AGGREGATES - Materials for Civil and Construction Engineers - page 171

■

5.5.6 Bulk Unit Weight and Voids in Aggregate

The bulk unit weight of aggregate is needed for the proportioning of port-
land cement concrete mixtures. According to ASTM C29 procedure, a rigid
container of known volume is filled with aggregate, which is compacted ei-
ther by rodding, jigging, or shoveling. The bulk unit weight of aggregate
is determined as

1 g b 2

W s
V

g b = (5.13)

where W s is the weight of aggregate (stone) and V is the volume of the

container.
If the bulk dry specific gravity of the aggregate (ASTM C127 or
C128) is known, the percentage of voids between aggregate particles can be
determined as follows:

1 G sb 2

V s
V *

W/g s
W/g b

g b
g s

g b
G sb # g w

%V s = 100 = * 100 = * 100 = * 100

%Voids = 100 - %V s (5.14)

where

V s = volume of aggregate

weight of aggregate
unit weight of aggregate
weight of water

g s = unit

g b = bulk

g w = unit

Sample Problem 5.2

Coarse aggregate is placed in a rigid bucket and rodded with a tamping rod to deter-
mine its unit weight. The following data are obtained:
Volume of
Weight of empty
Weight of bucket filled with dry rodded coarse

1/3 ft 3 bucket =

bucket = 18.5 lb

aggregate = 55.9 lb

a. Calculate the dry-rodded unit weight
b. If the bulk dry specific gravity of the aggregate is 2.630, calculate the percent
voids in the aggregate.

Solution

112.3 lb/ft 3 a. Dry - rodded unit weight = 1 55.9 - 18.5 2 /0.333 =

112.3
2.630

b. Percent volume of particles =

62.3 *

100 = 68.5 %

*

Percent voids = 100 - 68.5 = 31.5 %