6. Calculate the absolute volumes. For one cubic yard of air-entrained concrete, the
volume of the air can be determined by simply multiplying the air content by 27. For this
mixture, the air content from Step 1 above is 5 percent; therefore, the volume of air is
0.05 x 27 = 1.35 cubic feet.
For the cement, water, and coarse aggregate, the absolute volumes can be calculated
using the following equation:
where:
W =
G =
62.4 =
weight of the material
specific gravity of the material
weight of water per cubic foot
By substitution into this formula, the absolute volumes of the cement, water, and coarse
aggregate are calculated as follows:
l
l
l
7.
Volume of cement (W = 610 pounds and G = 3.15)
= 610 ÷ (3.15 x 62.4) = 3.10 cubic feet
Volume of water (W = 305 pounds and G = 1)
= 305 ÷ (1 x 62.4) = 4.89 cubic feet
Volume of coarse aggregate (W = 1,735 pounds and G = 2.60)
= 1,735 ÷ (2.60 x 62.4) = 10.69 cubic feet
Determine the fine aggregate content. To determine the weight of the fine aggregate
needed for a cubic yard of the oncrete, you first need to add together the volumes obtained
in Step 6 above. The resulting sum is then subtracted from 27 cubic feet to obtain the volume
of the fine aggregate in a cubic yard of the concrete. This is shown as follows:
Cement
= 3.10 cubic feet
Water
= 4.89 cubic feet
Coarse aggregate
= 10.69 cubic feet
Air
= 1.35 cubic feet
= 20.03 cubic feet
Absolute volume of
= 27 – 20.03
fine aggregate
= 6.97 cubic feet
Now, having calculated the volume of the fine aggregate and having been given its
specific gravity, you can use the formula shown in Step 6 above to solve for the weight of
the fine aggregate as follows:
Weight of fine aggregate = 6.97 x 2.65 x 62.4
= 1,152 pounds
17-7
