1. SOIL PHASE RELATIONSHIPS, INDEX PROPERTIES AND
CLASSIFICATION
Lecture
2
Geol 4111– Fundamentals of soil and rock mechanics
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2. SOIL PHASE RELATIONSHIPS:
Soil mass is generally a three phase system.
It consists of solid particles, liquid and gas.
For all practical purposes, the liquid may be considered to be
water (although in some cases, the water may contain some
dissolved salts) and the gas as air.
The phase system may be expressed in SI units either in terms of
mass-volume or weight-volume relationships.
The inter relationships of the different phases are important since
they help to define the condition or the physical make-up of the
soil.
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4. Wa= weight of air ~ 0
Ww= weight of water
Ws= weight of solids
W= total weight of the soil
Ma= mass of air ~0
Mw= mass of water in the voids
Ms= mass of solids
M= Total mass of the soil
..eq..1 ….eq..2
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5. Volumetric Ratios
The volume relationships commonly used for the three phases in a soil
element are void ratio, porosity, and degree of saturation.
Void ratio (e) is defined as the ratio of the volume of voids to the volume of
solids.
Thus, ……………..eq..(3)
Porosity (n) is defined as the ratio of the volume of voids to the total volume, or
…………….eq…(4)
The degree of saturation (S) is defined as the ratio of the volume of water to the volume
of voids, or
…………….eq….(5)
S = 0%, the soil is completely dry,
-S = 100%, the soil is fully saturated
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7. Specific Gravity
•The specific gravity of solids, Gs, (excluding air and water) is expressed by
specific gravity of the materials making up the soil particles are close to the value
for quartz, that is
Gs =~ 2.65
For all the common soil forming minerals 2.5 < Gs < 2.9
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15. Example
In its natural state, a moist soil has a volume of 9.34 x10-3m3
and weighs 177.6 x10-3 kN. The oven-dried weight of the soil
is 153.6 x10-3 kN. If Gs = 2.67, calculate
a. Moisture content (%)
b. Moist unit weight (kN/m3)
c. Dry unit weight (kN/m3)
d. Void ratio
e. Porosity
f. Degree of saturation (%)
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21. Example
For a given sandy soil, emax 0.75 and emin 0.4. Let Gs 2.68. In
the field, the soil is compacted to a moist density of 17.6
KN/m3 at a moisture content of 12%.
• Determine the relative density of compaction.
• Solution
• First, calculate void ratio of sandy soil under in-situ state:
• ϒ=
1+𝑤 𝐺𝑠𝛾𝑤
1+𝑒
𝑒 =
𝐺𝑠𝛾𝑤 1+𝑤
𝛾
− 1,
• 2.68 ∗
9.8 1+0.12
17.6
− 1 =0.67
• 𝐷𝑟 =
𝑒𝑚𝑎𝑥−𝑒
𝑒𝑚𝑎𝑥−𝑒𝑚𝑖𝑛
=
0.75−0.67
0.75−0.4
= 0.229 = 22.9%
• The soil nature is loose
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34. 2/22/2021 35
For a given soil, the following are known:
• Percentage passing through No. 4 sieve 70%
• Percentage passing through No. 200 sieve 30%
• Liquid limit 33
• Plastic limit 12
Classify the soil using the Unified Soil Classification System. Give the
group symbol and the group name.
Example
Solution
The percentage passing No. 200 sieve is 30%, which is less than 50%. So it is a coarse-
grained soil. Thus
Hence, more than 50% of the coarse fraction is passing No. 4 sieve. Thus, it is a
sandy soil. Since more than 12% is passing No. 200 sieve, it is SM or SC.
For this soil, PI = 33 -12 = 21 (which is greater than 7). With LL = 33 and PI =
21, it plots above the A-line in Figure 5.3. Thus the group symbol is SC.
39. Following are some rules for determining the group index:
1. If Eq. (1) yields a negative value for GI, it is taken as 0.
2. The group index calculated from Eq. (1) is rounded off to the nearest whole
number (for example, GI = 3.4 is rounded off to 3; GI = 3.5 is rounded off to 4).
3. There is no upper limit for the group index.
4. The group index of soils belonging to groups A-1-a, A-1-b, A-2-4, A-2-5, and
A-3 is always 0.
5. When calculating the group index for soils that belong to groups A-2-6 and A-
2-7, use the partial group index for PI, or
…..2
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45. Assignment 1
• A sample of wet silty clay soil has a mass of 126 kg.
The following data were obtained from laboratory
tests on the sample:
• Wet density, pt = 2.1 g/cm3, G = 2.7, water content,
w = 15%.
? Determine (i) dry density, pd, (ii) porosity, n, (iii)
void ratio, e, and (iv) degree of saturation, S.
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Submission date: next Tuesday/21/10/19
