Index Properties of Soil

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.

3phase.wmf

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

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.

2phase-sat.wmf

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 

2phase-dry.wmf

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.

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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 = MV

Dry Density

  • It is the ratio of dry mass of soil to volume of soil

  • Symbol 🡪d

Saturated Density

  • It is the ratio of saturated mass of soil to volume of soil

  • Symbol 🡪sat

Submerged Density

  • It is the ratio of submerged mass of soil to volume of soil

  • Symbol 🡪'

  • '= sat- w= sat- 1 …….density of water 🡪w=1 g/cc

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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= WV

Dry Unit Weight

  • It is the ratio of dry weight of soil to volume of soil

  • Symbol 🡪d

Saturated Unit Weight

  • It is the ratio of saturated weight of soil to volume of soil

  • Symbol 🡪sat

Submerged Unit Weight

  • It is the ratio of submerged weight of soil to volume of soil

  • Symbol 🡪'

  • '= sat- w= sat- 9.81 …….unit weight of water 🡪 w=9.81 g/cc


Relation between density and unit weight

  • γ= ρ ×g= ρ ×9.81

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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 = MwMs= WwWs

  • Water content experiments formulae

    • Oven drying formula,  w=M-MsMs

    • Pycnometer method, w=M2-M1M3- M4 ×G-1G-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 = soil solidswater= sw

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 = soilwater= w

  • Specific gravity experiments formula

    • Pycnometer method, G=M2-M1(M3- M4)-(M3-M4)

    • 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


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 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 = VvVs

Porosity

  • It is the ratio of the volume of voids to the total volume of soil. 

  • Unit less quantity/ expressed as a percentage

  • n = VvV

Degree of saturation

  • It is the ratio of the volume of water to the total volume of voids. 

  • Unit less quantity/ expressed as a percentage

  • S = VwVv

  • 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 = VaVv

  • 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. 

  • Unit less quantity/ expressed as a percentage

  • na = VaV

Density Index/ relative density

  • DI=RD=ID = emax- eemax- 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= 11+e, 

Q: Prove that, e=  n1- n

Q: Prove that, e=  n1- n

Step 1)  three phase system

https://2.bp.blogspot.com/-0ZspKxTBf1c/UiAB1KYNrhI/AAAAAAAABW4/HSTq-MvCDDw/s320/Three+phase+diagram.jpg

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= VvVs

porosity = n= VvV

Step 3) Derivation:

e= VvVs

Add 1 on both the sides

1+ e=1+ VvVs

1+ e= Vs+ VvVs

From figure it is clear, Vs+ Vv=V

Hence,  1+ e= VVs

Rearranging V on right hand side,

1+ e= 1VsV

Here Vs = V-Vv

1+ e= 1V- VvV= 1VV- VvV= 11- VvV

From step 2

1+ e=  11- n

OR

1- n=  1 1+ e

Hence, 

Subtract 1 from both side, 

1+ e-1=  11- n-1

e=  1-(1-n)1- n

e=  n1- 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

Step 1)  three phase system

https://2.bp.blogspot.com/-0ZspKxTBf1c/UiAB1KYNrhI/AAAAAAAABW4/HSTq-MvCDDw/s320/Three+phase+diagram.jpg

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:

Degree of saturation = Sr= VwVv

Voids ratio = e= VvVs

Water content = w= wwws

Specific gravity = G= sw

Step 3) Derivation:

We have, Specific gravity = G= sw …………..(1)

Where,                s= WsVs           and            w= WwVw           

Putting these values in equation (1)

G= WsVsWwVw

Rearranging 

G= WsVs× VwWw

Consider volume and weight terms together, 

G= WsWw× VwVs

From basic formulae,

G= 1w× VwVs

Take w on left hand and Multiply-divide by Vv on right hand side

G w=  VwVs× VvVv

G w=  VwVv× VvVs

Again from basic formulae,

G w=  Sr× e

Rearranging 

Sr e= w G

Proved

Q: Derive relation between unit weight/density, dry unit weight/density and water content.

  • Q: prove that, d= 1+w

Step 1)  three phase system

https://2.bp.blogspot.com/-0ZspKxTBf1c/UiAB1KYNrhI/AAAAAAAABW4/HSTq-MvCDDw/s320/Three+phase+diagram.jpg

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:

Dry unit weight = d= WsV = WdV

Bulk unit weight = γ= WV

Water content = w= wwws

Step 3) Derivation:

We have, Bulk unit weight = γ= WV   …………..(1)

Where, W = Wv +Ws  =Wa +Ww +Ws = Ww +Ws ………………….Wa=0

γ= Ww+ WsV

γ= WwV + WsV

γ= WwV + d

Multiply-divide by Ws as shown below,

γ= WwV× WsWs + d

From basic formula,

γ= WwWs× WsV + d=w× d + d

Taking common factors

γ=w+1d

Rearranging 

d= 1+w

Proved

Q: Derive relation between degree of saturation, voids ratio, bulk unit weight and specific gravity

  • Q: prove that, = G+e S w1+ e

  •  Q: prove that, sat= G+e w1+ e

  • Q: prove that, d= G w1+ e

Step 1)  three phase system

https://2.bp.blogspot.com/-0ZspKxTBf1c/UiAB1KYNrhI/AAAAAAAABW4/HSTq-MvCDDw/s320/Three+phase+diagram.jpg

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:

Degree of saturation = Sr= VwVv

Voids ratio = e= VvVs

Water content = w= wwws

Specific gravity = G= sw

Dry unit weight = d= WsV = WdV

Bulk unit weight = γ= WV

Saturated unit weight = sat= WsatV


Step 3) Derivation:

We have, Bulk unit weight = γ= WV   …………..(1)

Where, W = Wv +Ws  =Wa +Ww +Ws = Ww +Ws ………………….Wa=0

γ= Ww+ WsVV+  Vs

Taking common from denominator (RHS)

γ= Ww+ WsVVVs+ 1 Vs= Ww+ Wse+ 1 Vs

γ= 1(1+e)Ww+ Ws Vs= 1(1+e)Ww Vs+ Ws Vs      

γ= 1(1+e)Ww Vs+ s

Multiply-divide by Vv & Vw as shown below,

γ= 1(1+e)Ww VsVvVv ×VwVw+ s

γ= 1(1+e)Ww VwVvVs ×VwVv+ s

γ= 1(1+e)w×e × Sr+ s

Taking common w from bracket

γ= 1(1+e)e × Sr+ sww

Then, 

γ= 1(1+e)e × Sr+ Gw

Rearrenging 

γ= G+e Sr w(1+e)

If Soil is dry, Sr = 0 and become d

d= G w(1+e)

If Soil is saturated, Sr = 1 and become sat

d= (G+e) w(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 Organic

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 75 micron and less than 4.75mm.  it is divided into three part coarse, medium and fine gravel as shown in table.

  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 Organic

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 than 0.075 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.


gradation curve.wmf

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 = D60D10

  • 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 = D302D60D10

  • 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

states of consistency.wmf

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:

    • Liquid limit (WL) - change of consistency from plastic to liquid state

    • 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-wIp  

  • Liquidity Index = IL = w - wPIp

  • Consistency Index + Liquidity Index = 1 🡪 IC + IL = 1

  • Flow index = slope of liquid limit curve (line) from logarithmic plot

Flow index = IF = w1- w2n2n1

Where w1 and w2 are water contents corresponding to number of blows n1 and n2 (in liquid limit test: experiment) respectively

  • Toughness Index = IT = IpIF

  • Activity of clays (Ac) = Ip% by weight of 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

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