Given conditions:

$\frac{W}{C}=0\mathrm{.}45$

Bulk volume of coarse aggregate per unit volume $(1y{d}^p)$ of concrete $=0\mathrm{.}67$

$4\mathrm{.}5\%$ Air

$7$ sack mix with $25\%$ Fly ash $)$; the weight of one sack $=94lb$

Find:

Weight of Cement:

Weight of Fly Ash: lb

Weight of Water $($before moisture correction$)$:

b

Weight of Coarse aggregate $($before moisture correction$)$:

Volume of Cement cubic $ft$

Volume of Fly ash cubic $ft$

Volume of water: cubic $ft$

Volume of air: cubic $ft$

Volume of coarse aggregate: cubic $ft$

Volume of Fine aggregate: cubic $ft$

Weight of Fine aggregate $($before moisture correction$)$ lb

Weight of coarse aggregate after moisture correction:

Weight of fine aggregate after moisture correction:

Weight of water after moisture correction