gdii lecture 1 [compatibility mode].pdf

8/9/2019 GDII Lecture 1 [Compatibility Mode].pdf

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BEng in Civil EngineeringSubject (CSE410)

Geotechnical Design II

岩土工程設計II

Jian-Hua YIN 殷建華Office: ZS909, Tel: 2766-6065Email: [email protected]


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Outline of Lectures by JH YIN:Lecture 1: Consolidation of soils (1-D, 2-D and 3-D)

Lecture 2: Pile foundation

Lecture 3: Soil nailed slopeLecture 4: Excavation and soil reinforcement

Lecture 5: Ground modification

Essential References:(1) Lecture notes.

(2) Das, Braja M. (2007). Principles of Foundation Engineering (6

th

edition),Thomson, United States (ISBN 0-534-40752-8) (or more updated version).

(3) Craig, R.F. (2004). Soil Mechanics, 7th edition (6thor 5th edition), Spon

Press, London and New York (ISBN 04-415-32702-2) (or more updated

version).


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Review of Soil Mechanics

(a) Terzaghi’s theory of one-dimensional consolidation

A total of 8 assumptions are made in the theory:

1 The soil is homogenous (ok for each layer )

2 The soil is fully saturated (ok for soil underwater )

3 The solid particles and water are incompressible (ok )

4 Compression and flow are one-dimensional (vertical) (ok )

5 Strains are small (ok for most civil engineering problems)6 Darcy’s law is valid at all hydraulic gradients (ok for water and

common civil engineering problems)

7 The permeability k and volume compressibility mv

are constants

through the process (approximate for a layer and small pressure)8 There is a unique relationship, independent of time, between void

ratio and effective stress (approximate for a layer and silty soils, no

good for soft clay)


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z v

dz z

vvdvv z z z z

z

uk

z

uu z

k

z

hk kiv e

w

w

e s

z z

)(

)headtotalstatic(constant sw

s hu

z

dz z

uk

z

uk dz

z

vvdvv e

w

e

w

z z z z 2

2

The condition of continuity: net water

coming out rate = volume compression rate

t

V dxdydz

z

uk vdxdydvvdxdy e

w

in z out z z

2

2

wv

ve

vee

w

ev

ev

e

w

e

w

ev

e sv

vvvv

m

k c

z

uc

t

u

z

uk

t

um

dxdydz t

umdxdydz

z

uk

t

V dxdydz

z

uk

dxdydz

t

umdxdydz

t

uum

dxdydz t

mdxdydz t

mdxdydz

t t

dxdydz

t

V

;

)]([

)(

2

2

2

2

2

2

2

2

''

cv

is the coefficient

of consolidation

dxdydz V V V

dxdydz V

vv

,

dxdydz V v dxdydz V


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2

2

z

u

ct

u ev

e

wvv m

k c

The initial value of excess pore water pressure (initial condition):

The boundary condition of excess pore water pressure:

To obtain an analytical solution using the method of “separation

of variables”:


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for ui=constant

)12(

2

m M


7/132

for ui=constantTwo sides are

free drainage

One side is impermeable


8/132

i

e z

u

uU 1

i

d

ed

i

d

e

H

i

H

ei

H

v

H

ev

H

v

H

ve

f

t

u

dz u

du

dz u

dz u

dz uu

dz m

dz m

dz

dz

s

sU

2

0

21

2

0

0

0

0

'0

'1

0

'0

'

0

1

0 12

1

)(

)(

)(

)exp(2

1 2

0

2 v

m

m

T M M

U

Good approximate solution for the average degree of consolidation U :

for ui=constant:

d

i

d

e

H

i

H

ei

f

t

dz u

dz u

dz u

dz uu

s

sU 2

0

2

0

0

0

1

)(

If ui is not constant:

f t sU s )12(2

m M


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848.0%90

196.0%50

v

v

T U

T U


10/132

Two sides are free drainage

The bottom side is impermeable

d is the maximum

water drainage

distance


11/132


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t B:t A=4:1

d

d

t 50

固結

次固結

蠕變

(creep)

lab EOP o

creept

t C

h

h

,

log

lab EOP t ,

tcoefficiensecondaryis

1log1

1

log 00 e

C

t

e

et

C e


13/132

X

1.15X

t 90

固結

次固結

蠕變(creep)

)(1

9

10;

0

0

90

0

100

0

0

0

p s

f

s p

f

s p

f

s

r r r

aa

aar

aa

aar

aa

aa

r

EOP t


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Why a clayey soil creeps?Creep is continuing deformation under constant loadCreep occurs under effective stress action

Creep is due to –viscous adsorbed water (double layers) on clay

plates

–viscous re-arrangement and sliding of clay plates –viscous deformation of clay plates (small)

–viscous deformation of clay skeleton (structure/frame

like)

Adsorbed water is NOT free water which

cannot flow freely under gravity.


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Under

particle

contact

force

(effective

stress)

Creep movement !


16/132


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In-situ value of cv

and c h

Use large samples c h

is horizontal coefficient of consolidationvh

cc )4~2(


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Solution:


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20/132


21/132

(c) Correction for construction period

• Previous solution/chart is suddenly applied load• Common real load is ramp loading (construction loading

likes this) – linear increase and then constant

• How to find a solution to this or make a correction?


22/132

f cc

t

ct corr t

c

f t

t t corr t

c

st

t U t

t swhere

t t st s

t t For

st

U t

swhere

P

P t st s

t t For

)2

()2

(

)

2

()(

:

)2

()2

(

)2

()(

0

,

'

'

,

't P

t

t 2

t

corr t s ,

' P

corr t s ,

t

U from previous

solution/chart


23/132

Solution:

u

u

'

'

0' u

2

'

/43.29)81.93(

mkN u

kPa72.1443.2921'

kPa72.14

'

mm H m s vcf 110872.1494.0'

Suddenly applied loading case:

Layer is open, thus, d =4m, for t =5 years

73.0437.04

54.122

U d

t cT vv


24/132

Dam

Sand

This sand layer is an “artesian” layer with higher

water pressure above ground surface

This clay layer is a

“aquitard” layer


25/132

Suddenly applied loading:

%2626.0)2

42.1

()2(

0621.04

4.1)

2(

6.2811026.0

)2

42.1()

2

42.1(

31.2071.06.28

7.14

7.14

)2

42.1

()42.1(

242.10

2242.1

22

242.1

,

U

t

U

d

ct T

mmmm

sU s

mm

s s

yearst t For

t v

v

f t

t corr t

c

42.1t 71.02

42.1

2

t

6.28

)2

(

t s

31.20

)(,

t s corr t

kPa

kPat P

yearst t For

t

c

7.14'

44.107.14)(

242.10

242.1'


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Solution:

C c=0.32

(a) One layer 3'3

,' /2.108.920;/2.98.919 mkN mkN clayw sand sat sand


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3m

2'1

2'0 /8.179)203(8.119;/8.119)32.10()62.9()217( mkN mkN

C c=0.32

)/log();/log(

'0

'1

10'0

'110

eeC C ee cc

00 11 eee

ee o

v

mm H

e

C H

e

ee H s cov f 1826000

8.119

8.179log

855.01

32.0log

11'

0

'1

00

855.0)100

8.119(32.088.0)

100log(32.088.0 0

'

og ee


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f cc

t c

t corr t c

f t t

t corr t c

st

t U t

t swheret

t st st t For

st

U t

swhere P

P t st st t For

)2()2()2()(:

)2

()2

()2

()(:0

,

'

'

,

Good

equations

to use !

335.0875.06

5.226.1)

2(

;5.22

13

2 22

U

d

t t c

T yearst

t

cv

vc

mm st t U t t st s f cc

t corr t 61182335.0)2

()2

()(,

(b) Two layers

Method 1: Settlement is proportional to the thickness H For Layer H 1of 4.5m:

For Layer H 2 of 1.5m:

mm s f 5.13618265.4

1

mm s f 5.4518265.1

2

mm sU sU d

t t c

T f c

cv

v 6.1125.136825.0825.0622.025.2

5.226.1)

2(

1111221

mm sU sU d

t t c

T f c

cv

v 1.445.4597.097.040.15.1

5.226.1)

2(

2222222

mm s s s ccc 7.1561.445.11221


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mm sU sU d

t t c

T f c

cv

v 6.1188.143825.0825.0622.025.2

5.226.1)

2(

1111221

mm sU sU d

t t c

T f c

cv

v 8.380.4097.097.040.15.1

5.226.1)

2(

2222222

)7.156(4.1578.386.11821 mmmm s s s ccc

2'

1

2'

0

/15.172)203(15.112;/15.112)25.22.10()62.9()217( mkN mkN

mm H e

C H

e

ee H s cov f 8.1434500

15.112

15.172log

864.01

32.0log

111'

0

'

1

0

1

0

11

864.0)100

15.112log(32.088.0)

100log(32.088.0 0

'

ee

Method 2: Settlements for H 1 of 4.5m and H 2 of 1.5m:

2'

1

2'

0 /75.202)203(75.142;/75.142)25.52.10()62.9()217( mkN mkN

mm H e

C H eee H s cov f 0.401500

75.14275.202log

831.0132.0log

112'

0

'

1

0

2

0

12

831.0)100

75.142log(32.088.0)

100log(32.088.0 0

'

ee


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vU


31/132

(d) Vertical drains (垂直排水井)

• 1-D vertical consolidation of soft soils takes a long time:

“primary consolidation” takes 3 years to 100 years and

creep settlements takes even longer time.• This cases problems of continuing excessive

settlement/deformation of ground or super-structures

• How to solve this problem?

• Use vertical drains plus pre-loading (預加荷載) –

commonly used in Hong Kong

• Why?


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Pre-loading –

removed afterward


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Prefabricated band (vertical) drains (排水板)


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Drainage channels

Geo-textile


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Purpose of Vertical Drains

and Preloading

(a) Vertical Drains 排水板

• Speed up escaping/dissipation of excess porewater/ pressure due to preloading

(b) Preloading預壓• Making the soil over-consolidated and reducing both

post-construction “primary” and creep consolidation

More accurately speaking, reducing instant compressionand creep


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Bedrock or soil

Marine Deposits

Water Table

Pre-loading fill

Sand fill


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Lo g(Pressure) (or Stress)

Settlement/strain - large

Set t lem ent/st ra in - sm all

Large creep set tlem ent/s tra in

Sm all creep set tlem ent

Soil pressure/stress

befo re co nstruc tio n

Soil pressure/stress

after construction

Pre- loading to here

Illustration of settlement/strain reduction using pre-loading technique:

reduction in post-construction “primary” consolidation (instant)

settlement/strain and reduction in “secondary” consolidation (creep)

settlement/strain


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2

2

2

2

2

2

)1

( z

uc

r

u

r r

uc

t

u ev

eeh

e

)1)(1(1)1)(1()1( r vr v U U U U U U

22 4);(;);(

R

t cT T f U

d

t cT T f U hr r r

vvvv


39/132

)1)(1(1)1)(1()1( r vr v U U U U U U

d

hr r r

vvvv

r

Rn

R

t cT T f U

d

t cT T f U ;

4);(;);(

22

2R=D is different from other books


40/132

vU


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Solution:mm H m s v f 162106525.0

' mm s s s remain f montht 137251626

%8585.0162

1375.0

f

year t

s

sU

82.0)(;7.24

)2.0(4

5.09.7

4

?)(2.0;2.02/4.0:

222

r r r h

r

d d

T f U T ntrial nn R

t cT

nnnr Rmmr drainSand

82.00235.01

85.011

1

1117.00235.0

10

5.07.422

v

r vv

vU

U U U

d

t cT

v

r r v

U

U U U U U

1

11)1)(1()1(

?)(20;202/40: nnnrRmmrdrainSand


42/132

82.0)(;7.24

)2.0(4

5.07.4

4

?)(2.0;2.02/4.0:

222

r r r

h

r

d d

T f U T ntrial nn R

t cT

nnnr Rmmr drainSand

n Tr

Ur

5 0.988 0.98

7.5 0.439 0.94

10 0.247 0.74

20 0.0618

0.26

0

0.1

0.20.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 2 4 6 8 10 12 14 16 18 20 22

n

U r

)(

2.3564.0

8.1

564.0

8.192.0

;9

patten square

m R

S

mnr R

n

d

How to make correction for construction period?


43/132

f cc

t

ct corr t

c

f t

t t corr t

c

st

t U t

t swhere

t t st s

t t For

st

U t

swhere

P

P t st s

t t For

)2

()2

(

)2

()(

:

)

2

()

2

(

)2

()(

0

,

'

'

,

't P

t

t 2

t

corr t s ,

' P

corr t s ,

t

U from previoussolution/chart


44/132

有蠕變性質的土的固結沉降計算的簡化計算方法

A simplified method for calculation of

consolidation settlement of clayey soils with

creep


45/132

Settlement due to creep (not use of “secondary” consolidation)

C e

C e 01

lab EOP o

creept

t C

h

h

,

log

t coefficienionconsolidat " "secondarycalled soisC

e

C

t

e

et C

e

e

00 1log1

1

log

)log(time


46/132

For Hong KongMarine Clays:

C e=0.3% to 1%

C e

Wh l il ?


47/132

Why a clayey soil creeps?Creep is continuing deformation under constant loadCreep occurs under effective stress action (discussed

before)

Creep is due to –viscous adsorbed water (double layers) on clay

plates

–viscous re-arrangement and sliding of clay plates

–viscous deformation of clay plates (small)

–viscous deformation of clay skeleton (structure/frame

like)

Adsorbed water is NOT free water which

cannot flow freely under gravity.


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• Creep always exists under the action of

effective stresses (loading), independent ofthe excess pore water (or pore pressure).

• Therefore, creep has nothing to do with

the“primary”consolidation.

• And creep exists during and after “primary”

consolidation.• Creep rate depends on stress/strain state:

–Creep rate is large in a normally

consolidated state.

–Creep rate is small in an over-consolidated

state.

Bjerrum’s time line model apparent “pre consolidation


49/132

Bjerrum s time line model, apparent pre-consolidation

pressure”, ageing and “delayed compression” (Bjerrum 1967)

A

BC

Path A 路径1：A>B

Path B 路径2：A>C>B

The creep rate at B is dependent on the

stress-strain (or void ratio) state, not on theloading history (or path)B点的蠕变率只与应力-应变状态点有关，与怎样到达这点的历史或路径无关

Vertical effective stress (t/m2)

V o i d r a t i o

（ e

）


50/132

A

B

C

Path 1：A>B

Path 2: A>C>B

The creep rate at Point B is dependent on thestress-strain (or void ratio) state, not on theloading path.

Vertical effective stress (t/m2)

V o i d r a t i o

（ e

）

Th t th d f 1D ttl t l l ti f


51/132

field EOP field EOP o

e

f v

field EOP f v

secondary primarytotalA

t t for H t

t

e

C S U

t t for S U S S S

,,

,

""""

)log(1

Hypothesis A：This method is not correct since the creep

settlement in the “primary” consolidation is not included. This

method underestimates settlement (低估了沉降).

There are two methods for 1D settlement calculation of

soil with creep: Hypothesis A and Hypothesis B

Equation of Hypothesis A method：

H th i B Thi th d i t i th ttl t


52/132

Hypothesis B：This method is correct since the creep settlement

in the “primary” consolidation is included. This method givescorrect settlement.

Equations of Hypothesis B method：

strainvertical pressure porewater excessu

water of weight unit ty permeabilik t z

uk

conditioncontiniuty From

z e

w

z e

w

;

;;

)1(

:

2

2

A constitutive model (stress-strain relation) is needed for t

z

Yin and Graham (1989 1994) developed a one dimensional


53/132

Yin and Graham (1989, 1994) developed a one-dimensional

elastic visco-plastic (1D EVP) model:

/

'')

'()(exp

'

zo

z ep

zo z

o z

z z

V

Vt V

totalB z z e

zo

e s z ep

zo z

o

e s z

e s z

z

s

e s z z e s z z z

S get u for and Solve

uuV

Vt

t uu

uuV t

pressure porewater staticknownu

uuuuuuknownis

,,,)2()1(

)2())

()(exp

)()(

1

;;;

'

/

'

''


54/132

YIN J-H. and GRAHAM J.(1989). Viscous elastic plastic

modelling of one-dimensional time dependent behaviour of clays.

Canadian Geotechnical Journal, 1989,26:,199 - 209.

YIN J H. and GRAHAM J. (1994). Equivalent times and elasticvisco-plastic modelling of time-dependent stress-strain

behaviour of clays. Canadian Geotechnical Journal, 1994,31: 42

- 52.

YIN J H. and GRAHAM J. (1996). Elastic visco-plastic modelling

of one-dimensional consolidation. Geotechnique, 1996, 46(3):

515 - 527.

Maxwell rheology model:


55/132

Maxwell rheology model:

modulus sYoung E

E E

Spring

z e

z z e

z

'

;

:''

t coefficienviscous

dashpot Viscous

z v

z

v

z z

'' ;

:

model plasticviscoelasticanmodel veconstituti A

E t

E

addition strainsconnectionSeries

z z z

z z v

z

e

z z

:

:)(

''

''

Yin and Graham’s (1989, 1994) 1-D Elastic


56/132

/

'

'

'

'

)()(exp

zo

z ep

zo z

o z

z z

V

Vt V

Yin and Graham s (1989, 1994) 1 D Elastic

Visco-Plastic (1-D EVP) model:

Linear elastic spring (non-linear for 1-D EVP model)

Linear visco dash-pot (non-linear for 1-D EVP model)

z '

z ' Maxwell’s Rheoloical Model:

''

z z z

E

Non-linear elastic strain rate; non-linear visco-plastic strain rate

Linear elastic strain rate; linear visco-plastic strain rate


57/132

Time (min)

ototalB H S /

)(kPaue

ie uu

o H

Hypothesis B (Simplified by Yin – 殷簡化計算方法):


58/132

ionconsolidat "primary" of end at settlement final S

ionconsolidat of degreeaverageU

t any for S S U

S S S

f

v

creep f v

creep primarytotalB

0)(

""

Hypothesis B (Simplified by Yin 殷簡化計算方法):

For details of this simplified method, please see:

殷建华（2011）. 从本构模型研究到试验和光纤监测技术研发。岩土工程学报，

第33卷第1期, 1~16。

Yin， Jian-Hua (2011). From constitutive modeling to development of laboratory testingand optical fiber sensor monitoring technologies. Chinese Journal of Geotechnical

Engineering. Vol.33 No.1, 1~16.

: Point to1 Point 4


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'

z Log

e

cC

eC ),('

zp zp tp

z 5

),( 2'

2 z z

),( 3'

3 z z

),( 4'

4 z z ),( 5'

5 z z

1 2

3

45

Normal

consolidation

Unload/ reload

Over-consolidation '2

tp

z

6 ),( 6'

6 z z

tp

z

tp

z 5

z

or

Figure 1. Relation and states of log(stress)-

void ratio (or strain) from 1D straining

H σ

σ

log e1

C

H σ

σ

log e1

C

s ' zp

'

z4

o

c

' z1

'

zp

o

e

f

H

e

C s

: Point to1 Point

z

z

o

e f '

1

'

2log

1

2

H

e

C s

: Point to point dation Preconsoli

zp

z

o

c f

zp zp

'

'

4

'

log

1

4),(

0log1

)(56

0log1

)(64

'

6

'

5e

'

4

'

6e

H e

C :sreloading Point to Point

H e

C s

:unloading Point to Point

z

z

o

f

z

z

o

f

C c and C e can be determined from (a) compression with time 24 hours (t 24) of

duration or (b) compression at the end of primary consolidation in lab (t EOP,lab)

which is about a few minutes.

Settlement due to creep (not use of “secondary” consolidation)


60/132

Settlement due to creep (not use of secondary consolidation)

C e

C e 01

lab EOP o

creept

t C

h

h

,

log

t coefficienionconsolidat " "secondarycalled soisC

e

C

t

e

et C

e

e

00 1log1

1

log

)lg(t

In the simplified Hypothesis B method, the “equivalent time” te


61/132

In the simplified Hypothesis B method, the equivalent time t e

is used for creep:

o

e

et

t t C ee

00 log

0

o

e

o

e

o

ee

o

eetp

z

e1V

parameter creepat

timeequivalent t

t

t t C t

t t

V

C

t

t t

e

C

e

ee

000

00

0

logloglog11

If C c and C e can be determined from (a) compression with time 24 hours (t 24) of

duration: t o=24 hours=1day.

If C c and C e can be determined from (b) compression at the end of primary

consolidation in lab (t EOP,lab) which is about a few minutes: t o= t EOP,lab

(a) Final point is at a normally consolidated state :

For example Point 4 (Figure 1)


62/132

For example Point 4 (Figure 1)

0)log(

1 0

""

e

o

eoe f

v

creep primarytotalB

t for H

t

t t

e

C sU

s s s

oe t t t

:usewhyisThis4.Point back toisaboveThe

1day)t(log1

log1

)1

11log(

1log

1log

1

:coupling pressure porewater noIf

'

'

4

'

1

'

0

'

'

4

'

1

'

""

H e

C H

e

C

H t

e

C H

e

C H

e

C

s s s

zp

z

o

c

z

zp

o

e

e

zp

z

o

c

z

zp

o

e

creep primarytotalB

oe t t t

'

z Log

e

cC

eC ),('

zp zp tp

z 5

),( 2'

2 z z

),( 3'

3 z z

),( 4'

4 z z ),( 5'

5 z z

12

3

45

Normal

consolidation

Unload/ reload

Over-consolidation'2

tp

z

6 ),( 6'

6 z z

tp

z

tp

z 5

z

or

(b) Final point is at an over-consolidated state :

For example Point 2 (Figure 1)'L'2


63/132

p ( g )

0)log(1 20

""

e

eo

eoe f

v

creep primarytotalB

t for H t t

t t

e

C sU

s s s

o

ee

zp

z c zp z

t

t t

V

C

V

C 0

'

'

loglog

:)1994,1989GrahamandYin(

"timeequivalent"thetoAccording

e

c

e zp z

eC

cC

zp

z

e zp z C

C

zp

z C

V

C

V

o

e

zp

z

e

c

e

zp z

o

e

t t t

C

C

C

V

t

t t

)(1010

log)(log:abovetheFrom

'

')()log()(

0

'

'

0

'

'

o

C

C

zp

z C

V

oe t t t e

c

e zp z

)(10

'

')(

z Log

e

cC

eC ),('

zp zp tp

z 5

),( 2'

2 z z

),( 3'

3 z z

),( 4'

4 z z ),( 5'

5 z z

1 2

3

45

Normal

consolidation

Unload/ reload


tp

z

6 ),( 6'

6 z z

tp

z

tp z 5

z

or

C

C

zC

V

ttt ec

e zp z

)(10'

2)( 2

'

z Log

12



64/132

o

C

zp

z oe t t t

ee

)(10'

22

o

C

C

zp

z C

V

ooee

e

t t t t t t t

t

e

c

e zp z

2)(10

:)1b(in

'

'

2)(

2

2

0)log(1 2

2

0

""

e

eo

ee f

v

creep primarytotalB

t for H t t

t t

e

C sU

s s s

)1dayt(log1

)log(1

log1


'

1

'

2

2

2

0

'

1

'

2

""

H e

C H

t t

t t

e

C H

e

C

s s s

z

z

o

e

eo

ee

z

z

o

e

creep primarytotalB

e

cC

eC ),('

zp zp tp

z 5

),( 2'

2 z z

),( 3'

3 z z

),( 4'

4 z z ),(

5

'

5 z z

3

45

Normal

consolidation

Unload/ reload

tp z

6 ),( 6'

6 z z

tp

z

tp

z 5

z

or

(c) Final point is at an over-consolidated state from unloading :

For example Point 4 to Point 6 (Figure 1)


65/132

0)log(1 60

""

e

eo

eoe f

v

creep primarytotalB

t for H t t

t t

e

C sU

s s s

0log1 '

4

'

6

H e

C s

z

z

o

e f

o

C

C

zp

z C

V

oe t t t e

c

e zp z

)(10

'

' 6)(

6

6

o

C

C

zp

z C

V

ooee t t t t t t t e

c

e zp z

2)(10'

'

6)(

6

6

)1c(0)log(

1 6

6

0

""

e

eo

ee f

v

creep primarytotalB

t for H

t t

t t

e

C sU

s s s

'

z Log

e

cC

eC ),('

zp zp tp

z 5

),( 2'

2 z z

),( 3'

3 z z

),( 4'

4 z z ),( 5'

5 z z

12

3

45

Normal

consolidation

Unload/ reload


tp

z

6 ),( 6'

6 z z

tp

z

tp

z 5

z

or


66/132

o

C

C

zp

z C

V

oe

e

eo

eecreep

t t t

t for H t t

t t

e

C s

e

c

e zp z

)(10

0)log(1

'

'

6)(

6

6

6

0

6

(d) Final point is at an over-consolidated state from reloading :

For example Point 6 to Point 5 (Figure 1)


67/132

0)log(1

50

""

e

eo

eoe f

v

creep primarytotalB

t for H t t

t t

e

C sU

s s s

0log1 '

6

'

5

H e

C s

z

z

o

e f

H

H t

t t

e

C

t

t t

e

C H

t t

t t

e

C s

tp

z

tp

z

o

eoe

o

eoe

eo

eoecreep

)(

)log(1

)log(1

)log(1

5

5

0050

)log(1

);log(1

5

0

5

0 o

eoetp

z

o

eoetp

z t

t t

e

C

t

t t

e

C

'

z Log

e

cC

eC ),('

zp zp tp

z 5

),( 2'

2 z z

),(3

'

3 z z

),( 4'

4 z z ),( 5'

5 z z

12

3

45

Normal

consolidation

Unload/ reload


tp

z

6),(

6

'

6 z z

tp

z

tp

z 5

z

or

C

C

z C

V

ttt ec

e zp z

)(10'

5)( 5


68/132

o

zp

oe t t t

)(10'5

o

C

C

zp

z C

V

ooee t t t t t t t e

c

e zp z

2)(10'

'

5)(

5

5

)1d(0)log(1 5

5

0

""

eeo

ee

f v

creep primarytotalB

t for H t t

t t

e

C sU

s s s

)1dayt(log1

)log(1

log1


'

6

'

5

5

5

0

'

6

'

5

""

H e

C H

t t

t t

e

C H

e

C

s s s

z

z

o

e

eo

ee

z

z

o

e

creep primarytotalB

Extension of the above method:

(a) Have vertical drains


69/132

(a) Have vertical drains

The U v in (1a) … (1d) is replaced by the total U :

toncondolidaiof degreeaverageradial

r

r v

r v

U

)U )(1U (11U

)U )(1U (1U)(1

(b) Multi-layers (n layers)Appling superposition principle

n

1ii,

n

1ii,""

n

1ii, creep primarytotalBtotalB s s s s

layers! betweencontinuityconsider toneed:n

1i

i,""

primary s

2

2 ;60.0085.0)1log(933.0;60.04 H

t C T U for U T U for U T vvvvvvvv


70/132

2

2

22

2

2

22

2

;;

ln44

3

)ln(

)2

exp(1

e

r r

w

s

w

e

s

h

r r

r t C T

r

r S

r

r n

S n

S n

K

K

n

S

S

n

S n

n

m

m

T U

f cc

t c

t corr t c

f t t

t corr t c

st

t U t

t swheret

t st st t For

st

U t

swhere P

P t st st t For

)2

()2

(),2

()(:

)2()2(,)2()(:0

,

'

'

,

sr 2


71/132

rw=rd=radius of well

re=R=radius of column

rs=radius of smear zone

Smear zone: remoulded soil zone

pushed by mandrel


72/132

Band drain

F 0


73/132

f cc

t

ct corr t

c

f t

t t corr t

c

st

t U t

t swhere

t t st s

t t For

st

U t

swhere

P

P t st s

t t For

)2

()2

(

)2

()(

:

)2

()2

(

)2

()(

0

,

'

'

,

't P

t

t 2

t

corr t s ,

'

P

corr t s ,

t

U from previoussolution/chart

How to get U, Ur , Uv?(a) Use existing solutions, tables, figures,


74/132

( ) g , , g ,

equations(b) Use new internet based tools for calculation

of the coefficients of consolidation and

porewater pressure in 1-D and 2-D cases

• Tools are developed by JH Yin et al. at PolyU

• http://www.cse.polyu.edu.hk/~civcal(Reclamation molulus)

• Interactive calculations!

• 1-D (oedometer condition) - ramp loading and varying

with depth• 2-D axi-symmetric - ramp loading & vertical drain

Internet based consolidation analysis


75/132


76/132

Example 1 :

H is 10m with double drainage in 1D straining. Other parameters are in the


77/132

g g p

following table. Calculate final settlements at different states at 5 years. The load issuddenly applied.

07.0e

C 5.0c

C 018.0e

C

21

1eo

oeV

)/(

488.0

2 yr m

C v

)(1 dayt o

%1.0

001.030kPa

z1

'

z1

%54.0

0054.040kPa

z2

'

z2

%15.1

0115.0

60kPa

zp

'

zp

%68.8

0868.0

120kPa

z4

'

z4

%40.8

0840.0

100kPa

z5

'

z5

35%.7

0735.0

50kPa

z6

'

z6

Table 1 Parameters and stress-strain state of soil

(1a) Settlement calculation for the case of loading and final stress-strain state is

normally consolidated (loading from Point 1 to Point 4)


78/132

858m.010)60

120log

2

5.0

30

60log

2

07.0(

log1

log1 '

'

4

'

1

'

H

e

C H

e

C s

zp

z

o

c

z

zp

o

e f

0.352614.3/0979.04/T4U

0976.05/488.05d/tCT,5m2/10d

vv

22

vv

0.406m0.2935)0.858(0.3526

)(

creep f vtotalB s sU s

0.2935m10)1

1-53651log(2

0.018

)t

tlog(

1

5yearst

o

eo

0

"

H t

e

C s e"creep

' z Log

e

cC

eC ),('

zp zp tp

z 5

),( 2'

2 z z

),( 3'3 z z

),( 4'

4 z z ),( 5'

5 z z

12

3

45

Normal

consolidation

Unload/ reload

Over-consolidation '2

tp

z

6 ),( 6'6 z z

tp

z

tp

z 5

z

or

(1b) Settlement calculation for the case of loading and final stress-strain state is at an

overconsolidated state (loading from Point 1 to Point 2)


79/132

0437m.01030

40log

2

07.0log

1 '1

'

2

H e

C s

z

z

o

e f

o

C

C

zp

z C

V

oe t t t e

c

e

zp z

)(10

'

'

2)(

2

2

= )(163541)60

40(10 018.0

5.0

018.0

2)0115.00054.0(

day

0.352614.3/0979.04/T4U

0976.05/488.05d/tCT,5m2/10d

vv

22

vv

0.047m0.090)0.0437(0.3526

)(


0.090m10)163541

163545365log(

2

0.018

)t

tlog(

1

5yearst

e2o

e2

0

"

H t

t

e

C s e"creep

'

z Log

e

cC

eC ),('

zp zp tp

z 5

),( 2'

2 z z

),( 3'

3 z z

),( 4'

4 z z ),( 5'

5 z z

12

3

45

Normalconsolidation

Unload/ reload


tp

z

6 ),( 6'

6 z z

tp

z

tp

z 5

z

or

(1c) Settlement calculation for the case of unloading and final stress-strain state is

at an overconsolidated state (unloading from Point 4 to Point 6)

'


80/132

133m.01012050log

207.0log

1 '4

6

H e

C s z

z

o

e f

o

C

C

zp

z C

V

oe t t t e

c

e

zp z

)(10

'

'

6)(

6

6

=

)(10226.11)60

05(10 9018.0

5.0

018.0

2)0115.07350.0(

day

0.352614.3/0979.04/T4U

0976.05/488.05d/tCT,5m2/10d

vv

22

vv

0.047m0.00])133.0[(0.3526

)(


0.00m10)10226.11

10226.15365log(

2

0.018

)t

tlog(

1

5yearst

9

9

e6o

e6

0"

H t

t

e

C s e

"creep

'

z Log

e

cC

eC ),('

zp zp tp

z 5

),( 2'

2 z z

),( 3'

3 z z

),( 4'

4 z z ),( 5'

5 z z

12

3

45

Normal

consolidation

Unload/ reload


tp

z

6 ),( 6'

6 z z

tp

z

tp

z 5

z

or

(1d) Settlement calculation for the case of reloading and final stress-strain state is at

an overconsolidated state (reloading from Point 6 to Point 5)

'


81/132

1054m.01050

100log207.0log

1 '6

5

H e

C s z

z

o

e f

o

C

C

zp

z C

V

oe t t t e

c

e

zp z

)(10

'

'

5)(

5

5

= )(18.761)60

010(10 018.0

5.0018.02)0115.0840.0(

day

0.352614.3/0979.04/T4U

0976.05/488.05d/tCT,5m2/10d

vv

22

vv

0.081m0.1252)1054.0(0.3526

)(


0.1252m10)18.761

76.185365log(

2

0.018

)

t

tlog(

1

5yearst

e5o

e5

0

"

H

t

t

e

C s e"creep

'

z Log

e

cC

eC ),('

zp zp tp

z 5

),( 2'

2 z z

),( 3'

3 z z

),( 4'

4 z z ),( 5'

5 z z

12

3

45

Normal

consolidation

Unload/ reload


tp

z

6 ),( 6'

6 z z

tp

z

tp

z 5

z

or

Example 2:

Calculate the curves of settlement and time for the case of loading and final stress-


82/132

.infiniteisfieldthein, field EOP t

Here we assume U v=0.987 is the end of primary consolidation in the field.T v=1.668, t EOP,field =3118day.

strain state is normally consolidated (loading from Point 1 to Point 4) usingHypothesis A and simplied Hypothesis B method.

Solution:

field EOP

field EOP o

e f v

field EOP f v

ondary primarytotalA

t t for H t

t

e

C S U

t t for S U

S S S

,

,

,

"sec"""

)log(1

)1a(0)log(1 0

""

e

o

eoe f v

creep primarytotalB

t for H t

t t

e

C sU

s s s

Curves of settlement and time from the two methods are shown in the following figure.

Cv= 0.488 m^2/year

sf= 0.858 m

t=(day) TV= Uv Uv*sf S_creep S_totalB S_"secondary" S_totalA

1 5.34795E-05 0.008254 0.007082 0 0.007082 0 0.007082

2 0.000106959 0.011673 0.010015 0.027093 0.010331 0 0.010015

4 0.000213918 0.016508 0.014164 0.054185 0.015058 0 0.014164


83/132

8 0.000427836 0.023346 0.02003 0.081278 0.021928 0 0.02003

16 0.000855671 0.033016 0.028327 0.108371 0.031905 0 0.028327

32 0.001711342 0.046691 0.040061 0.135463 0.046386 0 0.040061

64 0.003422685 0.066031 0.056655 0.162556 0.067388 0 0.056655

128 0.00684537 0.093382 0.080122 0.189649 0.097832 0 0.080122

256 0.01369074 0.132062 0.113309 0.216742 0.141933 0 0.113309

512 0.027381479 0.186764 0.160244 0.243834 0.205783 0 0.160244

1024 0.054762959 0.264124 0.226619 0.270927 0.298177 0 0.226619

1825 0.0976 0.352606 0.302536 0.293514 0.406031 0 0.3025362737.5 0.1464 0.431853 0.37053 0.309362 0.504128 0 0.37053

4106.25 0.2196 0.528909 0.453804 0.32521 0.625811 0 0.453804

6159.375 0.3294 0.640382 0.549448 0.341058 0.767855 0 0.549448

9239.063 0.4941 0.760495 0.652505 0.356907 0.923931 0 0.652505

13858.59 0.74115 0.869826 0.746311 0.372755 1.070543 0 0.746311

20787.89 1.111725 0.94784 0.813247 0.388603 1.18158 0 0.813247

31181.84 1.6675875 0.98677 0.846649 0.404451 1.245749 0 0.846649

46772.75 2.50138125 0.99831 0.85655 0.420299 1.276139 0.015848213 0.87239870159.13 3.752071875 0.999923 0.857934 0.436148 1.294048 0.031696427 0.88963

105238.7 5.628107813 0.999999 0.857999 0.451996 1.309995 0.04754464 0.905544

157858 8.442161719 1 0.858 0.467844 1.325844 0.063392853 0.921393

236787. 1 12. 66324258 1 0.858 0. 483692 1. 341692 0. 079241067 0. 937241

355180.6 18.99486387 1 0.858 0.49954 1.35754 0.09508928 0.953089

532770.9 28.4922958 1 0.858 0.515389 1.373389 0.110937493 0.968937

799156.3 42.7384437 1 0.858 0.531237 1.389237 0.126785707 0.984786

1198735 64.10766555 1 0.858 0.547085 1.405085 0.14263392 1.0006341798102 96. 16149833 1 0.858 0. 562933 1. 420933 0. 158482133 1. 016482

2697153 144.2422475 1 0.858 0.578781 1.436781 0.174330346 1.03233

4045729 216.3633712 1 0.858 0.59463 1.45263 0.19017856 1.048179

6068594 324.5450569 1 0.858 0. 610478 1.468478 0.206026773 1.064027

9102890 486.8175853 1 0.858 0. 626326 1.484326 0.221874986 1.079875

13654335 730.2263779 1 0.858 0.642174 1.500174 0.2377232 1.095723

20481503 1095.339567 1 0. 858 0. 658023 1. 516023 0.253571413 1. 111571

30722255 1643.00935 1 0.858 0.673871 1.531871 0.269419626 1.12742


84/132

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

1 10 100 1000 10000 100000 1000000 10000000

T o t a l

s e t t l e m e n t ( m )

Time (day)

Simplified Hypothesis B method (Yin)

Hypothesis A method

Example 3 :

At a site in Hong Kong with a clay layer of 5 m thick. The bottom of the layer is

permeable. The water table in the worst case is at the ground level. A uniform


85/132

Solution :

year mC dayt C

kPaemkN C C

voe

zpo sat ec

/1,1%;1

;50;2,/5.16;04.0;4.0

2

'3

vertical pressure by fill is assumed to be applied suddenly in two stages. The firststage with pressure 45 kPa is applied and in 6 months later, the second stage of 155

kPa is applied. Calculate and plot the time-settlement curve up 50 years using Yin’s

simplified method (using the initial effective stress at the middle clay layer) since the

first loading. Calculate the final total settlements at time of 0.5, 10, and 50 years

since the first loading. Soil parameters are:

m5

m5.2)( year t

)(kPa p

kPa45

kPakPa 45155

year 5.0 year 2

year 5.1

2

2 ;60.0085.0)1log(933.0;60.04 H

t C T U for U T U for U T vvvvvvvv

0927.05725.61

log4.050

log04.0

s

50725.6145725.16;725.16)81.95.16(5.2

45:1

1f1

''''

m H

kPakPa pkPa

kPa Loading Stage

f

zp zi z zi

'

z Log


86/132

)(

128.00.03770927.0984.0;0824.00.03770927.0632.0(

984.0101;632.0101101

6.1

5.2

101;32.0

5.2

21:10;2

?0.03770927.0(:,5.0:

)t

log(1

0927.0(:,5.00:

4160.00377.00927.0319.0(

00754.0)1

1-5.03651log(

21

0.01)

t

tlog(

1

0.0377m5)1

1-5.03651log(

21

0.01)

t

tlog(

1

)t-t(t5year .0t

6.0319.014.3/08.04;080.05.2

5.01

5.0

00634.0725.16

50log

21

04.0log

1;01854.0

5021725.1621

1111

933.0

6.1085.0

933.0

32.0085.0

933.0

085.0

1

2222

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00

1111

1111

0

e0

0

1

0

e0

0

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1221

'

'

0

1

1

filedataand curve see

mm s sU s

U

H

t C

H

t C T years yearsTime

U s sU siscurve settlement timethe year t timeOther

H t

e

C U s sU siscurve settlement timethe year t timeOther

m s sU s

t

e

C

H t

e

C s

ok U H

t C T

yearstime At

e

C

creep f v B

T

v

vvv

vcreep f v B

evcreep f v B

creep f v B

ecreep

ecreep

vv

v

zi

zpe zp f

v

）

）

）

）（）

e

cC

eC

),( ' zp zp 1

2

3

Normal

consolidation

Over-consolidation

1 f s

2 f s

1creep s

2creep s

Stage 1

Stage 2

),( 2'

2 zp zp ),( 1'1 z z

026080007540018540

725.6145725.1645)81.95.16(5.2

155:2

111

'

1

f

z kPanew

kPa Loading Stage

'

z Log

O lid ti


87/132

02692.0725.61

336.71log

21

04.002608.0log

1

71.336100.632577.22310725.6150

10

-

1-logloglog

-log1log1-log)11

-log1

log1

log1

log1

log1

;log1

:,

02608.000754.001854.0

'

1

'

2

0

12

1645.004.04.0

2100634.002608.0

0.04-0.4

0.04-

0.04-0.4

0.4

1

'

1

''

2

01

'

1

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2

1

'

1

0

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0

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2

00

1'

1

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0

'

'

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0

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0

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2

'

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111

01

z

zpe z zp

C C

e

C C

C

z C C

C

zp zp

ec

zp z z

ec

e zp

ec

c zp

zp z z e

zpc

zpec

zp z

z

zpe

zp

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zp

zpc zp

z

zpe z

zp

zpc zp zp

z

zpe z zp

zp zp

creep f z

e

C

kPa

C C

e

C C

C

C C

C

e

C

e

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e

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e

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e

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e

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e

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are New

ec zp z

ec

e

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）（

（

e

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),('

zp zp 1

2

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Normal

consolidation

Over-consolidation

1 f s

2 f s

1creep s

2creep s

Stage 1

Stage 2

),( 2'

2 zp zp ),( 1'1 z z

N

kPanew

kPa Loading Stage

creep f z

z

02608.000754.001854.0

725.6145725.1645)81.95.16(5.2

155:2

'

111

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'

z Log Over-consolidation


88/132

m H t

e

C U s sU s

U H

t C T

yrst yrsand

m H t

e

C U s sU s

ok T U H

t C T

yrst timeloading yearstimetotal At

m

H e

C

e

C H S

kPa

are New

evcreep f v B

T

vv

v

evcreep f v B

vvv

v

zp

z c

z

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f f

z

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zp zp

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326.05)

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21

001.0326.0981.0)

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log(

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326.0(

981.0101101;52.15.2

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5)336.71

725.216log

3

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consolidation

1 f s

2 f s

1creep s

2creep s

Stage 1

Stage 2

),( 2'

2 zp zp ),( 1'1 z z

'

z Log Over-consolidationHtCUU

iscurve settlement timethe year t timeOther

Summary

e )l (09270(

:,5.00:

:

）


89/132

e

cC

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),('

zp zp 1

2

3

Normal

consolidation

1 f s

2 f s

1creep s

2creep s

Stage 1

Stage 2

m s s s

yrsand yrst At

m s s s

yrst At

H t

e

C U U

s sU s sU s s s

iscurve settlement timethe year t timeOther

H t e

C U s sU s s

B B B

B B B

evv

creep f vcreep f v B B B

evcreep f v B B

454.0326.0128.0

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log(

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90/132

0.04


91/132


92/132

'

i H

0.269 187.5x0.269

Empirical equations to fit a measured settlement-time curve in the

field and find the final settlement Sf (or )S


93/132

S

Hyperbolic equation method:

Find parameter :

Find final settlement:

2

22

1

1

1

)(

)(

t

t s s s s

t

t

s s s s

d d

d d

Empirical equations to fit a measured settlement-time curve in the

field and find the final settlement Sf (or )S


94/132

S

Hyperbolic equation method:

Find parameter :

Find final settlement:

)(;1

)(

1

)(:)()(

)(

1

)()()(

d d

d d d t

d d d t d t d

d d t

s s A B

s s s s

Band s s

A xt vs y s s

t Plot

Bx A yt s s s s s s

t

s s

t

s s

t

t

t s s s s

0 025

0.03

0.035

Data


95/132

y = 0.0149x + 0.0066

0

0.005

0.01

0.015

0.02

0.025

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Time (year)

t / ( s t - s d )

Data

Linear (Data)

d t s s

t

);(;1

;)(

1

)(

)(

1

)(

d d

d d

d d d t

s s A B

s s s s

Band s s

A

Bx A yt s s s s s s

t

0066.0)(

d s s A

0149.0)(

1

d s s B

0d s


96/132

)1)((

)1)((

)1)((

2

1

2

1

t

d d

t d d

t

d d t

e s s s s

e s s s s

e s s s s


97/132

time (years) Set (cm)

Set

(prediction 1) t/(st-sd)

Set (prediction

2)

0 0 0 0

0.226 22.45 22.26519974 0.010067 22.67391697

0.302 26.7 26.7 0.011311 27.207697440 425 30 35 32 22375862 0 014003 32 8629422

t

t s s s s d d t

)(


98/132

0.425 30.35 32.22375862 0.014003 32.8629422

0.488 33.34 34.48516325 0.014637 35.18080627

0.533 35.81 35.92092809 0.014884 36.65321111

0.61 39.08 38.09188759 0.015609 38.88074447

0.665 40.54 39.4571216 0.016404 40.28227883

0.754 44.09 41.39960169 0.017101 42.27737095

0.814 45.72 42.55338369 0.017804 43.462939030.89 46.95 43.86525718 0.018956 44.8114395

0.92 47.02 44.34270318 0.019566 45.3023439

0.998 48.02 45.49074971 0.020783 46.4830323

1.06 48.19 46.31805672 0.021996 47.33410735

1.168 48.63 47.60669616 0.024018 48.66017864

1.373 49.72 49.63014421 0.027615 50.7434113

1.402 49.88 49.88 0.028107 51.00073482

1.44 50.1 50.19572156 0.028743 51.325919592 53.70756979 54.94505495

2.5 55.71214989 57.01254276

3 57.13379008 58.47953216

3.5 58.19449265 59.57446809

4 59.01623055 60.42296073

4.5 59.67158239 61.09979633

6 61.02694237 62.5

6.5 61.34850754 62.83228613

7 61.62684428 63.11992786

8 62.08456815 63.59300477

9 62.4453037 63.96588486

10 62.7369245 64.26735219

t t

s s s s

s s

s s

s s s s

t

s s

s s

t

t

t

t

d

d

d d

d

1

1

12

1

12

1

1

12

1

1

1

1

1

)(

1

1

0d s

Dt= 1.1

Dt/t1= 3.642384 t

t s s s s d d t

)(


99/132

t1/Dt= 0.274545

(s1-sd)/(s2-s1) 1.151855

(s2-s1)/(s1-sd) 0.868165

alfa= 0.438742s_infinite= 65.48946

A= 0.0066

B= 0.0149alfa= 0.442953

s_infinite= 67.11409 t

t

s s

s s

s s

s s

s s s s

t

s s

s s

t

t

t

t

d

d

d d

d

1

1

12

1

12

1

1

12

1

1

1

1

1

)(

1

1

0d s

0 2 4 6 8 10 12

Tim e (year)


100/132

0

10

20

30

40

50

60

70

S e t t l e m e n t ( c m )

Data

Prediction 1

Prediction 2

0.035


101/132

y = 0.0149x + 0.0066

0

0.005

0.01

0.015

0.02

0.025

0.03

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Time (year)

t / ( s t - s d )

Data

Linear (Data)

(2-D problem)


102/132


103/132

See paper by:

Zhu, G.F. and Yin, J.-H. (2001). Consolidation of soil with vertical and horizontal

drainage under ramp load. Geotechnique, Vol.51, No.2, pp.361-367.


104/132

r w=r d=radius of wellr e=R=radius of column

r s=radius of smear zone

Smear zone: remoulded

soil zone pushed by

mandrel

sr 2


105/132

Band drain

No covered here


106/132


107/132


108/132

See Figures later on this approach


109/132

ln3

)ln(

)2

exp(1

2222

SS n K S nn

m

m

T U

h

r r


110/132

)!4

:(

;;

ln44

)ln(

22

2222

before R

t C T fromdifferent isthisnote

r

t C T

r

r S

r

r n

S n K nS S n

m

r r

e

r r

w

s

w

e

s


111/132

See proof later


112/132


113/132

o p3 p

2 p


114/132

o

1 p

2

00

0

33

0

22

0

11

00

0

33

0

22

0

110

0

33

0

22

0

11

332211332211321

333222111

332211

);(

)()(

;;

;;

Hp M S

p

pU

p

pU

p

pU U

US S p

pU

p

pU

p

pU Hp M

p

pU

p

pU

p

pU

Hp M U Hp M U Hp M U S U S U S U S S S S

S U S S U S S U S

Hp M S Hp M S Hp M S

v

v

vvvt t t t

t t t

vvv


115/132


116/132


117/132


118/132


119/132

Vacuum preloading

m


120/132

S S f S S f

m H 29.4

vm

vm

suv


121/132


122/132


123/132

24

02.002067.0

scm K K md r

zone smear awiththisdo Re

h s s s /105.0;5.02/

:

6

without


124/132


125/132


126/132


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132/132

Fully coupled consolidation modelling programs in CSE:

ABAQUS (3-D finite element program)

Plaxis (2-D finite element program)

gdii lecture 1 [compatibility mode].pdf

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