Consolidation

William J. Likos, Ph.D.Department of Civil and Environmental Engineering

University of Wisconsin-Madison

GLE/CEE 330 Lecture NotesSoil Mechanics

grout injection-(361 holes)

Ground water pumping

Higher compressibility

Construction (1173-1399)

(Burland et al., 1998)

CPT Tip Resistance Profiles for North and South Sides of Tower

Lead weights on North side (~1993)

Soil Extraction(1999-2001)

• 33 tons of soil were excavated from under the north side

• Moved further toward vertical by 17.72 in.

• Now exhibits a 5-degree tilt• Rate of subsidence reduced to

less than a couple of millimeters per year

Soil Settlement and Compression

1) Immediate Settlement• Elastic deformation, undrained compression (sands, gravels)

2) Primary Consolidation• Time dependent settlement in saturated soil as water is squeezed

from voids due to increase in vertical effective stress (clays)

3) Secondary Consolidation• Particle reorientation, creep, organic decay; does not involve

expulsion of water (highly plastic clays, organics)

4) Distortion Settlement• Lateral movements near edges of loaded area

Changes in Vertical Effective Stress

1) Fill Placement

2) External Loading

3) Water Table Changes

4) Combinations of 1, 2, and 3

fillfillzz H 0''

inducedzzz 0''

Hfillfill

P

z’

z’z’

z’(e.g., water table lowering)

Subsidence from Water Table LoweringExample:

Initial GWT

Final GWT5’

5’A

psf

pcfpcf

zz

psfpcfpcf

zz

f

f

wwf

i

i

wwi

888'

'54.62'10120'

'

576''104.62'10120'

'

USGS

= 120pcf

Primary Consolidation:Piston-Spring Analogy

1. The container is completely filled with water, and the hole is closed. (Fully saturated soil)

2. A load is applied onto the piston, while the hole is still unopened. At this stage, only the water resists the applied load. (Development of excessive pore water pressure)

3. As soon as the hole is opened, water starts to drain out through the hole and the spring shortens. (Drainage of excessive pore water)

4. After some time, the drainage of water no longer occurs. Now, the spring alone resists the applied load. (Full dissipation of excessive pore water pressure. End of consolidation)

• Spring is analogous to effective stress (stress carried by soil skeleton)• Initially, the pore water takes up the change in total stress so effective stress does not

change• As excess pore water pressure drains, the effective stress increases (skeleton takes up

load)• Consolidation is complete when excess pressure dissipates and flow stops

• So consolidation is TIME DEPENDENT because it is a pressure dissipation (flow) process!• Depends on hydraulic conductivity (k) and length of drainage path (Hdr)

FlowP

00

'' vv

uu

0''0

0

v

vv

v

v

uuu

uuuA

P

v

sou

'

vvv

uu

0

0

''

kPaukPau

kPa

AAA

A

A

69'678.681.9

1368.45.1621917.18

Example 11.2 (Coduto, 1999)

Before Fill:

Short Term After Fill:

kPau

kPauuuukPa

AAA

AAA

A

69'16555.1967

23455.19136

00

Long Term After Fill:

kPaukPauu

kPa

AAA

AA

A

167'67

23455.19136

0

Consolidation (Oedometer) Testing1-D consolidation test:

• Undisturbed saturated soil (clay, silt) – representative of in-situ stratum• Typical specimen size: h = 1”, diam. = 2”-3”• Specimen confined in rigid ring (no lateral deformation, “plane strain”)• Drainage allowed on top and bottom via porous stones• Apply increment of load and measure 1-D compression with time

Trimming Procedures

(Bardet, 1997)

Assumptions1) All compression occurs due to change in void ratio

• i.e., the grains do not compress• Thus, we can relate change in void ratio (e) to change in volume

ethenVifVVe vs

v

2) All strains are vertical (1-D)

0

0

0

1

1

ee

ee

VV

VV

VVVVV

VV

VVwhereVV

HH

s

s

s

v

sv

sv

v

t

v

v

t

H0

He0

Procedures for Incremental Consol Testing1) Trimming2) Specimen set up and initialization (seating load, ’v0)3) Apply an increment of vertical load (’v = P/A)4) Record H with time, compute e with time5) Monitor until volume change ceases (~24 h)6) Repeat 3-5 to generate load-compression curve

time

e0

e1

e2

en

1st increment

2nd

nth

~24h

e0

e1

e2

en

log ’v

’v0 ’v1 ’v2 ’vn

We will use this data to predict magnitude of consol settlement

We will use this data to predict rate of consolidation

aka “e-log p” curve

The “e-log p” curve

San Francisco Bay Mud (Holtz and Kovacs, 1981)

Loading

Unoading

Loading

Unloading

Virgin (NC)Compression

Cc

Cs

Cr

’p

OC Compression

A

• Divide e-log p curve into linear segments• OC = Overconsolidated (stiff response)• NC = Normally Consolidated (soft response)

• OC and NC portions separated by ’p

• ’p = “maximum past pressure”

• Cr = Recompression Index (slope of OC)

• Cc = Compression Index (slope of NC)

• Cs = Swell Index (slope of unload response)

sr

rc

DCs

CBc

BAr

CCCC

eC

eC

eC

'log

'log

'log

B

CD

0.1 ~ Cc ~ 2.6

0.01 ~ Cr ~ 0.5

(Bardet, 1997)

NCCc

Cs

Cr

’p

OC

A

Maximum past pressure (’p ) quantifies The “stress history” of the soil – it is the largest magnitude of effective stress the soil has been consolidated to in the past.

Overconsolidation Ratio (OCR) quantifies the magnitude of a soil’s existing state of stress relative to its maximum past stress.

'' pOCR

B

CD

Stress History

If OCR = 1, then ’p = ’ and the soil is “normally consolidated” (soft response – virgin compression)

If OCR > 1, then ’p > ’ and the soil is “over consolidated” (stiff response – it has been precompressed)

Sources of Overconsolidation:• Extensive erosion• Past glacial activity• Removed structures• Risen water table• Evaporation

)10(009.0 LLC c

Disturbance & Empirical Correlations For NC or undisturbed specimens:

)10(007.0 LLC c

For OC or disturbed specimens:(Disturbance decreases virgin compressibility)

In general:

2.0~1.0c

r

CC

Disturbance “erases” stress history

Load-Rebound BehaviorMaximum past stress (aka preconsolidationStress) is a plastic yield stress.

If unloading occurs at a stress less than ’p, then the soil rebounds elastically.

If unloading occurs at a stress greater than ’p, then a new ’p results and the soil rebounds plastically.

A

BC

eA

ec

’A = ’C

’B’p

Similar to elastic-plastic response you learned about in Mechanics of Materials

yield stress

plastic deformation

Casagrande Construction for ’p

(Coduto, 1999)

Calculating Consolidation Settlement

fillFill

Soft Clay

Rock

1) Site characterization to quantify thickness of compressible layer.

2) 1-D laboratory consolidation testing to determine stress history and compression indices (’p, Cc, Cr, Cs). Will also need cv for rate prediction.

3) Divide compressible layer into sublayers.

4) Calculate initial (preconstruction) effective vertical stress at midpoint of each sublayer (’i).

5) Calculate final post-construction effective vertical stress at midpoint of each sublayer (’f).

6) Calculate consolidation settlement for each sublayer (H) and sum for total settlement.

p

fc

i

pr

eHC

eHCH

''

log1'

'log

1 00

1

2

3

4

H1

H2

H3

H4

Where did that equation come from?

NCCc

Cs

Cr

’p

OC

’i ’f

e0

ef

ee comes from compression in the OC range and in the NC range.

For OC portion:

For NC portion:

So,

0eee f

i

proc

iproc

Ce

Ce

''

log

'log'log

p

frnc

pfcnc

Ce

Ce

''

log

'log'log

p

fc

i

pr

f

pc

p

frncoc

eHC

eHC

eeHHH

CCeee

''

log1'

'log

11

''

log''

log

000

What if ’f < ’p ?

NCCc

Cs

Cr

’p

OC

’i

’f

e0

efIn other words, what if we put a load on a highly overconsolidated deposit? (maybe significant past glacial activity).

Consolidation will result solely from recompression.

Settlement will be relatively small because response is stiff.

Equation must be modified:

i

fr

eHCH

''

log1 0

What if ’i = ’p ?

NCCc

Cs

Cr

’i ’p

OC

’f

e0

ef

In other words, what if we put a load on a normally consolidated (NC) deposit?

Consolidation will result solely from virgin compression.

Settlement will be relatively large because response is soft.

Equation must be modified:

p

fc

eHCH

''

log1 0

1''

pOCR

Example – Fill PlacementProp. A B

Cc 0.25 0.20

Cr 0.08 0.06

e0 0.66 0.45

’p 101 kPa 510 kPa

•See Coduto for solution using 7 layers•Let’s try using two layers (A and B)

B

A4.5 m

9 m

kPa

kPa

Af

AiAf

Ai

Ai

233'

5.83.2060''60'

5.281.95.21923.18'

Compare to max past stress:

e

Log ’23310160

Need both terms

AiApAf '''

Prop. A BCc 0.25 0.20

Cr 0.08 0.06

e0 0.66 0.45

’p 101 kPa 510 kPa

mH

mmH

eHC

eHCH

A

A

p

fc

i

prA

59.049.0097.0101233log

66.01925.0

60101log

66.01908.0

''

log1'

'log

1 00

Compare to max past stress:

e

Log ’

361

510188

Only needCr term

BpBf ''

Now for layer B:

kPa

kPa

Bf

BiBf

Bi

Bi

361'

5.83.20188''188'

1681.995.1971923.18'

mHHHso

mH

mH

eHCH

BAtot

B

B

i

frB

8.0

21.0188361log

45.011806.0

''

log1 0

Compare to solution for 7 layers…. Htot = 0.83 m

http://www.prenhall.com/coduto/html/Geotechnical/Software.htm

Rate of Consolidation Settlement• Recall that consolidation is volume change due to pore water being squeezed out• Dissipation of excess pore pressure• So consolidation takes time!!! – depends on:

• Hydraulic conductivity (k)

• Drainage boundaries (max length of drainage path, Hdr)

fillFill

Soft Clay

Rock

k

Hdr

z

Static Pressure

us

+

+ =

z

Excess Pressure

u

z

u

TransientProfile

ImmediateLong Term

Consider a case of double-drainage (like an oedometer test)

Top porous disk

SaturatedClay

Bottom porous disk

H0

0u

0u

),( tzfu z

u

0t

1t

2tpt

• So there is a hydraulic gradient from middle of sample to boundaries• The magnitude of the gradient decreases with time• So flow rate decreases with time• So rate of volume change decreases with time

AdLdhkQ t Volume

Time

We want to be able to plan for this!

Drainage Path Length, HdrA

dLdhkQ t

SaturatedClay

Flow rate (rate of consolidation) depends on k and dht/dL

Hdr = H0

Impervious Rock,Aquitard, etc.

Ground surface, sand layer, etc.

Single Drainage

SaturatedClay

Hdr = ½ H0

Pervious layer, sand, etc.

Ground surface, sand layer, etc.

Double Drainage

Significantly decreases time for consolidation

Sand Drains at Kansai Intnl. Airport

1-D Consolidation Theory

SaturatedClay H0

0u

0u

),( tzfu z

u0t

1t

2tpt

We seek a solution for excess pressure a function of location and time

Dissipation of excess pressure is a diffusion process governed by 1-D PDE

),( tzfu 2

2

zuc

tu

v

cv = “Coefficient of Consolidation”

(determined from lab or field testing)

2

2

zuc

tu

v

'

)1(

ddea

aekc

v

wvv

Units of cv = L2/T (e.g., m2/year)

“Coefficient of compressibility”

e

’

av

or in terms of strain.…..

'

ddm

mkc

v

wvv

“Coefficient of

volume compressibility”

’

mv

Solution of the Consolidation EquationAssumptions:

1. Darcy’s Law is valid2. Soil solids and fluid are incompressible3. Sr = 100%4. Linear compressibility (const av or mv)

2

2

zuc

tu

v

SaturatedClay H0

0u

0u

z2

2

zuc

tu

v

Boundary Conditions: 0,

0,0

0

tHzu

tzu

Initial Conditions: 00, utzu

0

0 2

sin2,m

TM

dr

eHMz

Mutzu

Solution is infinite Fourier series:

2

122 dr

v

HtcTmM

(“Time Factor”)

How Can We Apply This Solution?Consider Kansai Airport….How much consolidation for given time?

2dr

v

HtcT

Define “Percent Consolidation” U

%100SSU t St = consol settlement at any time t

S∞ = consol settlement at end of primary

0

2

221m

TMeM

U

Note that U only depends on time (T) – Percent

consolidation is independent of load!!!

Example 1: How long for 90% consolidation?

90.0U

From figure, T = 0.848

yearst

daysdayft

t

cHt

HtcT

v

dr

dr

v

6.4

1696/05.0'10848.0

848.0

848.0

2

2

2

2

Claycv = 0.05 ft2/day 20’

Sand

If S∞ = 4.0’, how long for 2’ of settlement?

%)50(5.0'4'2

SSU t

From figure, T = 0.196

yearst

daysdayft

t

cHt

HtcT

v

dr

dr

v

1.1

394/05.0'10196.0

196.0

196.0

2

2

2

2

Claycv = 0.05 ft2/day 20’

Sand

If bottom boundary is impervious, how long for 90% consolidation?

From figure, T = 0.848

Claycv = 0.05 ft2/day 20’

Impervious Rock

90.0U

!!!!!!6.18

6784/05.0'20848.0

848.0

848.0

2

2

2

2

yearst

daysdayft

t

cHt

HtcT

v

dr

dr

v

Compute total consolidation settlement and time for 95% consolidation

OCR = 1, so clay is NC.

pi '' Clayk = 10-6 cm/se0 = 1.2Cc = 0.40G = 100 pcfOCR = 1

50’

Impervious Rock

20’Prop. fill= 100pcf

'5.49402940log

2.11'504.0

''

log1

294020100940'940254.6225100'

S

eHCHS

psfpsf

iA

fA

o

c

fA

iA

Analyze using Point A at midpoint of clay

A

25’

Not given cv, so need to calculate…

Clayk = 10-6 cm/se0 = 1.2Cc = 0.40G = 100 pcfOCR = 1

50’

Impervious Rock

20’Prop. fill= 100pcf

dayft

pcflbftdayscmftscmc

solbft

psfH

H

ddm

mkc

v

v

wvv

/009.14.62/105.4

/400,865.30/1/10

105.42000

'50'5.4

''

225

6

250

A

25’

If U = 95%, T = 1.129

'28.4'5.495.0

7.7/009.1'50129.1

2

22

SUS

yearsdayftc

HTt

t

v

dr

(Bardet, 1997)

Log-Time Method for Determining cv

time

e0

e1

e2

en

plot deformationvs. log time

1) Select some point near U = 50% (tb, db) (this is an estimate of d50)2) Find ta such that tb = 4ta 3) Calculate (db – da) and find d0 = da – (db – da)4) Find d100 graphically with two tangent lines5) Calculate actual d50 as ½(d0 + d100); find corresponding t50

6) Calculate cv using t50 and time factor T

50

2

250

50

197.0

197.0

tHc

HtcT

drv

dr

v

(Bardet, 1997)

Square-Root-Time Method for Determining cv

1) Extrapolate linear portion backward to find d0

2) Measure length of segment AB (linear portion) 3) Draw AC such that AC = 1.15(AB)4) Draw line through d0 and C to find d90 and 5) Calculate cv using t90 and time factor T

90

2

290

90

848.0

848.0

tHc

HtcT

drv

dr

v

90t

*Preferred method in practice (don’t need to wait for t100)

(Bardet, 1997)

Secondary Compression

• Additional time-dependent compression after primary consolidation• Not due to dissipation of excess pore pressure (expulsion of water)• Relatively small amount of volume change• Creep, particle reorientation, organic decomposition

ss

ss

ttC

ttCe

10

10

log

log

C = secondary compression index

C = mod. secondary compression index

ts = time of start of secondary comp.

t = time

Secondary Compression

nwC 0001.0