Page 190 - Geotechnical Engineering Soil and Foundation Principles and Practice

Soil Density and Unit Weight
Soil Density and Unit Weight 185

where in the block diagram V ¼ 1. Then
ð9:4Þ

w ¼ W s þ W w
Solving eq. (9.2) for W w gives
w
W w ¼ W s ð9:5Þ
100
where w is the moisture content in percent. Substituting in eq. (9.4) gives
ð

w ¼ W s þ W s w=100ð Þ ¼ W s 1 þ w=100Þ ¼
d 1 þ w=100Þ ð9:6Þ
ð

w

a ¼ ð9:7Þ
1þw=100
In other words, the dry unit weight of a soil is equal to its wet unit weight divided by
1 plus the moisture content expressed as a decimal fraction. This formula is used so
often that it should be practiced and committed to memory.

Example 9.4
Use eq. (9.7) to convert the wet unit weight in Example 9.1 to a dry unit weight.

Answer:

W 124 3

d ¼ ¼ ¼ 105 lb=ft
1þw=100 1 þ 0:183
The same formula, which is so astounding for its simplicity, also may be used to convert
3
wet to dry density: 1.99/(1 þ 0.183) ¼ 1.68 Mg/m , or unit weights in SI: 19.5/1.183 ¼
3
16.5 kN/m .
9.4 SATURATED AND SUBMERGED UNIT WEIGHT

9.4.1 Saturated Unit Weight
The soil unit weight as measured gives only part of the story because it is the unit
weight after the soil becomes saturated with water that may be most critical. This
can be calculated by use of the block diagram since at saturation the voids are
completely filled with water. The saturated unit weight is frequently encountered
in landslides.

Example 9.5
Calculate the saturated unit weight for Example 9.1 assuming that the specific gravity of
the mineral portion equals 2.70.
Answer: The simplest procedure is to write known quantities into a block diagram and go
from there. The known quantities are written as shown in Fig. 9.4. The wet unit weight or
weight per unit volume, 124 lb, was measured, and the dry unit weight, or weight of the

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