part (a), and the apparent value 2.710 is close to the G_se = 2.712 recovered there — the two halves of this question describe the same aggregate.

Final Results.