Permeability Test for Granular Soils (Constant-Head Method) - Soil Properties: Testing, Measurement, and Evaluation - page 261

Volume of specimen (15.5)(81.07) 1,256.6 cm 3

With M 1 and M 2 known, the mass of the soil specimen (air dried) can be
determined, and then its unit weight can be found:

Mass of soil specimen (air dried) 3,200.0 1,102.5 2,097.5 g

2,097.5
1,256.6

Unit weight of soil specimen
(air dried)

62.4

104.2 lb/ft 3

The water content of the air-dried soil can be computed as follows:

Mass of water 299.41 295.82 3.59 g

Mass of oven-dried soil 295.82 59.39 236.43 g

3.59
236.43

Water content =

* 100 = 1.52%

The dry unit weight of the soil specimen can then be found:

104.2
1.52 100 100 102.6 lb

ft 3 Dry unit weight

>

To determine the volume of solids and of voids to be used in computing
the void ratio, it is convenient to consider 1 ft 3 of dry soil. This cubic foot
of soil consists of only solid material and air. Because the weight of air
is negligible, the weight of the solid material will be 102.6 lb. The vol-
ume of solid material can be computed by dividing this weight by the
unit weight of the solid material. The unit weight of the solid material
can be obtained by multiplying specific gravity of the solids by the unit
weight of water (62.4 lb/ft 3 ). Hence,

102.6
( 2.71 )(62.4)

Volume of solids 0.6067 ft 3

Subtracting this volume from 1.0 ft 3 gives the volume of voids. Hence,

Volume of voids 1 0.6067 0.3933 ft 3

The void ratio e , which is the ratio of volume of voids to volume of solids,
can now be computed:

0.3933
0.6067

e 0.648