@JozveBartarOfficial

32 Chapter 2: Origin of Soil and Grain Size

Figure 2.16 Scanning electron micrograph showing the fabric of montmorillonite (Courtesy of
David J. White, Iowa State University, Ames, Iowa)

surface of the clay. Some partially hydrated cations in the pore water are also attracted to
the surface of clay particles. These cations attract dipolar water molecules. All these pos-
sible mechanics of attraction of water to clay are shown in Figure 2.19. The force of attrac-
tion between water and clay decreases with distance from the surface of the particles. All

Ϫ ϩϩϩ Ϫ

Ϫ ϩϪ ϩϩϩ

Ϫ ϩϩϪϪ Ϫ Concentration of ions

Ϫ ϩϩϩ Ϫ Cations

Ϫ ϩϩϩϪϩϪ Anions
Distance from the clay particle
Ϫϩ ϩϩϩϩ
(b)
Ϫ ϩϪ

ϩϪϩϪϩ
Surface of clay particle

(a)

Figure 2.17 Diffuse double layer

2.3 Clay Minerals 33

Oxygen

Hydrogen Hydrogen

105°

Figure 2.18 Dipolar character of water

the water held to clay particles by force of attraction is known as double-layer water. The
innermost layer of double-layer water, which is held very strongly by clay, is known as
adsorbed water. This water is more viscous than free water is.

Figure 2.20 shows the absorbed and double-layer water for typical montmorillonite
and kaolinite particles. This orientation of water around the clay particles gives clay soils
their plastic properties.

It needs to be well recognized that the presence of clay minerals in a soil aggregate has
a great influence on the engineering properties of the soil as a whole. When moisture is
present, the engineering behavior of a soil will change greatly as the percentage of clay
mineral content increases. For all practical purposes, when the clay content is about 50%
or more, the sand and silt particles float in a clay matrix, and the clay minerals primarily
dictate the engineering properties of the soil.

ϩϪ ϩϪ
Ϫ Dipolar water molecule

ϩϪ ϩ ϩ
ϩ Ϫϩ Ϫ

Ϫ ϩ Ϫϩ ϩ ϩϪ
ϩ

Ϫ Cation ϩ ϩ
ϩ

ϩ Ϫϩ Ϫ ϪDipolar water molecule
ϩϪ
Ϫ

ϩ ϩ ϩϪ
ϩ ϩϪ ϩ

ϪϪ ϩ
ϩϪ ϩϪ

ϩ ϩ
ϩ

Ϫϩ

Ϫ
Clay particle

Figure 2.19 Attraction of dipolar molecules in diffuse double layer

34 Chapter 2: Origin of Soil and Grain Size

200 Å
10 Å

200 Å

Typical montmorillonite particle, 1000 Å by 10 Å
(a)

400 Å
10ϩ Å

1000 Å

10ϩ Å
400 Å

Typical kaolinite particle, 10,000 Å by 1000 Å

(b)

Montmorillonite crystal Adsorbed water
Kaolinite crystal Double-layer water

Figure 2.20 Clay water (Redrawn after Lambe, 1958. With permission from ASCE.)

2.4 Specific Gravity (Gs)

Specific gravity is defined as the ratio of the unit weight of a given material to the unit weight
of water. The specific gravity of soil solids is often needed for various calculations in soil
mechanics. It can be determined accurately in the laboratory. Table 2.4 shows the specific
gravity of some common minerals found in soils. Most of the values fall within a range of 2.6
to 2.9. The specific gravity of solids of light-colored sand, which is mostly made of quartz,
may be estimated to be about 2.65; for clayey and silty soils, it may vary from 2.6 to 2.9.

2.5 Mechanical Analysis of Soil 35

Table 2.4 Specific Gravity of Common Minerals

Mineral Specific gravity, Gs

Quartz 2.65
Kaolinite 2.6
Illite 2.8
Montmorillonite 2.65–2.80
Halloysite 2.0–2.55
Potassium feldspar 2.57
Sodium and calcium feldspar 2.62–2.76
Chlorite 2.6–2.9
Biotite 2.8–3.2
Muscovite 2.76–3.1
Hornblende 3.0–3.47
Limonite 3.6–4.0
Olivine 3.27–3.7

2.5 Mechanical Analysis of Soil

Mechanical analysis is the determination of the size range of particles present in a soil,
expressed as a percentage of the total dry weight. Two methods generally are used to find
the particle-size distribution of soil: (1) sieve analysis—for particle sizes larger than 0.075
mm in diameter, and (2) hydrometer analysis—for particle sizes smaller than 0.075 mm in
diameter. The basic principles of sieve analysis and hydrometer analysis are described
briefly in the following two sections.

Sieve Analysis
Sieve analysis consists of shaking the soil sample through a set of sieves that have progressively
smaller openings. U.S. standard sieve numbers and the sizes of openings are given in Table 2.5.

Table 2.5 U.S. Standard Sieve Sizes

Sieve no. Opening (mm) Sieve no. Opening (mm)

4 4.75 35 0.500
5 4.00 40 0.425
6 3.35 50 0.355
7 2.80 60 0.250
8 2.36 70 0.212
10 2.00 80 0.180
12 1.70 100 0.150
14 1.40 120 0.125
16 1.18 140 0.106
18 1.00 170 0.090
20 0.850 200 0.075
25 0.710 270 0.053
30 0.600

36 Chapter 2: Origin of Soil and Grain Size
The sieves used for soil analysis are generally 203 mm (8 in.) in diameter. To con-

duct a sieve analysis, one must first oven-dry the soil and then break all lumps into small
particles. The soil then is shaken through a stack of sieves with openings of decreasing
size from top to bottom (a pan is placed below the stack). Figure 2.21 shows a set of
sieves in a shaker used for conducting the test in the laboratory. The smallest-sized sieve
that should be used for this type of test is the U.S. No. 200 sieve. After the soil is shaken,
the mass of soil retained on each sieve is determined. When cohesive soils are analyzed,
breaking the lumps into individual particles may be difficult. In this case, the soil may be
mixed with water to make a slurry and then washed through the sieves. Portions retained
on each sieve are collected separately and oven-dried before the mass retained on each
sieve is measured.
1. Determine the mass of soil retained on each sieve (i.e., M1, M2, · · · Mn) and in the

pan (i.e., Mp)
2. Determine the total mass of the soil: M1 ϩ M2 ϩ · · · ϩ Mi ϩ · · · ϩ Mn ϩ Mp ϭ ͚ M
3. Determine the cumulative mass of soil retained above each sieve. For the ith sieve, it

is M1 ϩ M2 ϩ · · · ϩ Mi

Figure 2.21 A set of sieves for a test in the laboratory (Courtesy of Braja M. Das, Henderson,
Nevada)

2.5 Mechanical Analysis of Soil 37

4. The mass of soil passing the ith sieve is ͚ M Ϫ (M1 ϩ M2 ϩ · · · ϩ Mi)
5. The percent of soil passing the ith sieve (or percent finer) is

͚M Ϫ 1M1 ϩ M2 ϩ p ϩ Mi 2
͚M
F ϭ ϫ 100

Once the percent finer for each sieve is calculated (step 5), the calculations are plot-
ted on semilogarithmic graph paper (Figure 2.22) with percent finer as the ordinate (arith-
metic scale) and sieve opening size as the abscissa (logarithmic scale). This plot is referred
to as the particle-size distribution curve.

Hydrometer Analysis

Hydrometer analysis is based on the principle of sedimentation of soil grains in water.
When a soil specimen is dispersed in water, the particles settle at different velocities,
depending on their shape, size, weight, and the viscosity of the water. For simplicity, it is
assumed that all the soil particles are spheres and that the velocity of soil particles can be
expressed by Stokes’ law, according to which

v ϭ rs Ϫ rw D2 (2.1)
18h

where v ϭ velocity

rs ϭ density of soil particles
rw ϭ density of water
h ϭ viscosity of water

D ϭ diameter of soil particles

100

80

Percent passing 60

40

20

0

10.0 5.0 1.0 0.5 0.1 0.05

Particle size (mm) — log scale

Figure 2.22 Particle-size distribution curve

38 Chapter 2: Origin of Soil and Grain Size
Thus, from Eq. (2.1),

D ϭ 18hv ϭ 18h L (2.2)
B rs Ϫ rw B rs Ϫ rw Bt (2.3)

where v ϭ Distance ϭ L .
Time t

Note that

rs ϭ Gs rw
Thus, combining Eqs. (2.2) and (2.3) gives

18h L (2.4)
D ϭ B 1Gs Ϫ 1 2 rw B t

If the units of h are (g и sec)/cm2, rw is in g/cm3, L is in cm, t is in min, and D is in mm,
then

D1 mm 2 18h 3 1g # sec2/cm24 L 1cm2
10 ϭ B 1Gs Ϫ 1 2 rw 1g/cm3 2 B t 1min 2 ϫ 60

or

30h L
Dϭ

B 1Gs Ϫ 1 2 rw B t
Assume rw to be approximately equal to 1 g/cm3, so that

L 1cm2 (2.5)
D 1mm2 ϭ KB t 1min2

where

30h (2.6)
Kϭ

B 1Gs Ϫ 1 2

Note that the value of K is a function of Gs and h, which are dependent on the temperature
of the test. Table 2.6 gives the variation of K with the test temperature and the specific grav-

ity of soil solids.





























































Problems 69

1.0

0.8

Void ratio range, emax Ϫ emin 0.6

emax Ϫ emin ϭ 0.23 ϩ 0.06
D50

0.4

0.2

0.0
0.1 1.0 10

Mean grain size, D50 (mm)

Clean sands (FC ϭ 0 – 5%)
Sands with fines (5 Ͻ FC Յ 15%)
Sands with clay (15 Ͻ FC Յ 30%, PC ϭ 5 – 20%)
Silty soils (30 Ͻ FC Յ 70%, PC ϭ 5 – 20%)
Gravelly sands (FC Ͻ 6%, PC ϭ 17 – 36%)
Gravels

Figure 3.9 Plot of emax Ϫ emin versus the mean grain size (Cubrinovski and Ishihara, 2002)

Problems

3.1 For a given soil, show that

␥sat ϭ ␥d + n␥w
3.2 For a given soil, show that

gsat ϭ gd ϩ a 1 e e b gw
ϩ

3.3 For a given soil, show that

gd ϭ eSgw
11 ϩ e2w

3.4 A 0.4-m3 moist soil sample has the following:

• Moist mass ϭ 711.2 kg

• Dry mass ϭ 623.9 kg

• Specific gravity of soil solids ϭ 2.68

Estimate:

a. Moisture content

b. Moist density

70 Chapter 3: Weight–Volume Relationships

c. Dry density
d. Void ratio
e. Porosity
3.5 In its natural state, a moist soil has a volume of 0.33 ft3 and weighs 39.93 lb.
The oven-dry weight of the soil is 34.54 lb. If Gs ϭ 2.67, calculate the moisture
content, moist unit weight, dry unit weight, void ratio, porosity, and degree of satu-
ration.
3.6 The moist weight of 0.2 ft3 of a soil is 23 lb. The moisture content and the specific
gravity of the soil solids are determined in the laboratory to be 11% and 2.7,
respectively. Calculate the following:
a. Moist unit weight (lb/ft3)
b. Dry unit weight (lb/ft3)
c. Void ratio
d. Porosity
e. Degree of saturation (%)
f. Volume occupied by water (ft3)
3.7 The saturated unit weight of a soil is 19.8 kN/m3. The moisture content of the soil
is 17.1%. Determine the following:
a. Dry unit weight
b. Specific gravity of soil solids
c. Void ratio
3.8 The unit weight of a soil is 95 lb/ft3. The moisture content of this soil is 19.2%
when the degree of saturation is 60%. Determine:
a. Void ratio
b. Specific gravity of soil solids
c. Saturated unit weight
3.9 For a given soil, the following are given: Gs ϭ 2.67; moist unit weight, ␥ ϭ 112 lb/ft3;
and moisture content, w ϭ 10.8%. Determine:
a. Dry unit weight
b. Void ratio
c. Porosity
d. Degree of saturation
3.10 Refer to Problem 3.9. Determine the weight of water, in pounds, to be added per
cubic foot of soil for:
a. 80% degree of saturation
b. 100% degree of saturation
3.11 The moist density of a soil is 1680 kg/m3. Given w ϭ 18% and Gs ϭ 2.73,
determine:
a. Dry density
b. Porosity
c. Degree of saturation
d. Mass of water, in kg/m3, to be added to reach full saturation
3.12 The dry density of a soil is 1780 kg/m3. Given Gs ϭ 2.68, what would be the
moisture content of the soil when saturated?
3.13 The porosity of a soil is 0.35. Given Gs ϭ 2.69, calculate:
a. Saturated unit weight (kN/m3)
b. Moisture content when moist unit weight ϭ 17.5 kN/m3

Problems 71

3.14 A saturated soil has w ϭ 23% and Gs ϭ 2.62. Determine its saturated and dry den-
sities in kg/m3.

3.15 A soil has e ϭ 0.75, w ϭ 21.5%, and Gs ϭ 2.71. Determine:
a. Moist unit weight (lb/ft3)
b. Dry unit weight (lb/ft3)
c. Degree of saturation (%)

3.16 A soil has w ϭ 18.2%, Gs ϭ 2.67, and S ϭ 80%. Determine the moist and dry unit
weights of the soil in lb/ft3.

3.17 The moist unit weight of a soil is 112.32 lb/ft3 at a moisture content of 10%. Given
Gs ϭ 2.7, determine:
a. e

b. Saturated unit weight
3.18 The moist unit weights and degrees of saturation of a soil are given in the table.

␥ (lb/ft3) S (%)

105.73 50
112.67 75

Determine:

a. e

b. Gs
3.19 Refer to Problem 3.18. Determine the weight of water, in lb, that will be in 2.5 ft3

of the soil when it is saturated.

3.20 For a given sand, the maximum and minimum void ratios are 0.78 and 0.43,
respectively. Given Gs ϭ 2.67, determine the dry unit weight of the soil in kN/m3
when the relative density is 65%.

3.21 For a given sandy soil, emax ϭ 0.75, emin ϭ 0.46, and Gs ϭ 2.68. What will be
the moist unit weight of compaction (kN/m3) in the field if Dr ϭ 78% and
w ϭ 9%?

3.22 For a given sandy soil, the maximum and minimum dry unit weights are 108
lb/ft3 and 92 lb/ft3, respectively. Given Gs ϭ 2.65, determine the moist unit
weight of this soil when the relative density is 60% and the moisture content

is 8%.

3.23 The moisture content of a soil sample is 18.4%, and its dry unit weight is 100
lb/ft3. Assuming that the specific gravity of solids is 2.65,

a. Calculate the degree of saturation.

b. What is the maximum dry unit weight to which this soil can be compacted

without change in its moisture content?

3.24 A loose, uncompacted sand fill 6 ft in depth has a relative density of 40%.

Laboratory tests indicated that the minimum and maximum void ratios of the

sand are 0.46 and 0.90, respectively. The specific gravity of solids of the sand

is 2.65.

a. What is the dry unit weight of the sand?

b. If the sand is compacted to a relative density of 75%, what is the decrease in

thickness of the 6-ft fill?

72 Chapter 3: Weight–Volume Relationships

References

AMERICAN SOCIETY FOR TESTING AND MATERIALS (2007). Annual Book of ASTM Standards,
Sec. 4, Vol. 04.08. West Conshohocken, Pa.

CUBRINOVSKI, M., and ISHIHARA, K. (1999). “Empirical Correlation Between SPT N-Value and
Relative Density for Sandy Soils,” Soils and Foundations. Vol. 39, No. 5, 61–71.

CUBRINOVSKI, M., and ISHIHARA, K. (2002). “Maximum and Minimum Void Ratio
Characteristics of Sands,” Soils and Foundations. Vol. 42, No. 6, 65–78.

LADE, P. V., LIGGIO, C. D., and YAMAMURO, J. A. (1998). “Effects of Non-Plastic Fines on
Minimum and Maximum Void Ratios of Sand,” Geotechnical Testing Journal, ASTM. Vol. 21,
No. 4, 336–347.

4 Plasticity and Structure of Soil

4.1 Introduction

When clay minerals are present in fine-grained soil, the soil can be remolded in the pres-
ence of some moisture without crumbling. This cohesive nature is caused by the adsorbed
water surrounding the clay particles. In the early 1900s, a Swedish scientist named
Atterberg developed a method to describe the consistency of fine-grained soils with vary-
ing moisture contents. At a very low moisture content, soil behaves more like a solid. When
the moisture content is very high, the soil and water may flow like a liquid. Hence, on an
arbitrary basis, depending on the moisture content, the behavior of soil can be divided into
four basic states—solid, semisolid, plastic, and liquid—as shown in Figure 4.1.

The moisture content, in percent, at which the transition from solid to semisolid state
takes place is defined as the shrinkage limit. The moisture content at the point of transition
from semisolid to plastic state is the plastic limit, and from plastic to liquid state is the liq-
uid limit. These parameters are also known as Atterberg limits. This chapter describes the
procedures to determine the Atterberg limits. Also discussed in this chapter are soil struc-
ture and geotechnical parameters, such as activity and liquidity index, which are related to
Atterberg limits.

Stress
Stress
Stress

Strain Strain Strain

Stress-strain diagrams at various states

Solid Semisolid Plastic Liquid Moisture
content
Shrinkage limit, S L Plastic limit, PL Liquid limit, L L increasing

Figure 4.1 Atterberg limits 73

74 Chapter 4: Plasticity and Structure of Soil

4.2 Liquid Limit (LL)

A schematic diagram (side view) of a liquid limit device is shown in Figure 4.2a. This
device consists of a brass cup and a hard rubber base. The brass cup can be dropped onto
the base by a cam operated by a crank. To perform the liquid limit test, one must place a soil

46.8 mm 54 mm
27 mm

Soil pat

(a)

50 mm

8
mm

27 11 mm 2 mm

(b)

Section

11 8 mm
mm Plan

2 mm

Figure 4.2 Liquid limit

12.7 mm test: (a) liquid limit device;

(b) grooving tool; (c) soil

pat before test; (d) soil pat

(c) (d) after test

4.2 Liquid Limit (LL) 75

paste in the cup. A groove is then cut at the center of the soil pat with the standard groov-
ing tool (Figure 4.2b). By the use of the crank-operated cam, the cup is lifted and dropped
from a height of 10 mm (0.394 in.). The moisture content, in percent, required to close a
distance of 12.7 mm (0.5 in.) along the bottom of the groove (see Figures 4.2c and 4.2d)
after 25 blows is defined as the liquid limit.

It is difficult to adjust the moisture content in the soil to meet the required
12.7 mm (0.5 in.) closure of the groove in the soil pat at 25 blows. Hence, at least three
tests for the same soil are conducted at varying moisture contents, with the number of
blows, N, required to achieve closure varying between 15 and 35. Figure 4.3 shows a
photograph of a liquid limit test device and grooving tools. Figure 4.4 shows photo-
graphs of the soil pat in the liquid limit device before and after the test. The moisture
content of the soil, in percent, and the corresponding number of blows are plotted on
semilogarithmic graph paper (Figure 4.5). The relationship between moisture content
and log N is approximated as a straight line. This line is referred to as the flow curve.
The moisture content corresponding to N ϭ 25, determined from the flow curve, gives
the liquid limit of the soil. The slope of the flow line is defined as the flow index and
may be written as

IF ϭ w1 Ϫ w2 (4.1)
log a N2 b
N1

Figure 4.3 Liquid limit test device and grooving tools (Courtesy of ELE International)

(a)

(b) Figure 4.4
Photographs showing
50 the soil pat in the liquid
limit device: (a) before
Flow curve test; (b) after test [Note:
45 The half-inch groove
closure in (b) is marked
Liquid limit ϭ 42 for clarification]
40 (Courtesy of Braja M.
Das, Henderson,
Nevada)

Moisture content (%) 35

Figure 4.5

Flow curve for

30 liquid limit

10 20 25 30 40 50 determination of

Number of blows, N (log scale) a clayey silt

76

4.2 Liquid Limit (LL) 77

where IF ϭ flow index
w1 ϭ moisture content of soil, in percent, corresponding to N1 blows
w2 ϭ moisture content corresponding to N2 blows

Note that w2 and w1 are exchanged to yield a positive value even though the slope of
the flow line is negative. Thus, the equation of the flow line can be written in a general

form as

w ϭ ϪIF log N ϩ C (4.2)

where C ϭ a constant.
From the analysis of hundreds of liquid limit tests, the U.S. Army Corps of

Engineers (1949) at the Waterways Experiment Station in Vicksburg, Mississippi, pro-
posed an empirical equation of the form

N tan b
25
LL ϭ wN a b (4.3)

where N ϭ number of blows in the liquid limit device for a 12.7 mm (0.5 in.) groove
closure

wN ϭ corresponding moisture content
tan b ϭ 0.121 (but note that tan b is not equal to 0.121 for all soils)

Equation (4.3) generally yields good results for the number of blows between 20 and 30.
For routine laboratory tests, it may be used to determine the liquid limit when only one test
is run for a soil. This procedure is generally referred to as the one-point method and was
also adopted by ASTM under designation D-4318. The reason that the one-point
method yields fairly good results is that a small range of moisture content is involved
when N ϭ 20 to N ϭ 30.

Another method of determining liquid limit that is popular in Europe and Asia
is the fall cone method (British Standard—BS1377). In this test the liquid limit is de-
fined as the moisture content at which a standard cone of apex angle 30Њ and weight of
0.78 N (80 gf) will penetrate a distance d ϭ 20 mm in 5 seconds when allowed to drop
from a position of point contact with the soil surface (Figure 4.6a). Due to the
difficulty in achieving the liquid limit from a single test, four or more tests can be con-
ducted at various moisture contents to determine the fall cone penetration, d. A semi-
logarithmic graph can then be plotted with moisture content (w) versus cone
penetration d. The plot results in a straight line. The moisture content corresponding to
d ϭ 20 mm is the liquid limit (Figure 4.6b). From Figure 4.6(b), the flow index can be
defined as

IFC ϭ w2 1 % 2 Ϫ w1 1 % 2 (4.4)
log d2 Ϫ log d1

where w1, w2 ϭ moisture contents at cone penetrations of d1 and d2, respectively.

78 Chapter 4: Plasticity and Structure of Soil

Weight, W ϭ 0.78 N
30°
Soil

d
40 mm

55 mm (a)
50

Moisture content, (%) 40
Liquid limit

30 Figure 4.6
10 20 40 60 80 100 (a) Fall cone test (b) plot of moisture

Penetration, d (mm) content vs. cone penetration for

(b) determination of liquid limit

4.3 Plastic Limit (PL)

The plastic limit is defined as the moisture content in percent, at which the soil

crumbles, when rolled into threads of 4.2 mm ( 1 in.) in diameter. The plastic limit is the
8

lower limit of the plastic stage of soil. The plastic limit test is simple and is performed

by repeated rollings of an ellipsoidal-sized soil mass by hand on a ground glass plate

(Figure 4.7). The procedure for the plastic limit test is given by ASTM in Test

Designation D-4318.

As in the case of liquid limit determination, the fall cone method can be used to

obtain the plastic limit. This can be achieved by using a cone of similar geometry but with

a mass of 2.35 N (240 gf). Three to four tests at varying moisture contents of soil are con-

ducted, and the corresponding cone penetrations (d) are determined. The moisture content

corresponding to a cone penetration of d ϭ 20 mm is the plastic limit. Figure 4.8 shows

the liquid and plastic limit determination of Cambridge Gault clay reported by Wroth and

Wood (1978).

4.3 Plastic Limit (PL) 79
2

4

Figure 4.7 Rolling of soil mass on ground glass plate to determine plastic limit (Courtesy of
Braja M. Das, Henderson, Nevada)

W ϭ 0.78 N
70

Liquid limit

Moisture content, (%) 60

Plastic limit Cone weight,
W ϭ 2.35 N

50

40 2 5 10 20 Figure 4.8
1 Cone penetration, d (mm) Liquid and plastic limits for
50 Cambridge Gault clay deter-
mined by fall cone test

The plasticity index (PI) is the difference between the liquid limit and the plastic limit
of a soil, or

PI ϭ LL Ϫ PL (4.5)

Table 4.1 gives the ranges of liquid limit, plastic limit, and activity (Section 4.6) of
some clay minerals (Mitchell, 1976; Skempton, 1953).

80 Chapter 4: Plasticity and Structure of Soil

Table 4.1 Typical Values of Liquid Limit, Plastic Limit, and Activity of Some Clay Minerals

Mineral Liquid limit, LL Plastic limit, PL Activity, A

Kaolinite 35–100 20–40 0.3–0.5
Illite 60–120 35–60 0.5–1.2
Montmorillonite 100–900 50–100 1.5–7.0
Halloysite (hydrated) 50–70 40–60 0.1–0.2
Halloysite (dehydrated) 40–55 30–45 0.4–0.6
Attapulgite 150–250 100–125 0.4–1.3
Allophane 200–250 120–150 0.4–1.3

Burmister (1949) classified the plasticity index in a qualitative manner as follows:

PI Description

0 Nonplastic
1–5 Slightly plastic
5–10 Low plasticity
10–20 Medium plasticity
20–40 High plasticity
Ͼ40 Very high plasticity

The plasticity index is important in classifying fine-grained soils. It is fundamental
to the Casagrande plasticity chart (presented in Section 4.7), which is currently the basis
for the Unified Soil Classification System. (See Chapter 5.)

Sridharan, et al. (1999) showed that the plasticity index can be correlated to
the flow index as obtained from the liquid limit tests (Section 4.2). According to
their study,

PI 1 % 2 ϭ 4.12IF 1 % 2 (4.6)

and

PI 1 % 2 ϭ 0.74IFC 1 % 2 (4.7)

In a recent study by Polidori (2007) that involved six inorganic soils and their respec-
tive mixtures with fine silica sand, it was shown that

PL ϭ 0.04(LL) + 0.26(CF) + 10 (4.8)

and

PI ϭ 0.96(LL) Ϫ 0.26(CF) Ϫ 10 (4.9)

where CF ϭ clay fraction (Ͻ2 ␮m) in %. The experimental results of Polidori (2007)
show that the preceding relationships hold good for CF approximately equal to or greater
than 30%.

4.4 Shrinkage Limit (SL) 81

4.4 Shrinkage Limit (SL)

Soil shrinks as moisture is gradually lost from it. With continuing loss of moisture, a stage
of equilibrium is reached at which more loss of moisture will result in no further volume
change (Figure 4.9). The moisture content, in percent, at which the volume of the soil mass
ceases to change is defined as the shrinkage limit.

Shrinkage limit tests (ASTM Test Designation D-427) are performed in the labora-
tory with a porcelain dish about 44 mm (1.75 in.) in diameter and about 12.7 mm (21 in.)
high. The inside of the dish is coated with petroleum jelly and is then filled completely
with wet soil. Excess soil standing above the edge of the dish is struck off with a straight-
edge. The mass of the wet soil inside the dish is recorded. The soil pat in the dish is then
oven-dried. The volume of the oven-dried soil pat is determined by the displacement of mer-
cury. Because handling mercury may be hazardous, ASTM D-4943 describes a method of
dipping the oven-dried soil pat in a melted pot of wax. The wax-coated soil pat is then
cooled. Its volume is determined by submerging it in water.

By reference to Figure 4.9, the shrinkage limit can be determined as

SL ϭ wi 1 % 2 Ϫ ¢w 1 % 2 (4.10)

where wi ϭ initial moisture content when the soil is placed in the shrinkage limit dish
⌬w ϭ change in moisture content (that is, between the initial moisture
content and the moisture content at the shrinkage limit)

However,

wi 1 % 2 ϭ M1 Ϫ M2 ϫ 100 (4.11)
M2

where M1 ϭ mass of the wet soil pat in the dish at the beginning of the test (g)
M2 ϭ mass of the dry soil pat (g) (see Figure 4.10)

Vi ∆

Volume of soil Vf

Shrinkage limit Plastic limit Liquid limit i

Moisture content (%)

Figure 4.9 Definition of shrinkage limit