1

Compressibility &

Consolidation

CHAPTER 2

Settlement

If a structure is placed on soil surface, then the soil will undergo an elastic and plastic deformation.

In engineering practice, the deformation or reduction in the soil volume is seen as settlement or heave depending on either the load is increased or decreased.

Components of Total

SettlementS = Si + Sc + Ss

Si = Immediate SettlementSc = Consolidation SettlementSs = Secondary compression

2

Components of Total

SettlementImmediate settlement (Elastic settlement)

� This compression is usually taken as occurring immediately after application of the load.

Primary consolidation

� The process of compression due to extrusion of water from the voids in a soil as a result of increased loading.

Secondary consolidation

� Soil compression and additional associated settlement continue at a very slow rate, the result of plastic readjustment of soil grains.

Compression of soil

The compression of soil is due to� Elastic Deformation

� Deformation of soil grains (small and negligible)

Deformation of solid particles �

secondary compression

� Compression of air and water in the voids (water is incompressible)

� Squeezing out of water and air from the voids

� Expulsion of air from pores � Compaction

� Dissipation of pore-water � Consolidation

Immediate Settlement

Cohesive Soil

If saturated clay is loaded rapidly, the

soil will deform with virtually no

volume change due to without any

dissipation of water from the soil.

.

3

Immediate/Elastic

Settlement

General

( )αν1E

B q S 2o

i −=

Immediate/Elastic

Settlement

� A rigid foundation of size 3m x 3m carrying a uniform load of 1800 kN is lying on a compressible (medium soft) clay layer of infinite thickness. The undrained elastic modulus of clay (Eu) is 40Mpa. Poisson ratio of the clay is 0.5.Determine the average immediate settlement under this foundation.

Immediate/Elastic

Settlement

( )(0.82)0.5140000

(3) )(1800/(3x3 S 2

i −=

( )Csν1E

B q S 2o

i −=

m0092.0Si =

4

Immediate Settlement of Footing

on Sand

∑−=2B

0 s

zo21i Δz

EI

q)(qCCS

−−=

qq

q0.51C

o1

Consolidation Settlement

� Non-linear & irreversible

� Time dependent

CONSOLIDATION SETTLEMENT

When saturated soil is loaded � Load is taken by the water � water dissipate � Load is taken by the soil skeleton � Deformation

5

Oedometer Test

Water

Oedometer Test

Porous stone

Porous stone

Soil Specimen

Load

Confining ring

Ref: BS 1377 (Par 5)

Oedometer Test� Soil sample is placed in a metal ring (Figure 4).

� Purpose of the disks are to allow water to flow vertically into andout of the soil sample.

� Pressure is applied to the soil sample and dial readings(deformation) and corresponding time observation are made.

� Normally, this is done over a 24-hour period.

� Then graph a graph is prepared using these data with tie onlogarithmic scale (Figure 5).

� The procedure is repeated after the sample reached 100% ofconsolidation. The applied load then will be double.

� For each graph, the void ratio (e) and coefficient ofconsolidation (cv) that correspond to the specific appliedpressure (p) are determined.

6

Assumptions

� Saturated clay layer

� Laterally confined

� One dimensional flow

� Constant permeability

� Can be measured by oedometer Test

DATA ANALYSIS

Determination

of Cv

Determination of Cv

� Coefficient of consolidation (cv) indicates how rapidly (or slowly) the process of consolidation takes place.

� Log time method – Cassagrande

� Square root time method- Taylor

7

Determination of Cv, Log time

method

Most of the time it is difficult to get this straight line

1. Plot a graph relating dial reading (mm) versus log time2. Produce a straight line for primary consolidation and secondary

consolidation part of the graph. The two lines will meet at point C. 3. The ordinate of point C is D100= the deformation corresponds to U =

100%4. Choose time t1 (point A), t2 = 4t1 (point B), t3 = 4t2 etc. The difference in

the dial reading is equal to x. 5. An equal distance x set off above point A fixes the point D0 = the

deformation corresponds to U =0%. Notes that D0 is not essentially equal to the initial reading may be due to small compression of air within the sample.

6. The compression between D0 and D100 is called the primary consolidation.7. A point corresponding to U = 50% can be located midway between D0 and

D100. The value of T corresponds to U = 50% is 0.196. 8. Thus where Hd = half the thickness of the specimen

50

2196.0

t

HC

dv =

Determination of cv, Square root of time

method

8

1. Extent the straight line part of the curve to intersect the ordinate (t = 0) at point D. The point shows the initial reading (Do). The intersection of this line with the abscissa is P.

2. Take point Q such that OQ = 1.15 OP.

3. The intersection of line DQ and the curve is called point G

4. Draw horizontal line from G to the ordinate (D90). The point shows the value of √t90. The value of T corresponds to U = 90% is 0.848.

5. Thus

Hd is half the thickness of specimen for a particular

pressure increment.90

2848.0

t

HC d

v =

How do we get the factor

1.15

848.090 Fd =

197.059

59

50 Fd =

15.11525.1197.05

9848.0

59

50

90 ≈==F

F

d

d

SQUARE-ROOT OF TIME VS

LOG TIME METHOD

�Generally square root of time method is better�Square root of time methods usually gives

lower Cv

�Square root of time method is easier to program in computer

�k computed from Cv almost always less then measured value; slower compression

�To decide which is correct � compare the kvalue

9

SQUARE-ROOT OF TIME VS

LOG TIME METHOD

� The square root of time method works wellbased on the assumption of NO secondaryconsolidation

� Use strictly for vertical drainage� Cannot be used for some soils such as peat,

k changes very much when subjects to changein effective stress.

� Square root of time method is OK because Terzaghi theory does not account for secondary compression anyway

Data Analysis – Stress Strain curve

0.40

0.45

0.50

0.55

0.60

0.65

0.70

0.75

0 200 400 600 800 1000 1200 1400 1600 1800

pressure (p)

void

rat

io (

e)

Results of consolidation testing are presented by plotting:

� the void ratio (e) at the end of each increment period against the

corresponding effective stress (linear scale) or e-p' curve

21

21

'' σσσ −−

=∆∆

=eee

av

−−

+=

+=

21

1v ''1

1

1m

σσee

ee

a o

oo

v

Problems:The value of mv is not constant, and only true for a certain range of stress.

10

0.40

0.45

0.50

0.55

0.60

0.65

0.70

0.75

1 10 100 1000 10000

pressure (p)

void

rat

io (

e)

�the void ratio (e) at the end of each increment per iod against the corresponding effective stress (log scale) or e-log p' curve

Cc

Cr

1

2

21

loglog

σσσeee

Cc−=∆

∆=

Empirical formula & Correlation with other properties

Empirical formula Soil type

Cc = 0.009 (LL –10)Cc = 0.007 (LL –10)Cc = 1.15 (eo – 0.35)Cc = 0.30 (eo – 0.27)Cc = 1.15 x 10-2 ωnCc = 0.75 (eo – 0.50)

All clay (undisturbed)Remolded soilAll clay (undisturbed)An organic soilOrganic soilLow plasticity soil

Empirical Formulas, correlation of Cc with liquid limit, void ratio, and water content

Based on Critical State model (Wroth & Wood (1978)

Variation of Cc (normalized by 1+eo) with natural water content (Lambe&Whitman, 1979)

200

PIGC sc =

Correlation between Cc

and Cr� Recompression index (C r) can be estimated

as 0.05 – 0.20 Cc� Critical State concept

Cr = Cc (1-A)

for G s = 2.7, and A = 0.8 � Cr = PI/370

11

OVERCONSOLIDATED SOIL

Over-consolidated soil is the soil that has undergone recompression in oedometer test.

Over-consolidation is judged if pre consolidation p ressure σ’c obtained from e-log p’ curve based on oedometer tes t is greater than the currently existing overburden pres sure (σ’o). OR Cc obtained from test < Cc = 0.009 (LL-10) OR LI is between 0 to 0.6

The pre-consolidation pressure should be estimated based on e-log p curve

o

cOCR

'

'

σσ=

1. Choose by eye the point of maximum curvature on the consolidation curve (point A)

2. Draw a horizontal line from point A

3. Draw a line tangent to the curve at point A

4. Bisect the angle made by step 2 and 3

5. Extent the straight line portion of the virgin compression curve up to where it meets the bisector line obtained in step 4. The point of intersection of these two lines is the

approximate value of the pre-consolidation pressure (point B)

12

Over-consolidation is judged if pre

consolidation pressure σ’c obtained from

e-log p curve based on oedometer test is

greater than the currently existing

overburden pressure (σ’o).

o

cOCR

'

'

σσ=

Insitu e - log p’ curve

Due to effect of sampling and preparation, the specimen in the Oedometer test will be slightly disturbed

� decrease in the slope of thecompression line

Cc actual > Cc measured

The field virgin compression line should be estimated based on the e-log p curve

Estimation of

field consolidation curve

In-situ curveSlightly disturbed

13

Obtaining the field consolidation line

1. Produce two horizontal lines at eoand 0.42 eo

2. Produce a vertical line at pressure equal to σ’o or σ’c. This line should intersect the horizontal line at eo at point 1.

3. Plot compression line (Cc) based on the oedometer test. The line will intersect the horizontal line at 0.42 eoat point 2.

4. Draw a line from point 1 to point 2. The slope of this line is the field compression index (Ccf).

Obtaining the field consolidation line

Settlement Calculation

14

CONSOLIDATION SETTLEMENT:

ONE-DIMENSION

Consider a layer of saturated clay of thickness H, due to construction, the total vertical stress in an elemental layer of thickness dz at depth z is increased by ∆σ

dz

z

H

Sc

∆σeo

e1

σ’o σ’1

∆σ

Initial stress + stress

increment

Effective overburden

pressure σ’o

Applied pressure calculated using

stress distribution (elastic) ∆σ’

o

o

occ

σ'

Δσσ'

e

HCS

++

= log1

If we use mv, then

Sc = mv ∆σ’ H or Sc = av/(1+eo) ∆σ’ H

Settlement of Normally Consolidated Soil

15

c

o

oc

o

c

orc

σ'

Δσσ'

e

HC

σ'

σ'

e

HCS

++

++

= log1

log1o

o

orc

σ'

Δσσ'

e

HCS

++

= log1

Settlement of Over-consolidated Soil

DATA ANALYSISSettlement-Time Relationship

Rate of Consolidation

Terzaghi 1-D consolidation theory

Assumptions:

1. Compression and flow are 1-dimensional2. The compressible layer is homogenous and

saturated3. Soil particles and water is incompressible4. Darcy’s law is valid5. Strain due to external load is small and

within the range of elasticity6. The coefficient mv and k remain constant

throughout the process7. There is a unique relationship, independent

of time, between void ratio and effective stress

16

b

d

uniform load

excess pore pressure distribution

sandwater table

clay

dh

h

dz

dx

zd= H/2

d= H/2

H

a

c

uo = ∆σ

∆σ'zu’z

Distribution of excess pore water pressure in a clay layer subjected to uniform load

PRIMARY CONSOLIDATION �

The period of consolidation where the volume change of soil is due to the drainage of water driven by

excess pore water pressure � Basic Equation:

where

t

u

z

uCv

∂∂=

∂∂

2

2

wv γm

kCv=

General solution to basic consolidation equation is given by Taylor (1948)

where Z and T are non-dimensional parameters. Z is geometry factor, which is equal to z/H, T is Time factor, in which

( )∑∞=

=

−=n

n

TfZf0

2112 )()('σ'σu

22dwd

v

H

t

mv

k

H

tCT

γ==

General solution to basic consolidation equation is given by Taylor (1948)

where Z and T are non-dimensional parameters. Z is geometry factor, which is equal to z/H, T is Time factor, in which

( )∑∞=

=

−=n

n

TfZf0

2112 )()('σ'σu

22dwvd

v

H

t

m

k

H

tCT

γ==

17

Degree of Consolidation (U)

ii

iz u

u

u

uu

Δσ

σ'σ'

σ'σ'

σ'σ'U −=−=−=

−−= 11

12

1

Initially U = 0, U =100% when void ratio is equal to ef.

∑=

−=0

211n

z (T)f(Z)fU

fz ee

eeU

−−=

1

1

The degree of consolidation can be expressed in the form of void ratio or change in effective stress or thechange in p.w.p.

Relationship between U and Tv is presented in the form of Table, Graph or formula

Relationship among U, T, and depth �

Isochrones

Single drainage vs Double

drainage

Pervious

Pervious

Pervious

Soil layer

Soil layer

H

H

drainage

drainage

Drainage, Hd = H

Hd = H/2Impervious

18

ConsolidationPercentage

U average

Tv

Double drainage

10%20%30%40%50%60%70%80%90%95%100%

0.10.20.30.40.50.60.70.80.90.951.0

0.0080.0310.0710.1260.1960.2870.4030.5670.8481.136

α

Cassagrande: U = 0% to 60% � Tv = π/4 (U%/100)2

U > 60% � Tv= 1.781 – 0.933 log(100%-U%)

Relationship between Tv and U

Relationship between Tv

and U

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0 10 20 30 40 50 60 70 80 90 100

Percent consolidation (%)

Tim

e F

act

or,

Tv