Consolidation &Consolidation & SettlementsSettlements

updated April 23, 2007

Terms and DefinitionTerms and Definition

SettlementSettlementtotal vertical deformation at soil surface resulting from the load

Consolidation (volume change velocity)Consolidation (volume change velocity)rate of decrease in volume with respect to time

Compressibility (volume change flexibility)Compressibility (volume change flexibility)volume decrease due to a unit load

Contraction (temperature expansion)Contraction (temperature expansion)change in volume of soil due to a change in temperature

SwellingSwellingvolume expansion of soil due to increase in water content

ShrinkageShrinkagevolume contraction of soil due to reduction in water content

IntroductionIntroductionSoilsSoils

Considered elastic materialsVisco-elastic materials (time dependent in stress-t i )strain response)

But, visco-elastic only applicable to material that are linearSoil is highly nonlinear materialsSoil have a ‘memory’ non conservative materialPresent theory can’t handle that.Simplification

IntroductionIntroduction (cont’d)

When stressed soil deformWhen stressed soil deform

Stressed released deformation remains

Soil deformation :Soil deformation :Distortion (change in shape)

Compression (change in volume)

Both

Component of SetlementComponent of SetlementSettlementSettlement total vertical deformation at soil surface resulting from the load

Soil Movement:Downward

load increase or lowering water table

Up ardUpward temporary or permanent excavation

Points of interest:Points of interest:How muchHow fast

] settlement occurs

Component of SetlementComponent of Setlement (cont’d)

Total Setlement : St = Si + Sc + Ss

Si : immediate / distortion settlement Initial compressionSi : immediate / distortion settlementelastic theory, analog to deformation of column. 3D loading distortion in soil. compression modulus & volume of stressed soil

k

Initial compression

unknowndesign for shallow foundation

Sc : consolidation setlementPrimary consolidation

orm

atio

n

Sc : consolidation setlementtime dependent processoccurs in saturated fine-grained soillow coefficient of permeabilityS ttl t t d d f t Secondary consolidation

def

o

Settlement rate depend of pore water pressure

Ss : secondary compression (time dependent)time dependent

Secondary consolidation

Time (log scale)

time dependentoccurs at constant effective stressno subsequent changes in pore water pressures

Compressibility of SoilCompressibility of Soil

Compressibility (volume change flexibility) is the volume decrease due to a unit load

Assumption in settlement : 100% saturated and 1D (vertical) soil deformation

When soil is loaded it will compress because of:Deformation of soil grains (small, can be neglected)Compression of air and water in the voidspSqueezing out of water & air from the voids

Compressible soil mostly found below water table considered fully saturated

As pore fluid squeezed out:Soil grain rearrange themselves stable & denser configurationDecrease in volume surface setlement resulted

How fast? depend on permeability of soil

How much rearrangement & compression? depend on the rigidity of soil skeletonC i f d i t tlCompression of sand occurs instantlyConsolidation of cohesive soil is very time depend process

Consolidation of ClayConsolidation of Clayat equilibrium (t = 0)

System is analog to soil layer at equilibrium with weight of all soil layer (overburden) above it.X

valve closedpo (overburden load)

In equilibrium, valve is closed.

Piston is loaded, compresses a

Piston

spring in chamber.

Hydrostatic pressure = uo

spring

uo

water

spring ≈ soil skeleton

water ≈ water in poresp

valve ≈ pore sizes in soil / permeability

Consolidation of Clay yunder load Δp (0 < t < ∞)

Soil is loaded by increment ΔpSoil is loaded by increment Δp.

Valve initially closed.

valve closed initially

po + Δp X

Pressure (Δp) is transferred to the water.Piston

As water is incompressible and valve still closed, no water is out, no deformation of piston.

uo + Δuspring

Pressure gauge read : Δu = Δp where Δu is excess hydrostatic pressure.

uo + Δu

water

To simulate a fine grained cohesive soil, where permeability is low valve can be opened

spring ≈ soil skeleton

water ≈ water in soil void is low, valve can be opened.

Water slowly leave chamber.valve ≈ pore sizes in soil

Consolidation of Clay yat Equilibrium (t = ∞)

To simulate a fine grained cohesive soil, where permeability is low valve can be opened

=valve open

po + Δp is low, valve can be opened.

Water slowly leave chamber

po + ΔpS

As water flows out, load (Δp) is transferred to the spring.

Piston

spring

At equlibrium, no further water squeezed out, pore water pressure back to its hydrostatic condition.

water

uo

condition.

Spring is in equilibrium with load po + Δp

spring ≈ soil skeleton

water ≈ water in soil void

Δu = 0

Settlement ‘s’ existvalve ≈ pore sizes in soil

Setlement process:Setlement process:Initially all external load is transferred into excess pore y pwater (excess hydrostatic pressure)

No change in th effective stress in the soil

Gradually, as water squeezed out under pressure gradient, the soil skeleton compress, take up the load, and the effective stress increaseand the effective stress increase.

Eventually, excess hydrostatic pressure becomes zero y, y pand the pore water pressure is the same as hydrostatic pressure prior to loading.

Wh il i l d d t t l l t th itWhen soil is loaded to a stress level greater than it ever ‘experienced’ in the past, the soil structure is no longer able to sustain the increased load, and start to breakdown.

P lid ti P P

Voi

d ra

tio e

Preconsolidation Pressure - Pc:Maximum pressure experienced by soil in the past

Normal Consolidation: OCR = 1 Effective Consolidation Stress p’o

Pc

when the preconsolidation pressure is equal to the existing effective vertical overburden pressure Pc = p’opresent effective overburden pressure is the maximum pressure that soil has been subjected in the past

p’o

Pc

Over Consolidation: OCR > 1when the preconsolidation pressure is greater than the existing effective vertical overburden pressure Pc > p’opresent effective overburden pressure is less than that which soil h b bj t d i th t

Pc = p’o

Pc

has been subjected in the pastIt also said soil is in preconsolidated condition

OCR (over consolidation ratio) = 'c

o

P

pPc > p’o

p’o

Under Consolidation: OCR < 1when the preconsolidation pressure is less than the existing effective vertical overburden pressure Pc < p’o,

tl d it d il l i ll ll

op

Pc

e.g : recently deposited soil geologically or manually. p’o

Pc < p’o

Mechanism causing preconsolidationMechanism causing preconsolidationBrumund, Jonas, and Ladd (1976)

Change in Total Stress due toR l f b dRemoval of overburdenPast StructuresGlaciation

Ch iChange in pore water pressureChange in water table elevationArtesian pressureDeep pumping; flow into tunnelDessication due to surface dryingDessication due to plant life

Environmental changes such as pH, temperature and salt concentration

Chemical alteration due to ‘weathering’, precipitation, cementing agents, ion exchange

Consolidation Test data PlotsArithmetic scale Log scale

(a) (a)

trai

n ε

(%)

(a)

trai

n ε

(%)

(a)

Ver

tical

S

mv = coefficient of volume change

Ver

tical

S

Cce = modified compression index

Effective consolidation stress p’o (kPa) Effective consolidation stress p’o (kPa)

o (e

)

(b)

o (e

)

Cc = compression

(b)

Voi

d R

atio

av =coefficient of compressibility

Voi

d R

atio

Cc compression index

Effective consolidation stress p’o (kPa) Effective consolidation stress p’o (kPa)

Stress-strain history of a sedimentary clay during deposition sampling and reloading in the

1

Field virginO deposition, sampling and reloading in the

laboratory by the consolidation test:

OA represents the relationship between void ratio and the log effective stress of a particular element in the ground 0.9

Field virgin compression curve in situ

Rebound due to sampling

A

B

C g p gduring deposition. The process consolidates the element to point A. This point represents the in situ e vs log p’ocoordinates of the normally consolidated clay element.

When the boring is made and soil is sampled, overburden

C

C’Increasing

sample disturbance When the boring is made and soil is sampled, overburden

stressed are removed by the sampling operation and the samples rebounds or swells along curved AB.

When the sample is transferred from sampling tube into consolidometer ring and then reloaded in the

0.8

rati

o,

e

Laboratory consolidation

test curve

consolidometer ring and then reloaded in the consolidation test, the curve BC is obtained.

About point C, the soil structure start to break down and if the loading continuos the laboratory virgin compression curve CD is obtained

0.7

Vo

id r

curve CD is obtained

Eventually, the field curve OAD and lab curve BCD will converge beyond point D (approximately 0.4eo according to Terzaghi and Peck, 1967)0.6

ReconsolidationE

If the sampling operation was poor quality and mechanical disturbance to the soil dtructure occurred, curve BC’D would result upon reloading of the sample in the consolidometer.

ReboundD

F

The proconsolidation pressure is much more difficult to define when sample disturbance has occurred.

0.5

1 10 100

Pressure, p (log scale)

F

How to determine P ?How to determine Pc?(Cassagrande, 1936)

1. Choose point with minimum radius point. A

2 Dra hori ontal line from point2 6

2.8Pc possibility range

E D

2. Draw horizontal line from point A

3. Draw line tangent to the curve t i t A2 2

2.4

2.6

12

5

6A B

C

at point A

4. Bisect the angle made by step 2 and 3

1 8

2

2.2

d r

atio

, e

1

3

4

α

α

5. Extend the straight line portion of the virgin compression curve up to where it meets the 1 4

1.6

1.8

Vo

id 3

pbisector line obtained in step 4

6. Point of intersection step 4 and 5 is the (most probable) 1

1.2

1.4

( p )presonsolidation stress

point B

1

1 10 100

Pressure, p (log scale)

Pc

Settlement Calculation:Normally consolidated clay

voidsΔH = sΔe

eo

soil + water

solids

Hf

voids

solids

ef

1

Ho

1

H =

2 4

2.6

2.8

1vo o o o

L H s eor

L H H eε Δ Δ Δ

= = =+ 2

2.2

2.4

rati

o,

e

Cc

1 o v oo

es H H

e

e e e ee

εΔ= =

+

Δ1.4

1.6

1.8

Vo

id r Cc

1

1 2 1 2

22 1

1

'log ' log ' log ' log'

co

e e e eeC

pp p pp

− −−Δ= = =

Δ −1

1.2

1 10 100

P (l l )Effective Consolidation Stress p’

Pc

2

1

'log

1 'o

c co

H pS C

e p=

+

Pressure, p (log scale)Effective Consolidation Stress p’o

Settlement Calculation (cont’d):

For normally consolidated clay'H p p+ Δ

log1 '

'log

'

o oc c

o o

oc ce o

H p pS C or

e p

p pS C H

p

+ Δ=

++ Δ

=

p’1 = p’o, and p’2 include the additional stress Δp applied by the structure

when computing settlement using percentage vertical strain vs log effective pressure

For layered normally consolidated clay:

'H p p⎡ ⎤+Δ

op

In overconsolidated clay

log1 '

o oc c

o o

H p pS C

e p

⎡ ⎤+Δ= ⎢ ⎥+⎣ ⎦∑

In overconsolidated clay

'log

1 'o o

c ro o

H p pS C

e p

+Δ=

+p’1 = p’o, and p’2 =po+Δp < Pc

o op

'log log

1 ' 1 'o c o o

c r co o o o

H P H p pS C C

e p e p

⎛ ⎞ ⎛ ⎞+ Δ= +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

p’1 = p’o, and p’2 =po+Δp > Pc

Cr is the slope of rebound curve (swell index); Cr ≈ 20% to 10% Cc

o o o op p⎝ ⎠ ⎝ ⎠

Example:a. Perform Cassagrande construction and

find Pc = 121 kPaExample:The void ratio vs log effective pressure data shown in Fig Ex. 8.9. Determine: (a) the preconsolidation pressure P

c

b. Using point a and b, ea = 0.870, eb=0.655, p’a=100kPa, and p’b=300kPa.

(a) the preconsolidation pressure Pc

(b) the compression index Cc

(c) the modified compression index Cce2 2

1 1

0.870 0.6550.451

' ' 300loglog log

100' '

a be eeCc

p p

p p

−Δ −= = = =

Another way is to find Δe over one ‘log cycle’; for example log (1000/100) = log 10 = 1. Therefore Cc = Δe. I th fi th ti l l i t

1 1p p

In the figure the vertical scale is not sufficient for finding Δp = 1 log cycle, therefore it will be done in 2 steps:

Extend eaeb to one full log cycle on the a b g ysame graph, chose ec at the same pressure as eb. Draw the line eced parallel to eaeb.

Δe = Cc = (ea-eb)+(ec-ed)(0.870-0.655)+(0.9-0.664)= 0.451

c. The modified compression index Ccec. The modified compression index Cce

0.4510.242

1 1 0.865c

ceo

CC

e= = =

+ +

ExampleExample

Prior to placement of a fill covering a large area at a site thePrior to placement of a fill covering a large area at a site, the thickness of a compressible soil layer was 10m. Its original in situ void ratio was 1.0. Some time after the fill was constructed, measurements indicated that the average void ratio was 0 8measurements indicated that the average void ratio was 0.8. Estimate the settlement of the soil layer.

1 0 0 8eΔ −1.0 0.810m 1.0m

1 1 1.0oo

es H

e

Δ= = =

+ +o

Factors affecting the determination of Pc from laboratory test:g y1. Sample disturbance2. Load increment ratio (LIR)3 Load increment duration (LID)3. Load increment duration (LID)

1. Increasing sampe disturbance:Decreases the void ratio at any given value ofany given value of consolidation stressLowers the estimated value of Pc from the Cassagrande gmethodIncreases the compressibility at stresses less than PcDecreases the compressibility at stresses greater than Pc

2. Load Increment Ratio (LIR) denotes the ( )changes in consolidation stress divided by the initial consolidation stress.

LIR = p

p

Δ

op

3. Load increment duration denotes the total time tf allowed for consolidation prior to the application of the next load incrementthe application of the next load increment. Standard consolidation test often use a duration of 1 day for a each increment.

Plot preferencesPlot preferencesThe use of average vertical strain (ε) than void ratio (e) versus log effective t ( ’ ) i d d bstress (p’o) is recommended because:

Strains are easier to compute than void ratio

Differences in initial void ratio may cause samples to exhibit quite different plotsDifferences in initial void ratio may cause samples to exhibit quite different plots of void ratio versus stress but almost identical plots of strain versus stress

Settlements are directly proportional to strain, but use of Δe data also requires a knowledge of (1+e ) which introduces 2 variables Δe and (1+e ) This can onlyknowledge of (1+eo) which introduces 2 variables, Δe and (1+eo). This can only be determined at the end of test, not during the settlement test. The e vs log p’ocurve cannot be plootted during the test.

Strain plot are easier to standardize than void ratio plotsStrain plot are easier to standardize than void ratio plots.

Estimating field settlement is simple, percent compression can read directly from the graph,once a good estimate of in situ overburden pressure.

Field Consolidation CurveField Consolidation Curve Schertman (1955) Procedure

Normally consolidated SoilFind Pc using Cassagrande

Calculate eo (initial void ratio)

Draw a horizontal line from eo to Pc Point 1

Draw a horizontal line from 0.42 eo to the extension of laboratory virgin compression curve (L) Pointof laboratory virgin compression curve (L) Point 2

Draw a line from Point 1 to point 2 Field virgin consolidation curve (F)( )

Overconsolidated SoilCalculate eo and draw a line from eo to the existing overburden pressure p’o Point 1overburden pressure p o Point 1

Find Pc using Cassagrande.From point 1 draw a line paralel to rebound-reload curve to Pc Point 2

Next steps is similar to normally consolidated soil

Example 8 16Example 8.16 Holtz & Kovacs

The void ratio vs pressure data shown below. The initial void ratio is 0.725 and the existing vertical efective overburden

i 30 kPpressure is 30 kPa.

Void ratio 0.708 0.691 0.670 0.632 0.635 0.650 0.642 0.623 0.574 0.510 0.445 0.460 0.492 0.530

Pressure (kPa) 25 50 100 200 100 25 50 200 400 800 1600 400 100 25

Requiredq1. Plot the data as e vs log p’o2. Evaluate overconsolidation ratio3 Determine the field compression index using Schertmann procedure3. Determine the field compression index using Schertmann procedure4. If this consolidation test is representation of a 12m thick clay layer,

compute the settlement of this layer if an additional stress of 220 kPa were added

Solution 1. The data is plottedSolution2. The given value of p’o is plotted on

the graph, and from Cassagrande construction a value for Pc = 190 ckPa is found.

3. OCR = Pc/p’o = 190/130 = 1.46.The soil is slightly overconsolidated.g y

4. Using Schmertmann procedure for overconsolidated clay the values of Cr and Cc are 0.022 and 0.262

5. The settlement is:

'log log

1 ' 1 '

12 190 12 130 2200 022 log 0 262 log

⎛ ⎞ ⎛ ⎞+ Δ= +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

+⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

o c o oc s c

o o o o

H P H p pS C C

e p e p

m m0.022 log 0.262 log

1 0.725 130 1 0.725 190

0.025 0.484 0.509 0.5

+⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠= + = ≈m m m m

Time Rate of ConsolidationTime Rate of Consolidation

S

2%

tt

SU

S

Uπ

=

⎛ ⎞%0 to 60%,

4 100

60%, 1.781 0.933log(100 %)

tt v

t v

Ufor U T

for U T U

π ⎛ ⎞= = ⎜ ⎟⎝ ⎠

> = − −2

2 or v v dr

v v vdr v

t T HT c t

H c= =

Ut = average degree of consolidation (%)St = settlement of the layer at time tS = ultimate settlement of the layer due to primary consolidationTv = time factorHdr = average longest drainage path during consolidation

ffi i t f lid ticv = coefficient of consolidationtv = time for consolidation

Coefficient of consolidation (c )Coefficient of consolidation (cv)

ELog-of-time (Cassagrande)

d0

BC

D E

xx

1. Extend the straight line portion of primary and secondary consolidation curve to intersect at A. A is d100, the deformation at the end of consolidation

ncre

asin

g)

F

0.5(d0 + d100)

2. Select times t1 and t2 on the curve such that t2 = 4t1. Let the difference is equal to x

3. Draw a horizontal line (DE) such that the orm

atio

n (in d50

F

3. Draw a horizontal line (DE) such that the vertical distance BD is equal to x. The deformation of DE is equal to d0.

4. The ordinate of point F represents the deformation at 50% primary consolidation

Def

o

dA

deformation at 50% primary consolidation, and it abscissa represents t50

2

50

0.197 drv

Hc

t=

d100

50

Time (log scale)t50t2t1

Coefficient of consolidation (c )Coefficient of consolidation (cv)

Square-root-of-time (Taylor)

A

1. Draw a line AB through the early portion of the curve

2. Draw a line AC such that OC = 1.15 OB. The abscissa of D which is the

ncre

asin

g)

intersection of AC and the consolidation curve, gives the square-root-of-time for 90% consolidation.

2 orm

atio

n (in

D2

90

0.848 drv

Hc

t= D

efo D

Time (square root)B CO t90

Secondary ConsolidationSecondary Consolidatione e

CΔ Δ

22 1

1

log log log

e eC

tt tt

αΔ Δ

= =−

' '2

1

log where 1s

p

CtS C H C

t eα α

α= =+

ratio

e

void ratio at the end of primary consolidationpe =Voi

d

Δeep

ttTime (log scale) t2t1

ExampleExampleCalculate the settlementCalculate the settlement due to primary con-solidation for 5m clay

Surcharge = 50kPa

layer due to a surcharge of 50kPa applied at the

d l l Th l i

2m

S d

Sand50% saturation

Ground water table

ground level. The clay is normally consolidated.

Calculate the time rate of

5m SandGs=2.65, e=0.7

Calculate the time rate of settlement when cv is given as 0.85m2/yr 5m

ClayCc=0.45, eo=0.9 g y

Rock

c , o

γsat=15kPa

Rock

Solution

Submerged unit weight of

Solution

Calculation of Average effective Submerged unit weight of clay

Calculation of Average effective Overburden Pressure (po)

The moist unit weight of sand b th d t t bl

( )' 'clay sat clay wγ γ γ= −

So

above the ground water table

( )2.65 0.5 0.7 9.81

1 1 0.7s w w

sand

G Sr e

e

γ γγ+ ⋅ ⋅⋅ + ⋅ ⋅

= =+ + o sand sand clay

15 9.81 5.19 kPa

5p' 2 +3 ' + '

2γ γ γ

= − =

= ⋅ ⋅ ⋅

22.21kPa=y2

52 22.21+3 9.516+ 5.19 85.94 kPa

2= ⋅ ⋅ ⋅ =

Calculation of Settlement

Submerged unit weight of sand below the ground water table

' 'γ γ γ= − 'H + Δ

( )

( )

( )

1

1 12 65 1 9 81

sand sat sand w

s ws w ww

GG e

e e

γ γ γ

γγ γ γ

=

−⋅ + ⋅= − =

+ +− ⋅

'log

1 '

2.5 85.94 500.45 log

1 0 9 85 94

o oc c

o o

H p pS C

e p

+ Δ=

++

=+( )2.65 1 9.81

9.516 kPa1 0.7

= =+

1 0.9 85.940.592m 0.199m 0.792m 0.8m

+= + =

Time rate of settlementTime rate of settlement

( ) ( )

10%

2 22

; 0.8 10% 0.08m

2 5 0 008 2 5

tavg t c avg

c

sU s S U s

S

TT H

= = × = × =

Uavg TvSt

(m)t (yr)

( ) ( )2

10%

2.5 0.008 2.5; 0.06

0.85 0.85v dr

vv

TvT Ht yr t yr yr

c

⋅ ⋅= = = =

Time (yr)( )

10 0.008 0.08 0.06

20 0.031 0.16 0.23

30 0 071 0 24 0 52

0

0.1

0 1 2 3 4 5 6 7 8 9

Time (yr)

30 0.071 0.24 0.52

40 0.126 0.32 0.93

50 0.197 0.40 1.45

0.2

0.3t

(m)

60 0.287 0.48 2.11

70 0.403 0.56 2.96

80 0.567 0.64 4.17

0.4

0.5

Set

tele

men

t

90 0.848 0.72 6.24

95 1.163 0.76 8.55

100 ∞ 0.80 ∞

0.6

0.7

0.8100 0.80

ReferenceReferenceHoltz, R.D and Kovacs, W.D. (1981) An Introduction toHoltz, R.D and Kovacs, W.D. (1981) An Introduction to Geotechnical EngineeringDas, B.M. (1985) Principles of Geotechnical EngineeringTransportation Research Board Commision on Sociotechnical System (1976) Estimation ofSociotechnical System (1976) Estimation of Consolidation Settlement