Ge i-module3-rajesh sir

Geotechnical Engineering IGeotechnical Engineering - I

Compressibility

Dr. Rajesh K. N.Assistant Professor in Civil EngineeringAssistant Professor in Civil EngineeringGovt. College of Engineering, Kannur

Dept. of CE, GCE Kannur Dr.RajeshKN

Module III

Compaction: definition and objectives of compaction –definition and objectives of compaction –Proctor test and modified Proctor test –concept of Optimum Moisture Content and maximum dry density –zero air voids line –zero air voids line factors influencing compaction –effect of compaction on soil properties –field compaction methods - Proctor needle for field controlp

Consolidation: definition - Compressibility –p ycoefficient of volume change and compression index –Laboratory consolidation test –e-log p curves - pre-consolidation pressure –Terzaghi's theory of one dimensional consolidation –Time rate of consolidation –difference between consolidation and compaction


ConsolidationConsolidation• Compression: Decrease in volume due to a stress• Compression: Decrease in volume due to a stress

• Compressibility: Property of a soil mass by virtue of which compression happens in itcompression happens in it

• Consolidation: Decrease in volume and a consequent escape of pore water

• Swelling (Opposite process of consolidation): Increase in water content due to an increase in the volume of voids

• Principle of consolidation: Spring analogy (Terzaghi and Frohlich)


Consolidation may be due to:y

•External static loads from structures

•Self-weight of soil

•Lowering of groundwater tableg g


The total compression settlement of a clay strata consists of:

1. Immediate compression settlement2. Primary consolidation settlement: . a y co so dat o sett e e t: 3. Secondary consolidation settlement

•Immediate compression settlement: Settlement occurs almostsimultaneously with the applied load – due to immediate compressionsimultaneously with the applied load due to immediate compressionof soil layer under undrained condition

• Soil mass assumed as elastic• Soil mass assumed as elastic


• Primary consolidation settlement:

• Due to primary consolidation

• Primary consolidation:

• If the rate of compression is controlled solely by the resistanceof the flow of water under the induced hydraulic gradients, theprocess is called primary consolidation

• Reduction in volume due to squeezing out of water fromvoids – till equilibrium (i.e., till all pore pressure is carried bysoil solids)

• Explained by means of one dimensional consolidation theory.


•Secondary consolidation settlement:

• Due to secondary consolidation, occurring at a very slow ratey g y

• Starts after primary consolidation stops, i.e., after excess porewater pressure approaches zerop pp

• Secondary consolidation is assumed to proceed linearly withlogarithm of timeg


One dimensional consolidation

• Settlement of a structure many times happens due to thecompression of a soft clay layer between layers of sand or stiffer clay

• Adhesion between soft and stiff layers prevents lateral movementof soft layers

• Theory based on this assumption: Terzaghi’s theory of onedimensional consolidation

• In the laboratory, this condition is simulated by confinedcompression or consolidation test


Consolidation of laterally confined soil

AA

Virgin compression curve

Recompression

s ra

tio e

BCE i

Recompression

Voi

ds

DExpansion

0 100

logCe e C σσ′

= −′

E

log σ’ (log of normal pressure)P id ti f ld d i


Pressure-voids ratio curve for remoulded specimen

loge e C σ ′= −

e0 - initial voids ratio corresponding to0σ ′loge e C σ ′

= −0 100

logCe e Cσ

=′ e - final voids ratio corresponding to

CC – compression indexσ ′0 10

0

logCe e Cσ

=′

CC=( )10log

eσ

Δ′Δ

Slope of pressure-voids ratio curve( )10log σΔ

Coefficient of compressibility 0v

decrease in voids ratio e e eai i

− −Δ= = =

′ ′ ′Δff f p y

0v increase in pressure σ σ σ′ ′ ′− Δ

Coefficient of volume changechange in volumem =Coefficient of volume change vm

initial volume increase in pressure=

×

( )1 1 1v v v Δ( )( )

0

0

0

0 0 0

1 1 11 1 1

v v s vv

v s s

v v v e e e ame e ev v vσ σ σ

− − −Δ= = = =

′ ′ ′Δ + Δ + Δ ++


If laterally restrained1Hm −Δ

= H m H σ ′∴Δ = − ΔIf laterally restrained,0

vmH σ ′Δ 0vH m H σ∴Δ Δ

CONSOLIDATION SETTLEMENT

0vH m Hρ σ ′= Δ = Δ

00 0 10

0 0 0

log1 1

Ce e CH He e

σρσ′−

= =′+ +0 0 01 1e e σ+ +

Empirical formulae for compression index:

0.007( 10)c lC w= −Skempton’s formula:(for remoulded clays) wl - liquid limit in %wl liquid limit in %

0 009( 10)C w= −Terzaghi and Peck formula:


0.009( 10)c lC w=g(for normally consolidated clays)

wl - liquid limit in %

Problem X: A soil has a compression index of 0.28. At a stress of 120 kN/m2,the void ratio was 1 02 Calculate:the void ratio was 1.02. Calculate:

a) void ratio if the stress on the soil is increased to 180 kN/m2

b) total settlement of the stratum of 6m thickness

a) Void ratio if the stress on the soil is increased to 180 kN/m2

0 10logCe e C σ ′= −

′ 101801.02 0.28log120

= − 0.97=0 100

gC σ ′ 10g120

b) Total settlement of the stratum of 6m thickness

logCCH σρ′

b) Total settlement of the stratum of 6m thickness

0.28 1806 log 0 1460 100 0

log1

CHe

ρσ

=′+ 106 log

1 1.02 120=

+0.146 m=


Terzaghi’s Theory of One-dimensional Consolidationg y

rate of expulsion of excess pore waterRate of change of rate of expulsion of excess pore waterfrom a unit volume of soil during the same time interval

Rate of change of excess hydrostatic pressure

α

( )01v

v w v w

k ekcm aγ γ

+= = Coefficient of consolidation

2v w v wγ γ2cm s

Coefficient of consolidation combined effect of permeability and compressibility on the rate of volume change


Laboratory consolidation testy

• Consolidometer (Oedometer) - Devised by Terzaghi• Fixed ring type and floating ring type consolidometers• Fixed ring type and floating ring type consolidometers


• In a fixed ring type consolidometer, the brass ring is fixed and the t t i f t d dtop porous stone is free to move downwards

• In floating ring type, the brass ring is floating and both the top and bottom porous stones are free to move towards the middle

Measurements:

• Specimen allowed to consolidate under vertical pressure increments 10, 20, 50, 100, 200, 400, 800, 1000 kN/m2

• Each pressure increment kept until compression virtually ceases (say, 24 hrs.)

• Vertical compression measured at various time intervals for each pressure increment: 0.25 hrs, 1 hr, 2.25 hrs …, 24 hrs.


• Final compression for each pressure increment also recorded

Pressure-voids ratio curves

• Obtained from the observations of laboratory consolidation test

• Voids ratio of the sample under consolidation test is determined at the endVoids ratio of the sample under consolidation test is determined at the endof each load increment

h f l d h d

• If hs is the thickness of the solid matter of the sample and h is the thickness of the sample at the end of a load increment

1. Height of solids method

sh he −=Voids ratio at the end of the load increment

thickness of the sample at the end of a load increment,

s

eh

Voids ratio at the end of the load increment

s dV Wh = =hs can be determined as: shA GA

hs can be determined as:

Volume of solids, Dry weight of soil, s dV W


Specific gravity, Cross sectional area of sample G A

2. Change of voids ratio method

• Assuming full saturation at the end of the test, final voids ratio can be found as:

2. Change of voids ratio method

found as:

.f fe w G= Final voids ratio, Final water content= = f fe w

• Change of voids ratio under each pressure increment can be found as follows:

( )( )

( )( ) 1 1

vf v vf v s f

f fvf s vf s s

v v v v v e eH A H eH AH e ev v v v v

− − −Δ Δ Δ= = = = =

+ ++ +( ) ( ) f fvf s vf s s

( )1 fe He

+ Δ∴Δ =e

H∴Δ =

• Knowing Δe, working backwards from the known value of ef , the voids


fratio corresponding to each pressure can be found

• On a semi-log scale, pressure voids ratio curve (e-logp curve) is drawn

• e-logp curve should pass through A, the point corresponding toinitial voids ratio, if the test conditions are exactly as in the field

• This rarely happens, as the samples will atleast slightly differ fromthe site conditions


To find coefficient of consolidation

Square root of time Logarithm of time qfitting method

gfitting method

2T dCasagrande Taylorv

vT dc

t=

vT Time factor→2U⎧ ⎛ ⎞ , 60%

4 100v

U UT

U

π⎧ ⎛ ⎞ <⎪ ⎜ ⎟⎪ ⎝ ⎠= ⎨⎛ ⎞⎪

100.9332log 1 0.0851, 60%100U U⎛ ⎞⎪− − − >⎜ ⎟⎪ ⎝ ⎠⎩

100t

f

degree of compressionU ρρ

→ = ×


,2

iaverage drainage path H H for double dd rainage− Δ→ =

Square root of time fitting method

2vT d

From a plot of compression dial reading Vs square root of time

0 848 f 90% lid ti Tvvc

t=

2d

0.848 for 90% consolidation vT =

90

90

0.848vdct

∴ =


Logarithm of time fitting method

2T d

From a plot of compression dial reading Vs log of time

vv

T dct

=


250

50

0.197vdct

∴ =


Problem 1:



Problem 2: Two clay layers A and B are respectively 4m and 5m thick. Thei k f l A h 50 % lid i i 6 h C l l htime taken for layer A to reach 50 % consolidation is 6 months. Calculate the

time taken by the layer B to reach the same degree of consolidation. Thecoefficient of consolidation of layer B is half that of layer A. Both layers haved bl d idouble drainage.

50 6 monthst =Layer A50 m2

2Hd = =

250

50

0.197vdct

=

220.1976vc = ×

50 2

m/months0.13133=6

50 ?t =Layer B

50 m5 2.52

d = =50 ?t

m/months0.13133

2vc =

50 22 22 0.13133 2.50.197 0.197c = × ⇒ = ×2

50

0.197 0.1976 2vc

t⇒

50 18.75monthst∴ =


50 18.75monthst∴

Problem 3: A 2.5 cm thick soil sample of clay was taken from field forpredicting the time of settlement for a proposed building which exerts ap g p p guniform pressure 100 kN/m2 over the clay stratum. The sample was loaded to100 kN/m2 and proper drainage was allowed from top and bottom. It wasseen that 50 % of the total settlement occurred in 3 minutes. Find the timerequired for 50 % of the total settlement of the building if it is to stand on 6mthick layer of clay which extends from ground surface and is underlain bysand.

2500 197 d50

50

0.197vct

∴ =250

50 0.197 dt =50vc


Problem 4: A clay layer 3.6 m thick is sandwiched between layers of sand.Calculate the time the clay layer will take to reach 50% consolidation TheCalculate the time the clay layer will take to reach 50% consolidation. Thecoefficient of consolidation is 4x10-4 cm2/s.

250d

50 m3.6 1.82

d = =

50

50

0.197vdct

=

4 cm/s4 10vc −= ×

2

50 ?t =

2

50 4

0.197 1804 10

t −

×=

×s15957000= days185=


Based on consolidation history, a soil can be:y

• Pre-consolidated (over-consolidated) soil:– In the past, it has been subjected to a pressure in excess of its

present overburden pressure• Normally consolidated soil: Normally consolidated soil:

– It has never been subjected to an effective pressure greater than its present overburden pressure,

– It is completely consolidated by the existing overburden• Under-consolidated soil:

– not fully consolidated under the present overburden pressure – not fully consolidated under the present overburden pressure


Pre-consolidation pressure: The temporary overburden pressure to which a soil has been subjected and under which it got consolidated

Determination of pre-consolidation pressure

which a soil has been subjected, and under which it got consolidated

Undisturbed sample of clay is Casagrande

Undisturbed sample of clay is consolidated in lab pressure-voids ratio curve plotted (semilog)

Point A of maximum curvature (min radius) selected

AB – horizontalAC - tangentAD - bisector

Straight portion of virgin curve extended to meet AD at P


pσ ′ is the pre-consolidation pressure

Secondary consolidation

• After excess pore pressure is dissipated, change in voids ratio continues at a reduced rate – secondary consolidation

• Highly viscous water forced out, plastic readjustment, progressive fracture of some particles

• Terzaghi’s theory not applicable• Terzaghi s theory not applicable

• It is assumed that secondary consolidation is linearly increasing with log of time

210log te

tαΔ = −

1t

t2 Total elapsed time since load was applied

t1 Period required for primary consolidation to complete

– taken as time for 90% consolidation


α Coefficient representing rate of consolidation

Compaction


Compaction

• Rapid reduction in voids by expulsion of air deliberately produced by mechanical means in the field or in the laboratory

• Objectives: – increase in shear strength– reduction in permeability– reduction in settlement under loading

• In the field, compaction can be done by varying water content, amount of compaction and type of compaction

• Proctor (1933) showed that there is a definite relation between water content and degree of the attained dry density

• For a particular compactive effort, soil attains a maximum dry density at a water content known as ‘optimum moisture content’(OMC)


p ( )

Factors affecting compactiong p

• Water content

• Amount and type of compaction

• Type of soil: Well-graded coarse-grained soils have lower OMCthan fine-grained soils having larger specific surface

• Admixtures: can be incorporated in the soil mass to improve itscompaction properties


Measurement of Compaction

1d wγγ =+

In terms of dry density 1 w+

• Standard Proctor Test (AASHTO Test) R. R. Proctor, 1933( )

• Modified Proctor Test

,

Procedure for Proctor Tests• Sample with a known water content is filled in a mould by applying a p y pp y gspecific compactive effort

• Compaction(dry density) is measured

• For the same sample, the above steps are repeated for varying water contents

• From the curve drawn water content corresponding to maximum dry


• From the curve drawn, water content corresponding to maximum dry density(OMC) is found

STANDARD PROCTOR TESTSTANDARD PROCTOR TEST

Compaction Apparatus

• Weight of hammer = 2.5 kg, free fall of hammer= 30 cm, no. of l 3 f bl l 25

p pp


layers=3, no. of blows per layer = 25

MODIFIED PROCTOR TESTMODIFIED PROCTOR TEST

• Weight of hammer = 4.5 kg, free fall of hammer = 45cm, no. of l 5 f bl l 25layers=5, no. of blows per layer = 25

• More compactive effortMore compactive effort


Effect of moisture content and compactive effort on compaction (cohesive soils)compaction (cohesive soils)


• Optimum moisture content (OMC) is the water content at which soil attains• Optimum moisture content (OMC) is the water content at which soil attainsa maximum dry density for a particular compactive effort.

• Curve shifts to the left with increase in compactive effort


Comparison of moisture content and shear strength curves

•Highest shear strength attained at a moisture content lower than Optimum


g g pmoisture content (OMC) at which soil attains a maximum dry density

Zero air voids line• Line showing relationship between moisture content and dry

density of a soil having zero air voids ( all the voids are filled with water)

( )1 n Gγ−

)• Otherwise called 100% saturation line

( )11

a wd

n GwG

γγ =

+an % air voids

01

wa d

GnwGγγ= ⇒ =

+Equation of zero air voids line

• Imaginary lineIt i i ibl t l ll th t d i– It is impossible to expel all the entrapped air

– Hence, no compaction curve will cross the zero air voids line


• Air voids line can be drawn for any value of % air voids

wd

GwGγγ =⎛ ⎞

1r

wGS

⎛ ⎞+ ⎜ ⎟⎝ ⎠

• Sr = 1 corresponds to zero air voids (100 % saturation)


Compaction curve for sandp

• Films of water around particles can keep them away anddecrease density, initiallyy y

• More compactive efforts have less effect on cohesionless soilsthan on cohesive soils


Problem 3: In a standard proctor test, the following values were obtained.Volume of the mould was 940x10-6 m3 and specific gravity of the soil solidsg ywas 2.6.

Weight of wet sample (N) 16.5 17.25 17.75 17.9 17.75

%Water content % 19.1 20.5 22.3 22.5 24

i) Determine maximum dry density and OMCii) Determine percentage air voids, degree of saturation and air content at) p g , gmaximum dry densityiii) Draw zero air voids line and find theoretical max dry density at OMCiv) Draw 10% air voids line)

Water content % 19.1 20.5 22.3 22.5 24

18.35 18.89 19.04 18.89Wet density 316 5 10 kN−×Wet density

Wet weight

Volume

γ =

3

6

16.5 10940 1

17.5530 3

3

kNm

kN m

γ −=

=

××

15.23 15.45 15.54 15.2217.5531 0.191dγ =+

Dry density γ


14.738 3kN m=1d wγγ =+

Plot moisture content Vs dry density curve

Max si

tyMax

dry densitydr

yde

ns

i t t tOMC

moisture content

3⎫3max ( )

1622.6

kN/mMax dr

%

y densityfrom cur

O C ve

M

dγ ⎫=⎬

= ⎭


To get percentage air voids at maximum dry density (at OMC)

( )11

a wd

n GwG

γγ

−=

+( )1 2.6 9.81

161 0 226 2 6

an− ×=

+ ×1 wG+ 1 0.226 2.6+ ×

0.0041 0.41%an∴ = =

To get degree of saturation Sr at max dry density

. . %a

wd

GwGγγ =⎛ ⎞

To get degree of saturation Sr at max dry density

2.6 9.81160 226 2 6×

=⎛ ⎞×

1r

wGS

⎛ ⎞+ ⎜ ⎟⎝ ⎠

0.226 2.61rS

⎛ ⎞×+ ⎜ ⎟⎝ ⎠

0 989 98 9%S 0.989 98.9%rS∴ = =

Hence, air content at maximum dry density


, y y

1.0 0.989 0.0109 1.0988 %ca = − = =

To plot zero air voids line

1w

dG

Gγγ =

+Equation of zero air voids line

1 wG+q

Water content % 19.1 20.5 22.3 22.5 24

Dry density for zero air voids

16.64 16.15 16.09 15.71

1w

dG

wGγγ =

+

2.6 9.81 17.041 0.191 2.6dγ

×= =

+ ×

Max dry densityTheo. max dry density at OMC

zero air voids linede

nsity

Max dry density

dry

d

Dept. of CE, GCE Kannur Dr.RajeshKNmoisture contentOMC

To plot 10 % air voids line (90 % saturation line)

Water content % 19.1 20.5 22.3 22.5 24

Dry density for 10% air voids 2 6 9 81×voids

16.02 15.51 15.46 15.061

wd

GwGS

γγ =⎛ ⎞

+ ⎜ ⎟⎝ ⎠

2.6 9.810.191 2.61

0.916 44

dγ×

=×⎛ ⎞+ ⎜ ⎟

⎝ ⎠=rS⎝ ⎠ 16.44=

ensi

ty

10% i id li

dry

de 10% air voids line

Dept. of CE, GCE Kannur Dr.RajeshKNmoisture content

Equipments for field compactionq p p

• Smooth wheel roller

• Rubber tyred roller

• Sheepsfoot roller

• Vibratory rollerVibratory roller


Field control of compactionp

• To attain a desired density by compaction, periodic measurements ofmoisture content and dry density will be required during the compactionmoisture content and dry density will be required during the compactionprocess

• The methods for determining moisture content and dry density in the field,g y yseveral methods are available

– Core-cutter method

– Sand replacement method

Nuclear method– Nuclear method

– Proctor needle method


Proctor needle method

• Used for rapid determination of moisture contents in-situ.

• Equipment consists of a needle attached to a springloaded plunger, calibrated to read penetrationresistance in kg/cm2

• Needle can have a suitable bearing area for the soil(larger bearing area for cohesive soils)


• In the lab, sample is compacted in the mould and penetration resistancemeasured using Proctor needlemeasured using Proctor needle.

• Water content and dry density are found

• The above process is repeated on the same sample for varying moisturecontents.

• Thus laboratory calibration curve is drawn


•To get moisture content in the site, sample of wet soil is compacted intothe mould and penetration resistance read off from the Proctor needle

M i t t t di t th t ti i t bt i d•Moisture content corresponding to the penetration resistance obtainedfrom the laboratory calibration curve

•Method is fast It gives fairly accurate results for fine grained cohesive•Method is fast. It gives fairly accurate results for fine-grained cohesivesoils

•Presence of gravel in the soil makes the readings less reliablePresence of gravel in the soil makes the readings less reliable

•Not very accurate for cohesionless soils


Summaryy

Consolidation: definition - Compressibility –coefficient of volume change and compression index –Laboratory consolidation test –

l l de-log p curves - pre-consolidation pressure –Terzaghi's theory of one dimensional consolidation –Time rate of consolidation –diff b t lid ti d tidifference between consolidation and compaction

Compaction: definition and objectives of compaction definition and objectives of compaction –Proctor test and modified Proctor test –concept of Optimum Moisture Content and maximum dry density –zero air voids line –zero air voids line –factors influencing compaction –effect of compaction on soil properties –field compaction methods - Proctor needle for field control


field compaction methods Proctor needle for field control

Ge i-module3-rajesh sir

Engineering

Transcript of Ge i-module3-rajesh sir