CIV 307

SOIL MECHANICS

Soil Mechanics

• Origin of soil: Rock cycle and origin of soil

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Rock Cycles

Soils

(Das, 1998)

The final products due to weathering are soils

Major soil deposits in India

1.Alluvial deposits

2.Black cotton soils

3.Lateritic soils

4.Desert soils

5.Marine deposits

• Soil Particle sizeBased on the size of the particle soil classified

into 4 types:1.Clay: >0.002mm

2.Silt: 0.002mm to 0.075mm

3. Sand: 0.075mm to 4.75mm

4.Gravel: >4.75mm /08/10

Soil particle size classificationGrain size(mm)

Name of Organization Gravel Sand Silt Clay

MIT classification (Massachusetts Institute of Technology)

>2 2 to 0.06 0.06 to 0.002

<0.002

US dept. of Agriculture (USDA) >2 2 to 0.05 0.05 to 0.002

<0.002

AASHTO classification 76.2 to 2 2 to 0.075 0.075 to 0.002

<0.002

Unified soil classification system 76.2 to 4.75

4.75 to 0.075

0.075 to 0.002

<0.002

Specific Gravity (G)

• Specific gravity (G) : Ratio of the unit weight of the given material to unit weight of the water

• Sp. gravity = s / w

• Sp. gravity of sandy soil = 2.65 to 2.8

• Sp. gravity of Clay soil = 2.6 to 2.9

• Application: It is required to solve the geotechnical problems

Mechanical analysis of soil

• Objective: To determination of the size range of particles present in a soil

Two methods:1. Sieve analysis: for particles larger than 0.075mm

in diameter

2. Hydrometer analysis: for particles smaller than 0.075mm in diameter

Sieve Analysis

• Principle: Finding grading of coarse grained soil through the set of sieves

• Apparatus/Procedure: Seives sizes• For Gravel : 80mm,40mm, 20mm,10mm, 4.75mm

• For Sands : 2mm, 1mm, 600, 425, 212, 150, 75

• Output result/Application: Percentage finer

Hydrometer Analysis

• Principle: Stokes law: v = (s-w),D2/18

V= particle velocity

• Apparatus/Procedure: Hydrometer placed into cylinder with a liquid of sodium hexameta phosphate

Output result / Application :s = density of solidsw = density of waterGs = Sp. Gravity of solids= Viscosity of liquidsL = distanceT = time

t

LwGSD .).1(

18

Particle size distribution curve

• From graph: Graph analysis

• Effective size: D10D10 = Particle size such that 10% of the soil is finer

than this sizeD60= Particle size such that 60% of the soil is finer

than this size• Uniform coefficient (Cu): D60/D10

Cu for sands - 6 - called well graded sandsCu for gravel – 4 – called well graded gravels

• Coefficient of curvature (Cc): (D30)2/(D60XD30)

• Cc for well graded soils lies between 1 to 3

Plastic size distribution curve

Uses of Particle size distribution curve

Usefull for coarse grained soils & fine grained soils

Others:

• 1. Classifying the coarse grained soils

• 2. Coeff. Permeability will depend on the particle size. It can determine accurately

• 3. useful to design the drainage filters

• 4. it provides an index to the shear strength of the soil.

• 5. Soil stabilization and pavement design

Particle shape

• 1. Bulky

• 2. Flaky

• 3. Needle shaped

Problem1: The results of sieve analysis of a soil are given below, total massofsample900gm Draw the particle size distribution curve hence determine the uniformity coefficient and the Cc

IS Sieve

20mm

10mm

4.75mm 2mm 1.0mm

0.6mm

4.25 212 150

75 pan

Mass of soil retained (gm)

35 40 80 150 150 140 115 55 35 25 75

Solution: Calculation for percentage finer N than different sizes are shown

IS Sieve Mass retained% retained = ((2)/900)x100

Cumulative % retained

% finer (N) = 100 - (4)

20mm 35 3.89 3.89 96.11

10mm 40 4.44 8.33 91.67

4.75mm 80 8.89 17.22 82.78

2.0mm 150 16.67 33.89 66.11

1.0mm 150 16.67 50.56 49.44

0.6mm 140 15.56 66.12 33.88

425 115 12.78 78.9 21.1

212m 55 6.11 85.01 14.99

150m 35 3.89 88.9 11.1

75m 25 2.78 91.68 8.32

PAN 75 8.32 100  

weight – volume relationships• Weight- volume relationship:• Three phase diagram(block diagram): Soil mass consists of 3

constituents

• 1. Solids(soil)

• 2. Water

• 3. Air

• Actually it can’t be segregated, a 3-phase diagram is an artifice used for easy understanding and convenience for calculations.

• When the soil absolutely dry (no water content) – 2-phase diagram.

• When the soil fully saturated (no air)- 2 phase diagram.

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Three Phases in Soils

S : Solid Soil particle

W: Liquid Water (electrolytes) A: Air Air

Phase diagram

Three Volumetric Ratios

• (1) Void ratio e (given in decimal, 0.65)

• (2) Porosity n (given in percent 100%)

• (3) Degree of Saturation S (given in percent 100%, 65%)

)V(solidsofVolume

)V(voidsofVolumee

s

v

)V(samplesoilofvolumeTotal

)V(voidsofVolumen

t

v

%100)V(voidsofvolumeTotal

)V(watercontainsvoidsofvolumeTotalS

v

w

e1

e

)e1(V

eVn

s

s

Engineering Applications (e)

•Typical values •Engineering applications:

–Volume change tendency–Strength

(Lambe and Whitman, 1979)

Simple cubic (SC), e = 0.91, Contract

Cubic-tetrahedral (CT), e = 0.65, Dilate

Link: the strength of rock joint

)itan(strengthShear n

i

1. Void ratio (e): Volume of voids to the volume of solids.

e = Vv/VsExpressed in decimal like 0.4 , 0.5..etcFor coarse grained soil smaller than that for fine grained soils.Some soils it may have a value even greater than unity.

2. Porosity (n): Volume of voids to the total volumen = Vv/V

Also called percentage of voids, expressed in percentage,The porosity of soil cannot exceed 100%Reason: Vv cannot be greater than V

Both porosity and void ratio are measures of the denseness or looseness of soil.

As the soil becomes more and more dense, e, n values will decrease

Relationship for ‘e’ and ‘n’:

1/n = V/Vv

1/n = Vs/Vv+Vv/Vv

1/n = e+1

n = e/(1+e),

e = n/(n-1)

3. Degree of saturation (s): Ratio of the volume of water to the volume of voids.

S = Vw/Vv Expressed in percentage (%) It is zero when the soil is absolutely dry It is 100% when the soil is fully saturated

4. Percentage air voids (na): Ratio of the volume of air to the total volume.

na = Va/V …….expressed in percentage (%)

5. Air content (ac): Ratio of the volume of air to the volume of voids.

ac = Va/Vv….expressed in percentage (%)

Both air content and percentage air voids are zero when the soil is saturated (Va = 0).

6. Water content (w): Ratio of the weight of water to the weight of solids in a given soil sample.

w = Ww/WsExpressed in percentage (%).

7. Unit weight (): Weight of soil per unit volume = W/V

From the drawing, = W / V, = Ws+Ww/V = Ws(1+Ww/Ws)/V = Ws(1+w)/V

8. Dry unit weight (d): d = Ws/V (Weight of solids per unit volume)

Relationship between moisture content (w, d and ),

d =/(1+w)

Weight Relationships

• (1)Water Content w (100%)

• • For some organic soils w>100%,

up to 500 %• For quick clays, w>100%

• (2)Density of water (slightly

varied with temperatures)

•(3) Density of soil•a. Dry density

•b. Total, Wet, or Moist density (0%<S<100%, Unsaturated)

•c. Saturated density (S=100%, Va =0)

•d. Submerged density (Buoyant density)

%100)(

)(

s

w

MsolidssoilofMass

MwaterofMassw

)V(samplesoilofvolumeTotal

)M(solidssoilofMass

t

sd

)V(samplesoilofvolumeTotal

)MM(samplesoilofMass

t

ws

)V(samplesoilofvolumeTotal

)MM(watersolidssoilofMass

t

wssat

wsat'

333w m/Mg1m/kg1000cm/g1

Relationship among unit weight, void ratio, moisture content and sp.gravity

e

wGsVWsd

e

wGsw

e

wGswwGs

V

WwWsVW

wGswWswWw

wGsWs

1

./

1

.).1(

1

.../

...

.

Considering the volume of soil solids is 1, If the volume of soil solids is 1, then the volume of voids numerically equal to void ratio, e. The weights of soil solids and water can be given as

Because the weight for the soil element under consideration is w.Gs.w, the volume occupied by water

Gsww

wGsw

w

WwVw .

..

Hence from the definition of degree of saturation, S = Vw/Vv = w. Gs/ e/08/10

• Various unit weight relationships: (Usefull for solveing the problem)

Moist unit weight () Dry unit weight(d) Saturated unit weight(sat)

wSnnwGs

wnwGs

SGsw

wGsw

e

weSGs

e

wGsw

..)1(.

)1)(1(.

.1

.).1(

1

)..(

1

.).1(

wnsatd

e

wesatd

nwGsd

e

wGsd

wd

.

1

.

)1(.

1

.

1

wnGsnsat

e

weGssat

)].).1[(

1

).(

• Problem:1 The mass of a soil sample having a volume of 0.0057m3 is 10.5kg, the moisture content (w) and the specific gravity of soil solids(Gs) were determined to be 13% and 2.68, respectively, Determine

a. Moist density, b. Dry density, d, c. Void ratio, e, d. Porosity, n, e. Degree of saturation, S (%)

Solution: = M/V, d =/ (1+w), e = (Gs.w/rd)-1, n = e/(1+e), S(%) = (w.Gs/e)x100

Problem. 2, A soil has void ratio = 0.72, moisture content = 12% and Gs= 2.72. Determine its(a) Dry unit weight (b) Moist unit weight, and the(c) Amount of water to be added per m3 to make it saturated

Problem.3, The dry density of a sand with porosity of 0.387 is 1600 kg/m3. Find the void ratio of the soil and the specific gravity of the soil solids.

Problem: A moist soil sample weighs 3.52N. After drying in an oven, its weight is reduced to 2.9N. The specific gravity of solids and the mass specific gravity are respectively 2.65 and 1.85. Determine the water content, void ratio and the degree of saturation. w = 10kN/m3

Solution: From given data: Weight of water Water content, w

= Gm.w

d = /(1+w)

d = G.w/(1+e)

n = e / (1+e)

S = w.G/e

Relative density

Application: In granular soils, the degree of compaction in the field can be measured according to the relative density.

Relative density (Dr) = (emax – e)/(emax – emin)

Expressed in percentage (%)

emax = void ratio of the soil in the loosest state

emin = void ration of the soil in the densest state

e = insitu void ratio of the soil

The term relative density is commonly used to indicate the insitu denseness or looseness of granular soil

Some values for granular soilRelative density (%) Description of soil deposit

0 - 15 Very loose

15 - 50 Loose

50 -70 Medium

70 - 85 Dense

85 - 100 Very dense/08/10

Problem:2- For a given sandy soil, emax = 0.82, emin = 0.42, Gs = 2.66. In the field the soil is compacted to a moist density of 1720 kg/m3 at a moisture content of 9%. Determine the relative density of compaction.

Solution: Calculate the ‘e’ value

= (1+w).Gs.w/(1+e)

Dr = (emax – e) / (emax – emin)

Problem:3 The laboratory test results of a sand are as follows, emax = 0.91, emin = 0.48, and Gs = 2.67. What would be the dry and moist unit weights of this sand when compacted at a moisture content of 10% to relative density of 65%? Ans: 16.07kN/m3, 17.68kN/m3

Problem: A fully saturated sample of soil has a volume of 25cc. And a weight 0f 40gm. After drying in the oven, its weight is 28gm. With the help of Phase diagram, calculate the, 1.Void ratio, 2. Water content 3. Porosity, 4. degree of saturation 5. Saturated unit weight. (Ans: e= 0.923, w = 42.9%, n = 48%, s =100%, sat = 1.6 g/cc

Problem: when a given soil sample of sand was tested in the laboratory, the void ratio in the loosest and densest possible states were 0.95 and 0.4 respectively . Calculate the 1. Relative density 2. Degree of saturation (Ans: e=, s = %, Dr = %)

Problem. Tests on fill reveal that one cubic metre of soil on the fill weighs 1624 kgs and after being dried weighs 1.40 tonnes. If the specific gravity of solida is 2.65, determine w, e, n, s of the soil mass from the first principles.

Consistency of soil – Atterberg Limits

• Liquid limit (LL): It is the moisture where the soil sample starts behaves like a liquid.

• Plastic limit (PL): It is the moisture where the soil sample starts behaves like a plastic

• Shrinkage Limit (SL): The moisture content at which the transaction from solid to semi solid state takes place.

Liquid Limit test

Objective: determination of liquid limit from clay silt soil

Apparatus: Liquid limit test device, grooving tools

Output: Flow index, IF = (w1 – w2) / log(N2/N1)

w1 = moisture content of soil, in percent, corresponding to N1 blows

w2 = Moisture content corresponding to N2 blows

Liquid limit test device and grooving tools

Graph: X – Number of blows, N

y – Moisture content,w (%)

Flow curve for liquid limit determination of clayey silt

Plastic Limit Test

Plastic limit: The moisture content in percent at which the soil crumbles, when rolled into threads of 3.2mm in diameter.

Test: Determination of plastic limit from the laboratory

Plasticity Index

Plasticity Index: The difference between the Liquid limit and plastic limit.

PI = LL – PL When either LL or PL cannot be determined, the soil is non

plastic. When the plastic limit is greater than Liquid limit, the

plasticity index reported as zero(not negative).

Liquidity index : LI = (w – PI)/(LL – PL)

w = water content of the soil in natural condition.

When the soil is at its liquid limit its liquidity index is 100% When the soil is at its plastic limit its liquidity index is zero

Consistancy Index (Ic):

W = Natural water content,

PI = plasticity index

Wl = Liquid limts

Activity: A = PI / (% of clay size fraction, by weight)

100.PI

wwlIc

Placticity chart

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Soil classification chart

Classification based on Plasticity chart

Silts and clays Liquid limits is 50% or less

ML = Inorganic silts with low plasticity

CL = Inorganic clays of low to medium plasticity

OL = Organic silts of low plasticity

Aline , PI = 0.73(LL – 20)

Silts and clays Liquid limits is greater than 50%

MH - Inorganic silts of high plasticity

CH – Inorganic clays of high plasticity

OH – Organic clays of medium plasticity

Pt - Peat or muck or highly organic soils

Problem: The weight of a moist soil sample is 20kg and its volume is 0.011m3. After drying in an oven, the weight of sample reduces to 16.5kg. Determine the w, gd, g, e, n, s.

Soil structureSingle grained

Loose state

Dense state

Structure of Sand

Honey combed

Structure of clays

Sediment Structure

Dispersion Nonsalt Flocculation

Salt Flocculation (Na, K )

Engineering classification of soils

AASHTO Classification system:

Soil classified into Seven major groups:

Soil classified under groups A-1, A -2, A -3 are granular material of which 35% or less of the particles pass through the no.200(0.075mm) sieve.

Soils of which more than 35% pass through the no.200 sieve are classified under groups

A -4, A – 5, A -6 & A -7/08/10

AASHTO classification…..

1. Grain size

a. Gravel : Particle passing through 75mm sieve and retained on the No.10(2mm) US sieve

b. Sand: Particle passing through the no.10(2mm) and retained on the No.200 (0.075mm) US sieve.

c. Silt and Clay: Particle passing the no.200 US sieve

2. Plasticity Chart:

Group Index (GI):

Significance: To evaluate the quality of soil as a highway material.

GI = (F200 – 35)[0.2+0.005(LL – 40)]+0.01(F200 – 15)(PI -10)

Where, F200 = Percentage passing through the No. 200 sieve

LL = liquid limit

PI = Plastic limit

1st term - (F200 – 35)[0.2+0.005(LL – 40)] – Partial group index determined from liquid limit

2nd term - 0.01(F200 – 15)(PI -10) – Partial group index determined from plastic limit

Group Index…….

Note: In general the quality performance of soil as a subgrade material is inversely proportional to the GI

1. If in the equation yields a negative value for GI, it is taken as 0.

2. The group index calculated is rounded off to the nearest whole number

3. There is no upper limit for the group index

4. The group index of soils belonging to groups A–1-a, A-1-b, A-2-4, A-2-5 and A-3 is always zero

Problem 1: The result of the particle size analysis of a soils as follows: a. % passing through no.10 sieve = 100b. % passing through no.40 sieve = 80c. % passing through no.200 sieve = 58liquid limit and plasticity index of the minus no.40 fraction of the soil are 30 and 10 respectively. Calculate the GI Solution:

GI = (F200 – 35)[0.2+0.005(LL – 40)]+0.01(F200 – 15)(PI -10)

Problem:2, Ninty five percent of a soil passes through the no.200 sieve and has a liquid limit of 60 and plasticity index of 40. Calculate th GI.

Solution:

GI = (F200 – 35)[0.2+0.005(LL – 40)]+0.01(F200 – 15)(PI -10)

Unified Soil Classification System

Main points:

1. The classification is based on material passing a 75 sieve.

2. Coarse fraction = percent retained above no.200 sieve = 100 – F200 = R200

3. Fine fraction = percent passing no. 200 sieve = F200

4. Gravel fraction = percent retained above no.4 sieve = R4

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Classification

• As per USCS soil divided into groups1. Coarse grained: Gravel or sand in nature with less

than 50% passing through the no.200 sieve. - It represents clear symbols – G (Gravel), S (sand)

2. Fine grained soils: Soils with 50% or more passing through the no.200 sieve. - The group symbols starts with ‘M’ (inorganic silts) - ‘C’ for inorganic clays, ‘o’ organic clays or silts - ‘Pt’ used for peat, muck and other highly organic

soils

Other symbols used for the classification:

W – well graded

P - poorly graded

L – Low plasticity (LL is less than 50%)

H – High plasticity (LL more than 50%)

Engineering classification of soil

Engineering classification of soil

AASHTO / USCS classification

1. Both systems are divide the soil into two major groups, that is coarse and fine grained soils as separated by sieve no . 200 (75)

2. a. According to the AASTHO system a soil is considered fine grained when more than 35% passes through the no.200 sieveb. USCS more than 50% passes through the no.200 sieve.

3. a. USCS Gravelly and sandy soils clearly separated,

b. AASHTO it is A-2

4. Clear symbols used in USCS (GW, SM, CH) , but in A in AASHTO classification

5. a. AASHTO system, the no.10 sieve is used to separate gravel from sand.

b. USCS no. 4 sieve is used, upper limits appropriate concrete and highway technology using no. 10(2mm) sieve.

6. a. USCS clearly organic soil classification (OL, OH, Pt) are given,

b. in AASHTO there is no classification for organic soils