EM 1110-2-1100 (Part III)
30 Apr 02
EXAMPLE PROBLEM III-1-4
FIND:
The volume concentration, the porosity, the voids ratio, the dry bulk density, and the saturated bulk
density of the densely packed aggregate given in example problem III-1-2.
GIVEN:
18.1 grams of dry beach sand exactly fills a small container having a volume of 10.0 cm3 after the
filled container is strongly vibrated. When this amount of sand is poured into 50.0 ml of water that is
subjected to a strong vacuum, the volume of the sand-water mixture is 56.8 ml.
SOLUTION:
It is important to recognize the difference in the volume of the grains themselves, 6.8 cm3
(= 56.8 - 50) (a milliliter is a cubic centimeter), and the volume of the aggregate (the grains
plus the void spaces), 10 cm3.
The volume concentration is the ratio of solid volume to total volume:
N = 6.8 cm3 / 10.0 cm3 = 0.68
The porosity is the ratio of void space to total volume, which, from Equation 1-13 is:
P = 1 - N = 1 - 0.68 = 0.32
The voids ratio is the ratio of pore space to solid space, which,
from Equation 1-13 is:
e = P/N = 0.32/0.68 = 0.47
The dry bulk density from Equation 1-14 is:
Dry Bulk Density = 0.68 * 2,660 kg/m3 = 1,800 kg/m3
The saturated bulk density from Equation 1-15 is:
Saturated Bulk Density = (0.68*2,660) +(0.32*1,000) kg/m3 = 2,100 kg/m3
These bulk density values are within the expected range as shown in Table III-1-6A.
III-1-32
Coastal Sediment Properties
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