Question 1 of 6: Aggregate Base Material — Moisture Content and Compacted Volume

Question 1: Aggregate Base Material — Moisture Content and Compacted Volume (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 wet mass to purchase from the stockpile for order 1; (b) the compacted volume produced by order 2 (same purchased mass, higher stockpile moisture), compared with order 1.

Approach. The dry unit weight is a soil-mechanics unit weight (force per volume, $g=9.81\ \text{m/s}^2$); convert the required dry WEIGHT to a dry MASS, then use the gravimetric moisture content $w=M_w/M_s$ to scale from dry mass to the wet (as-purchased) mass, and back again for order 2.

1. Part (a) — convert the required compacted volume to a dry mass. The dry weight of solids needed to fill $V$ at $\gamma_d$ is $$W_{dry}=\gamma_d V = (20\ \text{kN/m}^3)(800\ \text{m}^3)=16{,}000\ \text{kN}.$$ Converting weight to mass ($M=W/g$, with $W$ in newtons), $$M_{dry}=\frac{16{,}000{,}000\ \text{N}}{9.81\ \text{m/s}^2}=\boxed{1{,}630{,}989\ \text{kg}\ (\approx 1631\ \text{t})}.$$
2. Scale to the wet (purchased) mass using $w_1$. By definition $w=M_w/M_s$, so the as-purchased (wet) mass is $M_{wet}=M_{dry}(1+w)$: $$M_{wet,1}=1{,}630{,}989(1+0.08)=\boxed{1{,}761{,}468\ \text{kg}\ (\approx 1761\ \text{t})}.$$ This is the tonnage the contractor must purchase for order 1 to obtain exactly 800 m³ of compacted base.
3. Part (b) — order 2 purchases the same wet mass, $M_{wet,2}=M_{wet,1}=1{,}761{,}468$ kg, but the stockpile is now at $w_2=18\%$. The dry-solids mass actually delivered in that tonnage is lower, because a larger share of the same purchased weight is now water: $$M_{dry,2}=\frac{M_{wet,2}}{1+w_2}=\frac{1{,}761{,}468}{1.18}=\boxed{1{,}492{,}769\ \text{kg}}.$$
4. Convert the delivered dry mass back to a compacted volume at the same target $\gamma_d=20$ kN/m³. $$V_2=\frac{M_{dry,2}\,g}{\gamma_d}=\frac{(1{,}492{,}769)(9.81)/1000}{20}=\boxed{732.2\ \text{m}^3}.$$ Equivalently, since $M_{wet}$ is identical between the two orders, $V_2=V_1\cdot\dfrac{1+w_1}{1+w_2}=800\times\dfrac{1.08}{1.18}=732.2\ \text{m}^3$ — a direct ratio that needs no mass arithmetic at all.