The Atterberg limits can be used to distinguish between silt and clay, and to distinguish between different types of silts and clays. The water content at which the soils change from one state to the other are known as consistency limits or Atterberg's limit.

1. Module II
Clay Mineralogy & Plasticity Characteristics of Soil
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2. Content
Wasim Shaikh, AIKTC

3. Formation of Clay Minerals
• A soil particle may be a mineral or a rock fragment.
• A mineral is a chemical compound formed in nature
during a geological process, whereas a rock fragment
has a combination of one or more minerals.
• Based on the nature of atoms, minerals are classified
as silicates, aluminates, oxides, carbonates and
phosphates.
• Out of these, silicate minerals are the most important
as they influence the properties of clay soils.
• Different arrangements of atoms in the silicate
minerals give rise to different silicate structures.
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Arrangment of Atoms in
Silicate Minerals (Extra)

4. Basic Structural Units
• Soil minerals are formed from two basic structural
units: tetrahedral and octahedral.
• A tetrahedral unit consists of a central silicon atom
that is surrounded by four oxygen atoms located at
the corners of a tetrahedron. A combination of
tetrahedrons forms a silica sheet.
• An octahedral unit consists of a central ion, either
aluminium or magnesium, that is surrounded by six
hydroxyl ions located at the corners of an
octahedron.
• A combination of aluminium-hydroxyl octahedrons
forms a gibbsite sheet, whereas a combination of
magnesium-hydroxyl octahedrons forms a brucite
sheet. Wasim Shaikh, AIKTC

5. Two-layer Sheet Minerals
Kaolinite and halloysite clay minerals are the most common.
• Kaolinite Mineral:
• The basic kaolinite unit is a two-layer unit that is formed by
stacking a gibbsite sheet on a silica sheet.
• These basic units are then stacked one on top of the other to form a
lattice of the mineral.
• The units are held together by hydrogen bonds. The strong bonding
does not permit water to enter the lattice. Thus, kaolinite minerals
are stable and do not expand under saturation.
• Kaolinite is the most abundant constituent of residual clay deposits.
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6. Three-layer Sheet Minerals
• Montmorillonite and illite clay minerals are the most common.
• A basic three-layer sheet unit is formed by keeping one silica sheet each
on the top and at the bottom of a gibbsite sheet.
• These units are stacked to form a lattice as shown in fig.
• Montmorillonite Mineral:
• The bonding between the three-layer units is by van der Waals forces.
• This bonding is very weak and water can enter easily. Thus, this mineral
can weak against a large quantity of water causing swelling.
• During dry weather, there will be shrinkage.
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7. Illite Mineral
• Illite consists of the basic montmorillonite units but are bonded by
secondary valence forces and potassium ions, as shown.
• There is about 20% replacement of aluminium with silicon in the gibbsite
sheet due to isomorphous substitution.
• This mineral is very stable and does not swell or shrink.
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8. Consistency of Clay & Atterberg Limits
• The consistency of a fine-grained soil refers to its firmness,
and it varies with the water content of the soil.
• A gradual increase in water content causes the soil to
change from solid to semi-solid to plastic to liquid states.
• The water contents at which the consistency changes from
one state to the other are called consistency limits (or
Atterberg limits).
• The three limits are known as the shrinkage limit (WS),
plastic limit (WP), and liquid limit (WL) as shown. The
values of these limits can be obtained from laboratory tests.
• Two of these are utilised in the classification of fine soils:
• Liquid limit (WL) - change of consistency from plastic to
liquid state
• Plastic limit (WP) - change of consistency from
brittle/crumbly to plastic state Wasim Shaikh, AIKTC

9. Atterberg Limits
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Liquid Limit (LL or wL): ‘Liquid limit’ is defined
as water content at which the soil is
just about to pass from the plastic state into the
liquid state.
Plastic Limit (PL or wP): ‘Plastic limit’ is defined
as water content at which the soil tends to pass
from the plastic state to the semi-solid state of
consistency.
Shrinkage Limit (SL or ws): ‘Shrinkage limit’ is
defined as water content at which the soil tends to
pass from the semi-solid to the solid state.
It is that water content at which a soil,
regardless,of further drying, remains
constant in volume

10. Consistency Indices
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p
l
p w
w
I 

Plasticity Index (PI or Ip): It is the difference between liquid and plastic limits.
• It is the range of water content within which the soil exhibits plastic
properties.
Plasticity index for sands is zero.

11. Shrinkage Index
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S
P
S w
w
I 

• ‘Shrinkage index’ (SI or Is) is defined as the difference between the plastic and
shrinkage limits of a soil
• In other words, it is the range of water content within which a soil is in a semi solid
state of consistency.
Consistency Index: Consistency index’ or ‘Relative consistency’ (CI or Ic) is defined as the ratio of the
difference between liquid limit and the natural water content to the plasticity index of a soil
P
l
C
I
w
w
I


where w = natural water content of the soil (water content of a soil in the undisturbed condition in the ground).

12. Liquidity Index
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• ‘Liquidity index (LI or IL)’ or ‘Water-plasticity ratio’ is the ratio of the difference
between the natural water content and the plastic limit to the plasticity index:
P
P
L
I
w
w
I


1

 LI
CI
Classification of soil based on consistancy
State of soil

13. Particle size distribution
Semi Log Paper is required
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14. Laboratory Methods for the Determination of Consistency
Limits and Related Indices
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Determination of Liquid Limit
Determination of Plastic Limit
Determination of Shrinkage Limit

15. Determination of Liquid Limit
• The Liquid Limit (LL) is the water content at which the
soil changes from the liquid state to a plastic state.
• It is the minimum moisture content at which a soil flows
upon application of very small shear force.
• Liquid Limit can be determined using the Casagrande cup
method or a cone penetrometer.
• Liquid limit test of soil by Casagrande Apparatus, lS: 2720
(Part 5): 1985.
• https://www.youtube.com/watch?v=pM-w_cvk1nA
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16. Procedure lS: 2720 (Part 5): 1985.
• About 120 g of oven dry soil sample passing 425 micron IS sieve shall be mixed thoroughly with
distilled water in the evaporating dish or on the fiat glass plate to form a uniform paste.
• ln the case of clayey soils, the soil paste shall be left to stand for a sufficient time (24 hours) so as to
ensure uniform distribution of moisture throughout the soil mass.
• Make a 2 mm groove using 'A' type grooving tool. In case where grooving tool, Type A does not give
a clear groove as in sandy soils, grooving tool Type ‘B’ or Type ‘C’ should be use.
• The cup will be fitted and dropped by turning the crank at the rate of two revolutions per second
until the two halves of the soil cake come in contact with bottom of the groove along a distance of
approximately 12 mm. The number of drops is recorded.
• A representative slice of soil approximately shall be taken in a suitable container and its moisture
content is determined.
• The operations shall be repeated for at least three more additional trails by adding sufficient water to
bring the soil to a more fluid condition.
• In each case the number of blows will be recorded and the moisture content determined.
• The number of blows required to close the groove should be between 15 to 35 blows.
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17. Procedure lS: 2720 (Part 5): 1985.
• Formula to Calculate Liquid Limit of Soil:-
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.)
(
.
.
wt
Dry
solids
of
Wt
water
of
Wt
LL
w 

A flow curve will be plotted on a semi logarithmic graph representing water content on the
arithmetical scale and the number of drops on the logarithmic scale

18. Flow Curve
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)
/
(
log 2
1
10
1
2
N
N
w
w
I f


Flow Index= Slope of curve

19. Determination of Plastic Limit
• Objective:Determination of Plastic Limit is as important
as Liquid Limit so as to ascertain Plasticity Index, Ip of the
soil.
• Apparatus Required: Glass Plate, Palette knives, Air
Tight Containers, Spatula, Brass Rod & Porcelain
Evaporating Dish, Hot Air Oven, 425 micron IS Sieve and
Balance weight.
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• Porcelain Evaporating dish: about 12cm in diameter. Glass plate: 10mm thick and
45cm square or large. Brass Rod: 3 mm in diameter and 10 cm long. All the
equipments should be cleaned and dried before each test.
• Reference:IS 2720(Part 5):1985 Methods of test for soils: Determination of Liquid
and Plastic limit (second revision). Reaffirmed- May 2015.

20. Procedure
• Video Link
• Take about 20 gm of thoroughly mixed portion of the material passing through
425 micron I.S.sieve obtained in accordance with I.S. 2720 (Part 1)- 1983.
• Mix it thoroughly with distilled water in the evaporating dish till the soil mass
becomes plastic enough to be easily molded with fingers.
• Allow it to season for sufficient time (for 24 hrs) to allow water to permeate
throughout the soil mass.
• Take about 8 gm of this plastic soil mass and roll it between fingers and glass
plate with just sufficient pressure to roll the mass into a thread of uniform
diameter throughout its length. The rate of rolling shall be between 80 and 90
strokes per minute.
• Continue rolling till you get a thread of 3 mm diameter.
• Kneed the soil together to a uniform mass and re-roll.
• Continue the process until the thread crumbles when the diameter is 3 mm.

21. ACTIVITY OF CLAYS
• The presence of even small amounts of certain clay minerals can have significant
effect on the properties of the soil.
• The identification of clay minerals requires special techniques and equipment.
• The techniques include microscopic examination, X-ray diffraction, differential
thermal analysis, optical property determination and electron micrography.
• An indirect method of obtaining information on the type and effect of clay mineral in a
soil is to relate plasticity to the quantity of clay-size particles.
• Activity (A)’ is defined as the ratio of plasticity index to the percentage of clay-sizes:
• where c is the percentage of clay sizes, i.e., of particles of size less than 0.002 mm.
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c
I
A
p


22. Activity classification
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23. Sensitivity (St)
• It has been established that the strength of a clay soil is related to its structure.
• If the original structure is altered by reworking or remoulding or chemical changes,
resulting in changes in the orientation and arrangement of the particles, the strength
or the clay gets decreased, even without alteration in the water content.
• ‘Sensitivity (St)’ of a clay is defined as the ratio of the its unconfined compression
strength in the natural or undisturbed state to that in the remoulded state, without any
change in the water content:
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)
Remoulded
(
)
Undistubed
(
u
u
q
q
St 

24. Sensitivity classification
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25. Thixotropy
• When clays with a flocculent structure are used in construction, these may lose some
strength as a result of remoulding.
• With passage of time, however, the strength increases, though not back to the original
value.
• This phenomenon of strength loss-strength gain, with no change in volume or water
content, is called ‘Thixotropy’.
• The loss of strength on remoulding is partly due to the permanent destruction of the
structure in the in-situ condition, and partly due to the reorientation of the molecules
in the adsorbed layers.
• The gain in strength is due to the rehabilition of the molecular structure of the soil.
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26. A fine-grained soil has a liquid limit of 300% and a plastic limit of 55%. The natural water content of the
soil in the field is 80% and the clay content is 60%.
(a) Determine the plasticity index, the liquidity index, and the activity.
(b) What is the soil state in the field?
(c) What is the predominant mineral in this soil?
(d) If this soil were under a concrete slab used as a foundation for a building and water were to seep into
it from watering of a lawn, what would you expect to happen to the foundation?
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Given Data:
Liquid Limil (LL) wl = 300%
Plastic Limil (PL) wp = 55%
Natural water cont. w = 80%
Clay content c = 60%

27. Step a: Calculation of Plasticity index, liquididty index and Activity (A):
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p
l
p w
w
I 

%
245
55
300 


p
I
P
P
L
I
w
w
I


245
55
80 

L
I
%
1
.
0

L
I
Activity A is given by
c
I
A
p

60
245

A
08
.
4

A

28. Step b: What is the soil state in the field?
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State of soil depends up on Liquidity index of the soil given in the table below
The soil with LI is 0.1 is at the low end of the plastic state.

29. Step c: What is the predominant mineral in this soil?
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This can be determined by observing the activity of the soil
The predominant mineral is montmorillonite (most likely, Na-montmorillonite).
1
.
4

A

30. Step c: Determine the consequences of water seeping into the soil
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• Seepage from lawn watering will cause the soil to expand (montmorillonite is an expansive soil).
Because the water content in the montmorillonite will not increase uniformly under the foundation, the
expansion will not be uniform. More expansion will occur at the edge of the slab because the water
content will be greater there. Consequently, the concrete foundation will curl upward at the edge and
most likely crack.
• Construction on expansive soils requires special attention to water management issues such as
drainage and landscape. Generally, plants and lawns should be at least 3 m away from the edge of the
foundation and the land should be sculpted to drain water away from the foundation.

31. A liquid limit test, conducted on a soil sample in the cup device, gave the following
results:
Number of blows: 10 19 23 27 40
Water content (%): 60.0 45.2 39.8 36.5 25.2
Two determinations for the plastic limit gave water contents of 20.3% and 20.8%.
Determine (a) the liquid limit and plastic limit, (b) the plasticity index, (c) the liquidity
index if the natural water content is 27.4%, and (d) the void ratio at the liquid limit if G
is 2.7. If the soil were to be loaded to failure, would you expect a brittle failure?
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Given Data:
PL1= 20.3 %
PL2=20.8%
w=27.4%
G= 2.7

32. Step I: Plot the data on semi log
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Step 2: Extract the liquid limit.
The water content on the liquid state line
corresponding to a terminal blow of 25 gives
the liquid limit.
%
38

l
w
Step 3: Calculate the plastic limit
Two values of plastic limits are alredy given; 20.3%
and 20.8%
Their avrg vale will give the plastic limit.
2
8
.
20
3
.
20 

P
w %
6
.
20

P
w

33. Wasim Shaikh, AIKTC
%
38

l
w
%
6
.
20

P
w
Step 4: Calculate PI.
p
l
p w
w
I 

6
.
20
38 

p
I
Step 5: Calculate LI.
4
.
17
6
.
20
4
.
27 

L
I
%
4
.
17

p
I
39
.
0

L
I
P
P
L
I
w
w
I


Given Data:
LL1= 20.3 %
LL2=20.8%
w=27.4%
G= 2.7

34. Wasim Shaikh, AIKTC
%
38

l
w
%
6
.
20

P
w
Step 6: Calculate the void ratio at liquid limit
Given Data:
LL1= 20.3 %
LL2=20.8%
w=27.4%
G= 2.7
G
Se
w 
At liquid limit soil is 100% saturated, hence S=1
S
wG
e 
Also, here w = wl
1
7
.
2
274
.
0 

e
03
.
1

e