Page 191 - Geotechnical Engineering Soil and Foundation Principles and Practice

Soil Density and Unit Weight
186 Geotechnical Engineering

Figure 9.4
A block diagram
makes
calculations easy
and most density
formulas
superfluous: fill in
the known values
and calculate the
unknowns.

solids per unit volume, 105 lb, was calculated. The difference is the weight of the water,
19 lb. The respective unit weights for the solids and for the water are entered into the
appropriate boxes. From the formula at the top, V s ¼ 105/(3.70)(62.4) ¼ 0.62, which may
be written on the diagram. As the total volume is 1.0, the volume of voids is V v ¼ 1.0 –
0.62 ¼ 0.38. If that volume is filled with water, its weight will be W w ¼ 0.38 62.4 ¼ 23.5 lb,
3
and the saturated unit weight will be W sat ¼ 105 þ 23.5 ¼ 128 lb/ft .
Question: What is the void ratio in the previous example?

Answer: The calculated values for V s ¼ 0.62 and V v ¼ 0.38 should be written at
the left on the block diagram. Then e ¼ 0.38/0.62 ¼ 0.61.

9.4.2 Submerged Unit Weight

Unit weight is greatly influenced by the position of the groundwater table, which
submerges and buoys up the soil. To illustrate this concept, imagine weighing a
bucket of sand and then re-weighing it after it is suspended by a rope in water. It
was Archimedes who argued that the reduction in weight equals the weight of the
water displaced. Once a submerged unit weight has been calculated, determining a
submerged unit weight is simplest of all. Simply subtract the unit weight of water:

sub ¼
sat
w ð9:8Þ

Example 9.6
What is the submerged unit weight from the preceding example?
3
Answer:
sub ¼ 128 – 62.4 ¼ 65.5 lb/ft .

Note that in order to calculate a submerged unit weight the saturated unit weight
must be calculated first, unless for some indefensible reason a soil that is
submerged is assumed to be unsaturated.

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