A cubic yard of concrete is to contain 6 bags of cement and 39 gallons of water. The absolute volumes of the fine and coarse aggregates

Question

required are 6.86 cu ft and 11.67 cu ft respectively. The bulk specific gravities are 2.65 for the fine aggregate and 2.68 for the coarse aggregate.

 

 

Question: Determine the required weights of gthe cement, water, dry fine aggregate and dry coarse aggregate per cubic yard of concrete.

Answer 1

1.)  First, calculate the weight of water in 39 gallons:  Using the density equation ρ_H2O = mass (lbs)/volume (ft³)

 

where ρ_H2O = 8.33 lbs/gal  and V = 39gal

 

the,

 

Mass = ρ_H2O  (Vol)= 8.33 lbs/gal (39gal) = 324.87 lbs of water present

 

2.)  Next, calculate the fine aggregate weight:

 

First calculate the Density (ρFA) from the given specific Gravity (SG) of the fine agg.   SG_FA =  ρ_FA / ρ_H2O  =

 

 SG_FA (ρ_H2O ) =ρ_{FA =} 2.65 (62.4 lb/ft³) = 165.36 lb/ft³

 

Now, using the given volume of 6.86ft³ for the fine aggregate, we can now calculate the weight of fine aggregate using:

 

m (weight lbs) = ρ_FA/volume = 165.36 lb/ft³/6.86ft³ = 24 lbs

 

At this point, I hope that you have an idea how to continue.  I regret that I have no additional time to spend on this at the moment.  However, if you do need additional help from me, please feel free to look me up.  I will, however, definitely return at a later time to see if anyone else has come to your assistance.
 
Good luck