Page 196 - Geotechnical Engineering Soil and Foundation Principles and Practice

Soil Density and Unit Weight
Soil Density and Unit Weight 191

Answer: The difference in weight is the weight of the water: 196 – 136 ¼ 59 g. The weight of
the dish, or tare weight, must be subtracted to obtain the dry weight of the soil:
136 – 15 ¼ 121 g. The moisture content is w ¼ 100 59/121 ¼ 49%.

Example 9.9
A cylindrical soil sample 75 mm in diameter by 100 mm long weighs 690 g. After oven-
drying it weighs 580 g. The specific gravity of the solid particles is G ¼ 2.70. What are the
(a) moisture content, (b) wet unit weight, (c) dry unit weight, (d) void ratio, (e) porosity, (f)
saturated unit weight, (g) submerged unit weight?
Answer: The first step is to write known or given quantities in a block diagram as shown at
the top in Fig. 9.6. The next step is to calculate and fill in the empty spaces as
shown sequentially by the arrows starting at the lower right in the lower part of the figure.
That is:
The weight of water equals the difference between the wet and oven-dry
weights, 690 – 580 ¼ 110 g.
The volume of solids equals the weight of solids divided by G s ¼ 580 7
3
2.70 ¼ 215 cm .
The volume of water equals the weight of water divided by G w ¼ 110 7
3
1.0 ¼ 110 cm .

Figure 9.6
Block diagram for
Example 9.9.

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