6. Calculate the absolute volumes. For one cubic yard of air-entrained concrete, thevolume of the air can be determined by simply multiplying the air content by 27. For thismixture, the air content from Step 1 above is 5 percent; therefore, the volume of air is0.05 x 27 = 1.35 cubic feet.For the cement, water, and coarse aggregate, the absolute volumes can be calculatedusing the following equation:where:W =G =62.4 =weight of the materialspecific gravity of the materialweight of water per cubic footBy substitution into this formula, the absolute volumes of the cement, water, and coarseaggregate are calculated as follows:lll7.Volume of cement (W = 610 pounds and G = 3.15)= 610 ÷ (3.15 x 62.4) = 3.10 cubic feetVolume of water (W = 305 pounds and G = 1)= 305 ÷ (1 x 62.4) = 4.89 cubic feetVolume of coarse aggregate (W = 1,735 pounds and G = 2.60)= 1,735 ÷ (2.60 x 62.4) = 10.69 cubic feetDetermine the fine aggregate content. To determine the weight of the fine aggregateneeded for a cubic yard of the oncrete, you first need to add together the volumes obtainedin Step 6 above. The resulting sum is then subtracted from 27 cubic feet to obtain the volumeof the fine aggregate in a cubic yard of the concrete. This is shown as follows:Cement= 3.10 cubic feetWater= 4.89 cubic feetCoarse aggregate= 10.69 cubic feetAir= 1.35 cubic feet= 20.03 cubic feetAbsolute volume of= 27 – 20.03fine aggregate= 6.97 cubic feetNow, having calculated the volume of the fine aggregate and having been given itsspecific gravity, you can use the formula shown in Step 6 above to solve for the weight ofthe fine aggregate as follows:Weight of fine aggregate = 6.97 x 2.65 x 62.4= 1,152 pounds17-7