Chapter 1-index properties-2 presentation and question answer

Soil Index Properties
Q: Explain with sketch phase diagrams for the partially saturated soil and dry soil.
Q: Explain soil as three phase system
Soil as three phase system
A soil mass consist of the solid particles and the voids in between them. These voids are
filled with air or/and water. So there is a three phase system, but when the voids are only
filled with air, or only filled with water then soil becomes a two phase system. Three phase
system can be represented with a diagram as shown below. When the voids are only filled
with water, it is said to be saturated.
SOIL
SOLIDS
WATER
AIR
W
W
W
W
W
V
V
V
V
V
V
a
w
s s
w
a
V
Figure 1 Soil as three phase system
In above figure,
Wa= weight of air
Ws = weight of soil solids
Ww= weight of water
Wv =weight of voids = Ww+Wa
W= weight of soil mass = Wv +Ws = Ww+Wa+Ws
Va= Volume of air
Vs = Volume of soil solids
Chapter 1 GT notes Page 1 of 31

Vw= Volume of water
Vv =Volume of voids = Vw+Va
V= Volume of soil mass = Vv + Vs = Vw+Va+ Vs
This three phase system converted in to two phase system when soil becomes saturated or
dry as below.
SOIL
SOLIDS
WATER V
V
V
V
s
w
V
W
W
W
W
V
w
s
Figure 2 Soil as two phase system- saturated soil
Wv =weight of voids = Ww+Wa = Ww
W= weight of soil mass = Wv +Ws = Ww+Ws
Vv =Volume of voids = Vw+Va = Vw
V= Volume of soil mass = Vv + Vs = Vw+ Vs
SOIL
SOLIDS
AIR V
V
V
V
s
a
V
W
W
W
W
V
a
s
Figure 3 Soil as two phase system- dry soil
Wv =weight of voids = Ww+Wa = Wa= 0
W= weight of soil mass = Wv +Ws = Ww+Ws= Ws

Vv =Volume of voids = Vw+Va = Va
V= Volume of soil mass = Vv + Vs = Va+ Vs
We have the mass and weight relationship;
Weight of soil = Mass of soil x 9.81
Hence all above weight related quantities can be expressed in terms of mass of soil.
---------
Q: Explain following index properties
Q: Define and explain following properties
Density
 It is the ratio of mass of soil to volume of soil
 Unit  gram per cubic centimeter (g/cc)
 ρ=
M
V
Dry Density
 It is the ratio of dry mass of soil to volume of soil
 Symbol ρd
Saturated Density
 It is the ratio of saturated mass of soil to volume of soil
 Symbol ρsat
Submerged Density
 It is the ratio of submerged mass of soil to volume of soil
 Symbol ρ
'
 ρ
'
=ρsat−ρw=ρsat−1…….density of water ρw =1g/cc
---------
Unit weight
 It is the ratio of weight of soil to volume of soil
 Unit  kilogram per cubic meter (kg/m3
) or kN/m3
 γ=
W
V
Dry Unit Weight
 It is the ratio of dry weight of soil to volume of soil
 Symbol γd

Saturated Unit Weight
 It is the ratio of saturated weight of soil to volume of soil
 Symbol γsat
Submerged Unit Weight
 It is the ratio of submerged weight of soil to volume of soil
 Symbol γ'
 γ'
=γsat −γw=γsat −9.81…….unit weight of water γw=9.81g/cc
Relation between density and unit weight
 γ=ρ×g=ρ×9.81
---------
Water content
 It is the ratio of mass of water in soil to mass of soil solids
 Unit less quantity/ expressed as a percentage
 w=
Mw
Ms
=
W w
Ws
 Water content experiments formulae
 Oven drying formula, w=
M−Ms
M s
 Pycnometer method, w=
[M2−M1
M3−M4
×(G−1
G )−1
]×100
 Where, M1 is mass of empty pycnometer, M2 is mass of pycnometer + most
soil, M3 is mass of pycnometer + most soil + water, M4 is mass of pycnometer
+water only, G is specific gravity
Specific gravity
 It is the ratio of density of soil solid to density of water at standard temperature.
 Unit less quantity/ expressed as a decimal
 G=
ρsoilsolids
ρwater
=
ρs
ρw
Apparent/mass/ bulk Specific gravity
 It is the ratio of density of soil to density of water at standard temperature.
 Unit less quantity/ expressed as a decimal

 Gm=
ρsoil
ρwater
=
ρ
ρw
 Specific gravity experiments formula
 Pycnometer method, G=¿
 Where, M1 is mass of empty pycnometer, M2 is mass of pycnometer + dry
soil, M3 is mass of pycnometer + most soil + water, M4 is mass of pycnometer
+water only
---------
Void Ratio
 It is the ratio of the volume of voids to the volume of solids. It is denoted by ‘e’.
 Unit less quantity/expressed as a decimal
 e =
Vv
Vs
Porosity
 It is the ratio of the volume of voids to the total volume of soil.
 Unit less quantity/expressed as a percentage
 n =
Vv
V
Degree of saturation
 It is the ratio of the volume of water to the total volume of voids.
 S =
Vw
V v
 S is maximum (100%) when soil is saturated
 S is minimum (0%) when soil is dry
Air content
 It is the ratio of the volume of air to the total volume of voids.
 Unit less quantity/expressed as a decimals
 ac =
Va
Vv
 ac is maximum (100%) when soil is dry
 ac is minimum (0%) when soil is saturated

Relationship between S and ac
S + ac = 1
---------
Percentage of air voids
 It is the ratio of the volume of air to the total volume of soil.
 na =
Va
V
Density Index/ relative density
 DI=RD=ID=
emax−e
emax−emin
 emax = voids ratio when soil is in loosest state
 emin = voids ratio when soil is in densest state
 e = voids ratio when soil is in natural state
 ID is minimum; equals to 0 (zero) when is at loosest state
 ID is maximum; equals to 1 (one) when is at densest state
 Hence ID varies between 0 to 1
Q: Derive relation between voids ratio and porosity
Q: prove that,1−n=
1
1+e
,
Q: Prove that, e=
n
(1−n)
Q: Prove that, e=
n
(1−n)
Step 1) three phase system

Where,
V, W- total volume & weight of soil
Va, Vw, Vs, Vv- volume of air, water,
solids and voids as shown
Ww, Ws, Wv- weight of water, solids,
voids respectively
Mw, Ms, Mv- mass of water, solids,
voids respectively
Step 2) basic formulae:
Voids ratio = e=
Vv
V s
porosity = n=
V v
V
Step 3) Derivation:
e=
Vv
V s
Add 1 on both the sides
1+e=1+
Vv
V s
1+e=
Vs+V v
Vs
From figure it is clear, Vs+V v=V
Hence,
1+e=
V
Vs
Rearranging V on right hand side,
1+e=
1
(V s
V )
Here Vs = V-Vv
1+e=
1
(V−V v
V )
=
1
(V
V
−
V v
V )
=
1
(1−
Vv
V )
From step 2

1+e=
1
(1−n)
OR
(1−n)=
1
1+e
Hence,
Subtract 1 from both side,
1+e−1=
1
(1−n)
−1
e=
1−(1−n)
(1−n)
e=
n
(1−n)
We have,
(1−n)=
1
1+e
Subtract from 1, from both side,
1−(1−n)=1−
1
1+e
(n)=
1+e−1
1+e
n=
e
1+e
Proved
Q: Derive relation between degree of saturation, voids ratio, water content and specific
gravity
Q: prove that, Sr e=w G
Where,
voids respectively

voids respectively
Degree of saturation = Sr=
V w
V v
Voids ratio = e=
Vv
V s
Water content = w=
ww
ws
Specific gravity = G=
γs
γw
Step 3) Derivation:
We have, Specific gravity = G=
γs
γw
…………..(1)
Where, γs=
W s
Vs
and γw=
W w
Vw
Putting these values in equation (1)
G=
(Ws
V s
)
(W w
Vw
)
Rearranging
G=(Ws
V s
)×(V w
Ww
)
Consider volume and weight terms together,
G=
(W s
Ww
)×
(V w
Vs
)
From basic formulae,
G=(1
w )×
(Vw
V s
)
Take w on left hand and Multiply-divide by Vv on right hand side
G w=
(Vw
V s
×
V v
V v
)
G w=
V w
V v
×
Vv
Vs

Again from basic formulae,
G w=Sr ×e
Rearranging
Sr e=wG
Proved
Q: Derive relation between unit weight/density, dry unit weight/density and water
content.
 Q: prove that, γd=
γ
1+w
Where,
voids respectively
voids respectively
Dry unit weight = γd=
W s
V
=
Wd
V
Bulk unit weight = γ=
W
V
Water content = w=
ww
ws
Step 3) Derivation:
We have, Bulk unit weight = γ=
W
V
…………..(1)
Where, W = Wv +Ws =Wa +Ww +Ws = Ww +Ws ………………….Wa=0
γ=
W w+W s
V
γ=
W w
V
+
W s
V

γ=
W w
V
+γd
Multiply-divide by Ws as shown below,
γ=
W w
V
×
W s
W s
+γd
From basic formula,
γ=
W w
W s
×
W s
V
+γd=w×γd+γd
Taking common factors
γ=(w+1)γd
Rearranging
γd=
γ
1+w
Proved
Q: Derive relation between degree of saturation, voids ratio, bulk unit weight and specific
gravity
 Q: prove that, γ=
[G+eS]γw
1+e
 Q: prove that,
γ
sat=¿
[G+e]γw
1+e
¿
 Q: prove that,
γ
d=¿
Gγw
1+e
¿
Where,
voids respectively
voids respectively
Degree of saturation = Sr=
V w
V v

Voids ratio = e=
Vv
V s
Water content = w=
ww
ws
Specific gravity = G=
γs
γw
Dry unit weight = γd=
W s
V
=
Wd
V
Bulk unit weight = γ=
W
V
Saturated unit weight = γsat =
W sat
V
Step 3) Derivation:
We have, Bulk unit weight = γ=
W
V
…………..(1)
Where, W = Wv +Ws =Wa +Ww +Ws = Ww +Ws ………………….Wa=0
γ=
W w+W s
VV +V s
Taking common from denominator (RHS)
γ=
Ww +W s
(VV
Vs
+1
)V s
=
W w+Ws
(e+1)Vs
γ=
1
(1+e) [Ww +W s
V s
]=
1
(1+e)[Ww
V s
+
W s
Vs
]
γ=
1
(1+e) [Ww
Vs
+γs
]
Multiply-divide by Vv&Vw as shown below,
γ=
1
(1+e) [Ww
Vs
×
V v
V v
×
Vw
Vw
+γs
]
γ=
1
(1+e) [Ww
Vw
×
V v
Vs
×
Vw
Vv
+γs
]
γ=
1
(1+e)
[γw ×e ×Sr+γs]
Taking common γw from bracket

γ=
1
(1+e) [e×Sr+
γs
γw
]γw
Then,
γ=
1
(1+e)
[e ×Sr+G ]γw
Rearrenging
γ=
(G+e Sr)γw
(1+e)
If Soil is dry, Sr = 0 and γ become γd
γd=
G γw
(1+e)
If Soil is saturated, Sr = 1 and γ become γsat
γd=
(G+e)γw
(1+e)
Proved
Q: Explain IS soil classification system (as per IS 1498-1970)
Classification Based on Grain Size:The range of particle sizes encountered in soils is very large: from
boulders with dimension of over 300 mm down to clay particles that are less than 0.002 mm. Some
clays contain particles less than 0.001 mm in size which behave as colloids, i.e. do not settle in water.
In the Indian Standard Soil Classification System (ISSCS), soils are classified into groups according to
size, and the groups are further divided into coarse, medium and fine sub-groups.
Soil type Soil
Fine grained soil Coarse grained soil
sub
classification
Highly
Organi
c
clay Silt Sand Gravel Cobble Boulder
Symbol O C M S G - -
Size
-- Less than
0.002mm
0.002 mm
to
0.075 mm
 Coarse: 2 to
4.75 mm
 Medium:
0.425 to 2 mm
 Fine: 0.075 to
0.425mm
 Coarse: 20
to 80mm
 Fine: 4.75
to 20mm
80mm
to
30cm
> 30cm

Symbol Meaning Symbol Meaning
G Gravel H Highly compressible (Liquid limit >50)
S Sand I Intermidiate compressible (35<Liquid limit <50)
M Silt L Low compressible (35> Liquid limit)
C Clay O Mixture of slit, clay and organic material
Description:
1. Coarse-grained soils are those for which more than 50% of the soil material by weight has
particle sizes greater than 0.075 mm. They are basically divided into either gravels (G) or sands
(S).
2. Fine-grained soils are those for which more than 50% of the material has particle sizes less than
0.075 mm. Clay particles have a flaky shape to which water adheres, thus imparting the property
of plasticity.
3. Gravel are those for which more than 50% of the soil material by weight has particle sizes
greater than 4.75 mm and less than 80mm. it is divided into two part coarse and fine gravel as
shown in table.
4. Sand are those for which more than 50% of the soil material by weight has particle sizes greater
than 75micron and less than 4.75mm. it is dividedinto three part coarse, medium and fine
gravel as shown in table.
I
S
c
l
a
s
s
i
f
i
c
a
t
i
o
n
c
o
a
r
s
e
S
o
i
l
G
r
a
v
e
l
G
W
G
P
G
M
G
C
S
a
n
d
S
W
S
P
S
M
S
C
F
i
n
e
s
o
i
l
C
H
M
H
O
H
C
I
M
I
O
I
C
L
M
L
O
L
P
e
a
t
(
P
t
)

5. Fine grained soil classified according particle size as clay (< 0.002mm) and silt (2 -75micron). As
the sizes are very less to do sieve analysis, the plasticity chart is used for subclassification as
shown in chart and table below.
6. Based on liquid limit values and compressiblity characteristics the soil is classified as ML, OL, CL,
etc.
7. Based on gradation of different soil elements further classsfication is made as shown in chart.
8. GW, GP, GM, GC, SW, SP, SM, SC- here W stands for well graded soil, P- stands for poorly graded
soil, M is silt and C is clay. Hence the name indicates the classifition and grading of soil.
9. Soil classification using group symbols is as follows:
Group
Symbol
Classification Group
Symbol
Classification
Gw Well-Graded Gravel Sw Well-Graded Sand
Gp Poorly-Graded Gravel Sp Poorly-Graded Sand
Gm Silty Gravel Sm Silty Sand
Gc Clayey Gravel Sc Clayey Sand
Ml Silt of Low Plasticity Cl Clay of Low Plasticity
Mi Silt of Intermediate Plasticity Ci Clay of Intermediate Plasticity
Mh Silt of High Plasticity Ch Clay of High Plasticity
Ol Organic Soil of Low Plasticity Pt Peat
Oi Organic Soil of Intermediate Plasticity
Oh Organic Soil of High Plasticity
Q: Explain Plasticity chart
 A plasticity chart, based on the values of liquid limit (WL) and plasticity index (IP), is
provided in ISSCS to aid classification.

 The 'A' line in this chart is expressed as IP = 0.73 (WL - 20).
 Depending on the point in the chart, fine soils are divided into clays (C), silts (M),
or organic soils (O).
 The organic content is expressed as a percentage of the mass of organic matter in a
given mass of soil to the mass of the dry soil solids.
 Three divisions of plasticity are also defined as follows.
Symbol Meaning Symbol Meaning
M Silt H Highly compressible (Liquid limit >50)
C Clay I Intermidiate compressible (35< Liquid limit <50)
L Low compressible (35> Liquid limit)
O Mixture of slit, clay and organic material
 The 'A' line and vertical lines at WL equal to 35% and 50% separate the soils into various
classes.
 For example, the combined symbol CH refers to clay of high plasticity.
Q: Explain Soil gradation system
 In the Indian Standard Soil Classification System (ISSCS), soils are classified into groups
according to size, and the groups are further divided into coarse, medium and fine sub-groups.
 The grain-size range is used as the basis for grouping soil particles into boulder, cobble, gravel,
sand, silt or clay.
Soil type Soil
Fine grained soil Coarse grained soil
sub
classification
Highly
Organi
c
clay Silt Sand Gravel Cobble Boulder
Symbol O C M S G - -
Size
-- Less than
0.002mm
0.002 mm
to
0.075 mm
 Coarse: 2 to
4.75 mm
 Medium:
0.425 to 2 mm
 Fine: 0.075 to
0.425mm
 Coarse: 20
to 80mm
 Fine: 4.75
to 20mm
80mm
to
30cm
> 30cm
 Gravel, sand, silt, and clay are represented by group symbols G, S, M, and C respectively.
 Physical weathering produces very coarse and coarse soils. Chemical weathering produce
generally fine soils.
 Coarse-grained soils are those for which more than 50% of the soil material by weight has
particle sizes greater than0.075 mm. They are basically divided into either gravels (G) or sands
(S).

 According to gradation, they are further grouped as well-graded (W) or poorly graded (P). If fine
soils are present, they are grouped as containing silt fines (M) or as containing clay fines (C).
 For example, the combined symbol SW refers to well-graded sand with no fines.
 Both the position and the shape of the grading curve for a soil can aid in establishing its identity
and description. Some typical grading curves are shown.
Partical size
Percentage
finer
0 0.01 0.1 1 10 100
Fine grained
soil
Coarse grained
soil
(Poorly graded)
soil
Uniformly graded
soil
Well graded
soil
Gap graded
10
30
60
100
D10 D30 D60
Figure 4 Soil gradation curves (experiment: sieve analysis)
Coefficient of uniformity (Cu)
 It is a measure of particle size range.
 This term identified in mechanical Sieve analysis (experiment)
 Cu =
D60
D10
 For uniform soil Cu =1
 For well graded soil Cu > 4 for gravels (along with Cc condition)
 For well graded soil Cu > 6 for sands (along with Cc condition)
Coefficient of curvature (Cc)
 It is a measure of shape of particle size.
 This term identified in mechanical Sieve analysis (experiment)
 Cc =
( D30)
2
D60×D10
 Here D10, D30& D60 are the “particle sizes (let’s say A, B, C)” such that 10%, 30% & 60%
of particles in soil are smaller/finer than that (A, B & C respectively)sizes respectively.

These values are identified from gradation curve (experiment: - mechanical Sieve
analysis)
 For well graded soil Cc should be within 1 to 3 (along with Cu conditions)
Q: Explain consistency of soils
Solid state
Semi solid state
Plastic state
Liquid state
very low shear
strength
Soil can be mould
without any crack
Soil cannot mould
without any crack
no change in volume
with change in
water content
T
o
ta
l
v
o
lu
m
e
o
f
s
o
il
m
a
s
s
Water content (%)
w w w
S P L
(Shrinkage limit) (Plastic limit) (Liquid limit)
Figure 5 states of soil consistency
 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:
o Liquid limit (WL) - change of consistency from plastic to liquid state
o Plastic limit (WP) - change of consistency from brittle/crumbly to plastic state

 The difference between the liquid limit and the plastic limit is known as the plasticity index (IP),
and it is in this range of water content that the soil has a plastic consistency. The consistency of
most soils in the field will be plastic or semi-solid.
Liquid limit (wL)
 It is minimum water content at which soil is in liquid state.
Plastic limit (wP)
 It is minimum water content at which soil can be remould (shaped) without any
crack.
Shrinkage limit (ws)
 It is maximum water content reduction in water content will not affect total volume
of soil mass.
Additional formulae
 Plasticity Index = Ip = wL- wP
 Consistency Index = Ic =
wL−w
I p
 Liquidity Index = IL =
w−wP
I p
 Consistency Index + Liquidity Index = 1  IC + IL = 1
 Flow index = slope of liquid limit curve (line) from logarithmic plot
Flow index = IF =
w1−w2
log10
(n2
n1
)
Where w1 and w2 are water contents corresponding to number of
blows n1 and n2 (in liquid limit test: experiment) respectively
 Toughness Index = IT =
I p
I F
 Activity of clays (Ac) =
I p
% by weightof clay
 Ac< 0.75  Inactive soil
 0.75 < Ac< 1.40  Normal soil
 Ac> 1.40  Active soil
Determination of index properties
Note: Refer laboratory manual for all experiments

Sr
.
Experiment/ title Apparatus/equipments
used/required in
experiment
Used for determination of
(index) properties
1 Oven dying method oven Water content
2 Pycnometer method pycnometer Water content
3 Calcium carbide
method
Moisture tester Water content
4 Sand bath method,
Alcohol method
- Water content (rough
methods)
5 Specific gravity Pycnometer, density
bottle
Specific gravity
6 Particle size
distribution/mechanical
Sieve analysis
IS sieves Soil classification/ gradation
(coarse analysis)
7 Wet sieve analysis/
Sedimentation
analysis/ pipette
method/ hydrometer
analysis
Pipette
hydrometer
Soil classification (fine analysis)
8 Soil Consistency tests (a)Casagrande tool,
ASTM tool
(c)shrinkage dish,
evaporating dish
(a)Liquid limit
(b)Plastic limit
(c)Shrinkage limit
9 Static cone penetration
test/method
penetrometer Liquid limit
10 (a)Sand replacement
method
(b) core cutter method
(a)Calibrating container
Sand pouring cylinder
(b) core cutter, dolly,
rammer
In-situ density/unit weight
Voids ratio
Density index
Soil Structure
 Single grained
 Honeycomb
 Flocculent

 Dispersed Coarse grained skeleton
 Cohesive matrix
Experiments
1) Oven drying experiment for Water content determination
2) Water content determination using pycnometer
3) Water content determination using Calcium carbide method
4) Specific gravity determination using pycnometer
5) Core cutter test
6) Liquid limit test
7) Plastic limit test
8) Particle size distribution/mechanical Sieve analysis
9) Wet sieve analysis/ Sedimentation analysis/ pipette method/ hydrometer analysis
Title Oven drying experiment for Water content determination
Fig.
Accessories oven maintained at a temperature of 110 ± 5oC, Weighing balance, Air-tight
container with lid
Procedure  Clean the container, dry it and weigh it with the lid (Weight ‘W1‘).
 Take the required quantity of the wet soil specimen in the container and
weigh it with the lid (Weight ‘W2‘).
Particle size (>90% passing through) Min. qty of
soil
0.425 mm 25
2.00 mm 50
4.75 mm 200
 Place the container, with its lid removed, in the oven till its weight
becomes constant (Normally for 24hrs.).
 When the soil has dried, remove the container from the oven, using tongs.
 Find the weight ‘W3‘ of the container with the lid and the dry soil sample.
Result/
conclusion
The water content
w=
w 2−w3
w 3−w1
×100
An average of three determinations should be taken.
****

Title Water content determination using pycnometer
Fig.
Accessories Pycnometer, Weighing balance with an accuracy of 1.0g, Glass rod
Procedure  Wash, clean and dry the pycnometer and note down its mass (W1) along
with brass cap and washer using weighing balance with an accuracy of 1.0
g.
 Now place a sample of wet soil about 200 to 400 g in pycnometer and
note down its mass (W2).
 Then add water to the soil in the pycnometer to make it about half full.
 Stir the soil using glass rod to remove air voids of the soil sample.
 Add some more water and after eliminating the entrapped air stop stirring
and fix the brass cap.
 More water is added through hole in brass cap until the water is flush with
the hole.
 Now take the mass of pycnometer (W3).
 Now empty and wash the pycnometer. Then fill it with only water and take
its mass (W4).
Result/
conclusion
The water content
w=
[(w 2– w 1
w3 – w 4 )(G−1
G )−1
]×100
***
Title Water content determination using Calcium carbide method
Fig.
Accessories Metallic pressure vessel, with a clamp for sealing the cup, along with a gauge
calibrated in percentage water content, weighing balance, Scoop, for
measuring the absorbent, Steel balls – 3 steel balls of about 12.5mm dia. and
1 steel ball of 25mm dia., One bottle of the absorbent (Calcium Carbide)

Procedure  Set up the balance, place the sample in the pan till the mark on the
balance arm matches with the index mark.
 Check that the cup and the body are clean.
 Hold the body horizontally and gently deposit the levelled, scoop-full of
the absorbent (Calcium Carbide) inside the chamber.
 Transfer the weighed soil from the pan to the cup.
 Hold cup and chamber horizontally, bringing them together without
disturbing the sample and the absorbent.
 Clamp the cup tightly into place. If the sample is bulky, reverse the above
placement, that is, put the sample in the chamber and the absorbent in
the cup.
 In case of clayey soils, place all the 4 steel balls (3 smaller and 1 bigger) in
the body along with the absorbent.
 Shake the unit up and down vigorously in this position for about 15
seconds.
 Hold the unit horizontally, rotating it for 10 seconds, so that the balls roll
around the inner circumference of the body.
 Rest for 20 seconds.
 Repeat the above cycle until the pressure gauge reading is constant and
note the reading. Usually it takes 4 to 8 minutes to achieve constant
reading. This is the water content (m) obtained on wet mass basis.
 Finally, release the pressure slowly by opening the clamp screw and taking
the cup out, empty the contents and clean the instrument with a brush.
Result/
conclusion
The water content on dry mass basis,
w=m/[100-m] * 100%
*****
Title Specific gravity determination using pycnometer
Fig.
Accessories Pycnometer, Weighing balance with an accuracy of 1.0g, Glass rod
Procedure  Wash, clean and dry the pycnometer and note down its mass (W1) along
with brass cap and washer using weighing balance with an accuracy of 1.0
g.
 Now place a sample of dry soil about 200 to 400 g in pycnometer and note
down its mass (W2).
 Then add water to the soil in the pycnometer to make it about half full.
 Stir the soil using glass rod to remove air voids of the soil sample.
 Add some more water and after eliminating the entrapped air stop stirring
and fix the brass cap.
 More water is added through hole in brass cap until the water is flush with

the hole.
 Now take the mass of pycnometer (W3).
 Now empty and wash the pycnometer. Then fill it with only water and take
its mass (W4).
Result/
conclusion
The specific gravity
G=
w2 – w1
(w2 – w 1)−(w 3−w4)
***
Title In-situ density by Core cutter test
Fig.
Accessories Cylindrical core cutter, Steel rammer, Steel dolley, Weighing balance, accuracy
1g., Palette knife, Straight edge
Procedure  Determine the internal diameter and height of the core cutter to the
nearest 0.25mm
 Determine the mass (M1) of the cutter to the nearest gram.
 Expose a small area of the soil to be tested. Level the surface, about
300mm square in area.
 Place the dolley over the top of the core cutter and press the core cutter
into the soil mass using the rammer. Stop the pressing when about 15mm
of the dolley protrudes above the soil surface.
 Remove the soil surrounding the core cutter, and take out the core cutter.
Soil would project from the lower end of the cutter.
 Remove the dolley. Trim the tip and bottom surface of the core cutter
carefully using a straight edge.
 Weigh the core cutter filled with the soil to the nearest gram (M2).
 Remove the core of the soil from the cutter. Take a representative sample
for the water content determination.
 Determine the water content.
Result/
conclusion
Observation: Dia. of core, height of core, volume of core, mass of soil
Bulk density of soil = Mass/ Volume

ρ=
M
V
Dry density of soil = Bulk density /(1+ water content)
ρd=
ρ
1+w
***
Title Liquid limit test
Fig.
Accessories Balance, Casagrendes Liquid limit device, Grooving tool, Mixing dishes,
Spatula, Electrical Oven, Squeeze Bottle
Procedure  Put 250 gm of air-dried soil, passed thorough 425 mm sieve, into an
evaporating dish.
 Add distilled water into the soil and mix it thoroughly to form uniform
paste.
 The paste shall have a consistency that would require 30 to 35 drops of
cup to cause closer of standard groove for sufficient length.
 Place a portion of the paste in the cup of Liquid Limit device and spread it
with a few strokes of spatula.
 Trim it to a depth of 1 cm at the point of maximum thickness and return
excess of soil to the dish.
 Using the grooving tool, cut a groove along the centre line of soil pat in the
cup, so that clean sharp groove of proper dimension (11 mm wide at top,
2 mm at bottom, and 8 mm deep) is formed.
 Lift and drop the cup by turning crank at the rate of two revolutions per
second until the two halves of soil cake come in contact with each other
for a length of about 13 mm by flow only, and record the number of
blows, N.
 Take a representative portion of soil from the cup for moisture content
determination.
 Repeat the test with different moisture contents at least four more times
for blows between 10 and 40.
Result/
conclusion
 Plot the relationship between water content (on y-axis) and number of
blows (on x-axis) on semi-log graph.
 The curve obtained is called flow curve.
 The moisture content corresponding to 25 drops (blows) as read from the

represents liquid limit.
 It is usually expressed to the nearest whole number.
 Flow index = slope of the flow curve
***
Title Plastic limit test
Fig.
Accessories Ground Glass Plate, Spatula or pill knife, Drying Oven, Metallic rod 3.2mm
diameter and 100mm long, Water Content Containers ,Balance
Procedure  Select a 20-g or more portion of soil from the material prepared for the
liquid limit test.
 Reduce the water content of the soil to a consistency at which it can be
rolled without sticking to the hands by spreading or mixing continuously
on the glass plate or in the mixing/storage dish
 select a 1.5 to 2.0 g from the plastic-limit specimen and form the selected
portion into an ellipsoidal mass.
 Roll the mass between the palm or fingers and the ground-glass plate with
just sufficient pressure to roll the mass into a thread of uniform diameter
throughout its length.
 The thread shall be further deformed on each stroke so that its diameter
reaches 3.2 mm, taking no more than 2 min.
 Normally 80-90 stroke per minute is recommended.
 count a stroke as one complete motion of the hand forward and back to
the starting position.
 The rate of rolling shall be declined for very fragile soils.
Result/
conclusion
Calculate the average of the two water contents (trial plastic limits) and round
to the nearest whole number; this value is the plastic limit, PL.
***
Title Particle size distribution/mechanical Sieve analysis

Fig.
Accessories Set of fine sieves, 2mm, 1mm, 600 micron, 425, 212, 150, and 75 micron, Set
of coarse sieves, 100mm, 80mm, 40mm, 10mm, and 4.75mm, Weighing
balance, Oven, Mechanical shaker, Tray
Procedure  Take the required quantity of the sample.
 Take mass of each sieves
 Arrange all sieves on descending order. Place pan at bottom.
 With all soil mass on top most sieves with lid, place whole assembly on
sieve shaker.
 Shake all assembly for 10min.
 Care shall be taken so as not to break the individual particles.
 The quantity of the material taken for sieving on each sieve shall be such
that the maximum mass of material retained on each sieve does not
exceed the specified value.
 After 10 min, determine the mass of the material retained on each sieve.
 Calculate the percentage of soil retained on each sieve on the basis of the
total mass of the sample
 Determine the percentage passing (% finer) through each sieve.
Result/
conclusion
 Percentage finer can be used to plot the particle size distribution curve
with particle size as abscissa on log scale and the percentage finer as
ordinate.
 Find D10, D30, D60 on graph
 Find Cc and Cu
 Interpret the soil gradation (as well as classification)

***
Title Hydrometer analysis
Fig.
Accessories
1. Hydrometer, Glass measuring cylinder (jar), 1000ml, Rubber bung for the
cylinder (jar), Mechanical stirrer, Weighing balance, accuracy 0.01g, Oven,
Deflocculating agent., Desiccator, Evaporating dish, Conical flask or beaker,
1000ml, Stop watch, Wash bottle, Thermometer, Water bath, 75 µ Sieve, Scale
Procedure The particle size (D) is given by:
D=M
√He
t
M=
[ 0.3η
g(G−1) ρw
]
In which,
n = viscosity of water in poise,
G = specific gravity of solids,
Pw = density of water (gm/ml); ,
He= effective depth,
t= time in minutes at which observation is taken, reckoned with respect to
the beginning of sedimentation.
The percentage finer than the size D is given by
N=
[ G
G−1]×
R
Ms
×100
Where,
R= corrected hydrometer reading, Ms= mass of dry soil in 1000ml
suspension.
Stoke’s Law,
ν=
D
2
γw (Gs−GL)
18η
Procedure of Hydrometer Test
Part – 1: Calibration of Hydrometer
 Take about 800ml of water in one measuring cylinder. Place the cylinder
on a table and observe the initial reading.
 Immerse the hydrometer in the cylinder. Take the reading after the

immersion.
 Determine the volume of the hydrometer (VH) which is equal to the
difference between the final and initial readings. Alternatively weigh the
hydrometer to the nearest 0.1g. The volume of the hydrometer in ml is
approximately equal to its mass in grams.
 Determine the area of cross section (A) of the cylinder. It is equal to the
volume indicated between any two graduations divided by the distance
between them. The distance is measured with an accurate scale.
 Measure the distance (H) between the neck and the bottom of the bulb.
Record it as the height of the bulb (h).
 Measure the distance (H) between the neck to each marks on the
hydrometer (Rh).
 Determine the effective depth (He), corresponding to each of the mark
(Rh) as
 [Note: The factor VH/A should not be considered when the hydrometer is
not taken out when taking readings after the start of the sedimentation at
½, 1, 2, and 4 minutes.]
 Draw a calibration curve between He and Rh. Alternatively, prepare a
table between He and Rh. The curve may be used for finding the effective
depth He corresponding to reading Rh.
Part – 2 : Meniscus Correction
 Insert the hydrometer in the measuring cylinder containing about 700ml
of water.
 Take the readings of the hydrometer at the top and at the bottom of the
meniscus.
 Determine the meniscus correction, which is equal to the difference
between the two readings.
 The meniscus correction Cm is positive and is constant for the
hydrometer.
 The observed hydrometer reading Rh’ is corrected to obtain the corrected
hydrometer reading Rh as
Part – 3 : Pretreatment and Dispersion
 Weigh accurately, to the nearest 0.01g about 50g air-dried soil sample
passing 2mm IS sieve, obtained by riffling from the air-dried sample
passing 4.75mm IS sieve. Place the sample in a wide mouthed conical
flask.
 Add about 150ml of hydrogen peroxide to the soil sample in the flask. Stir
it gently with a glass rod for a few minutes.
 Cover the flask with a glass plate and leave it to stand overnight.
 Heat the mixture in the conical flask gently after keeping it in an
evaporating dish. Stir the contents periodically. When vigorous frothing
subsides, the reaction is complete. Reduce the volume to 50ml by boiling.
Stop heating and cool the contents.
 If the soil contains insoluble calcium compounds, add about 50ml of

hydrochloric acid to the cooled mixture. Stir the solution with a glass rod
for a few minutes. Allow it to stand for one hour or so. The solution would
have an acid reaction to litmus when the treatment is complete.
 Filter the mixture and wash it with warm water until the filtrate shows no
acid reaction.
 Transfer the damp soil on the filter and funnel to an evaporating dish using
a jet of distilled water. Use the minimum quantity of distilled water.
 Place he evaporating dish and its contents in an oven and dry it at 105 to
110 degree C. Transfer the dish to a desiccator and allow it to cool.
 Take the mass of the oven dried soil after pretreatment and find the loss
of mass due to pretreatment.
 Add 100ml of sodium hexa-metaphosphate solution to the oven – dried
soil in the evaporating dish after pretreatment.
 Warm the mixture gently for about 10minutes.
 Transfer the mixture to the cup of a mechanical mixture. Use a jet of
distilled water to wash all traces of the soil out of the evaporating dish.
Use about 150ml of water. Stir the mixture for about 15minutes.
 Transfer the soil suspension to a 75 µ IS sieve placed on a receiver (pan).
Wash the soil on this sieve using a jet of distilled water. Use about 500ml
of water.
 Transfer the soil suspension passing 75 µ sieve to a 1000ml measuring
cylinder. Add more water to make the volume exactly equal to 1000ml.
 Collect the material retained on 75 µ sieve. Dry it in an oven. Determine its
mass. If required, do the sieve analysis of this fraction.
Part – 4 : Sedimentation Test
 Place the rubber bung on the open end of the measuring cylinder
containing the soil suspension. Shake it vigorously end-over-end to mix the
suspension thoroughly.
 Remove the bung after the shaking is complete. Place the measuring
cylinder on the table and start the stop watch.
 Immerse the hydrometer gently to a depth slightly below the floating
depth, and then allow it to float freely.
 Take hydrometer reading (Rh’) after 1/2, 1, 2 and 4 minutes without
removing the hydrometer from the cylinder.
 Take out the hydrometer from the cylinder, rinse it with distilled water.
 Float the hydrometer in another cylinder containing only distilled water at
the same temperature as that of the test cylinder.
 Take out the hydrometer from the distilled water cylinder and clean its
stem. Insert it in the cylinder containing suspension to take the reading at
the total elapsed time interval of 8minutes. About 10 seconds should be
taken while taking the reading. Remove the hydrometer, rinse it and place
it in the distilled water after reading.
 Repeat the step (7) to take readings at 15, 30, 60, 120 and 240minutes
elapsed time interval.
 After 240 minutes (4 hours) reading, take readings twice within 24 hours.
Exact time of reading should be noted.
 Record the temperature of the suspension once during the first 15minutes

and thereafter at the time of every subsequent reading.
 After the final reading, pour the suspension in an evaporating dish, dry it
in an oven and find its dry mass.
 Determine the composite correction before the start of the test and also
at 30min, 1, 2 and 4 hours. Thereafter just after each reading, composite
correction is determined.
 For the determination of composite correction (C), insert the hydrometer
in the comparison cylinder containing 100ml of dispersing agent solution
in 1000 ml of distilled water at the same temperature. Take the reading
corresponding to the top of meniscus. The negative of the reading is the
composite correction.
Result/
conclusion
Particle Size distribution curve can be plotted using particle size and
percentage fineness.

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