Chapter 5, Problem 5.14QP

Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to determine its unit weight. The following data are obtained:

Volume of bucket = $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ ft³

Weight of empty bucket = 20.3 lb

Weight of bucket filled with dry rodded coarse aggregate:

Trial 1 = 76.6 lb

Trial 2 = 75.1 lb

Trial 3 = 78.8 lb

a. Calculate the average dry-rodded unit weight

b. If the bulk dry specific gravity of the aggregate is 2.620, calculate the percent voids between aggregate particles for each trial