Paper Geotechical Modeling Short Couse by Wood

7/25/2019 Paper Geotechical Modeling Short Couse by Wood

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Soilmodelling

DavidMuirWood

UniversityofDundee,Scotlan

SouthEastAsia

OctoberNovember2010


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1. Introductiontomodelling:soilbehaviour(SBCSSM

2. Elasticmodelling

(SBCSSM

2,

GM

3)

3. Themostwidelyusedsoilmodel:MohrCoulomb(

4. ApplicationofMohrCoulombmodel(exercise)(GM

5. Camclay(SBCSSM5,GM3)

6. Camclay

graphical

calculations

(exercise)

(SBCSSM

7. Camclay:complianceformulation(GM3)

8. Stiffnessformulation:MohrCoulomb,Camclay(G

9. Selectionofsoilparameters(exercise)(GM3)

10. MohrCoulombimproved:strength,criticalstates,

11. Camclayimproved:nonlinearity,structure

12. Conclusion


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CUP (1990) Spon (2004) CUP


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SBCSSM: Soil behaviour and critical state soil mCambridge University Press 1990: ISBN 0-521-33

GM: Geotechnical modelling. Spon Press 2004: I23730-5

1D: Soil mechanics: a one-dimensional introductiCambridge University Press 2009: ISBN 978-0-52


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1.Introductiontomode

soilbeha

(SBCSSM1,GM


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Soilmodelling:SouthEastAsia:OctoberNovem

1. Introductiontomodelling:soilbDavidMuirWood

[email protected]


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Models:

Scientific understanding proceeds by way of constru

analysing models of the segments or aspects of reality

The purpose of these models is not to give a mirror im

not to include all its elements in their exact sizes and

rather to single out and make available for intensive i

those elements which are decisive. We abstract fromwe blot out the unimportant to get an unobstructed vi

important, we magnify in order to improve the range

our observation. A model is, and must be, unrealistic

which the word is most commonly used. Neverthele

sense, paradoxically, if it is a good model it providesunderstanding reality.

(Baran and Sweezy, 1968)


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Everything engineers (and scientistdo is concerned with modelling!

You have been engaged in modellifor years perhaps unconsciously!

examples of models in civil andgeotechnical engineering


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behaviour ofin uniaxial te

idealisation of behaviour of mild steel

forming basis for plastic design of steel

structures

structural engineering


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geolo


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classification model - particle sizes

particles modelled as equivalent spheres

sieving sedimentation


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theoretical model

useful for confirming results of other mode

if boundary conditions are somewhat simil


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empirical model

Bjerrum vane correction factor

effects of rate, anisotropy,


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model ofree fiel

behind w

model of dock with adjacent docks included

Mair and Muir Wood (2001)

dock structures under seismic loading

numeri


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numerical modelling

for example, finite element, finite difference

equilibrium compatibility of

stresses strain

stress:strain relationship

constitutive model

lectures primarily concerned with introduction to vand possibilities of constitutive modelling


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observed and idealised shearing behaviour of soilfor settlement andbearing capacity calculations

constitutive models trying to reproduce more of theactual nonlinearity ofpre-failure soil response

shear

stress

shearstrain


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stress and strain variables (

concentrate on axisymmetric conditions of triaxial tes

volume changes are important in soils

and affect mechanical properties

choose volumetric strain increment:

rap 2+=

axial strainincrement

radial strainincrement


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( ) 3/'2''p ra +=

in developing constitutive models concept of workw

need to link strain increment and stress variables

work conjugate pairs

choose volumetric stress variable: effective mean str

volumetric work work done in changing size is:

pp 'pW =


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A/F''q ra ==

choice of second pair of variables to

describe change of shape somewhatarbitrary

for convenience choose deviator stressfrom triaxial test

distortional stress variable: deviator stress


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( ) 3/2 raq =

work conjugate pair for change in shape

distortional strain increment:

distortional work work done in changing shape i

qq qW =

q = a for constant volume deformation p = 0


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qprraa Wq'p'2'W =+=+=

confirm that total work done in strain increment is:

=

=

=

=

r

a

r

a

q

p

r

a

13/1

13/1

q

'p

3/11

3/21

'

'

23/2

1

'

'

11

3/23/1

q

'p

convenient transformation matrices


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define stress ratio: = q/p'

mobilised friction

m

m

ra

ram

'sin3

'sin6

6

3

''

'''sin

=

+

=+

=

for conditions of triaxial compression


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distortion: change in sh

distortion: change in sh

compression: change in


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Pore pressure parameter

total and effecti

undrained defor

total stress is external perturbation arbitrary

effective stress is soil response constrained by c

properties p'indicates desire of soil to change

when sheared (or not) dilatancy


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Pore pressure parameter

pore pressure increment: u = pp'

slope of effective stress path: a = p'/q

then: u = p + aq

logical definition of pore pressure parameter linking

change with total stress change (compare Skempton

a is not a soil property function of history etc.

subtraction of arbitrary total stress change p helps in i

pore pressure change


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Soil behaviour

particle continuum duality

laboratory element testing

stiffness


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Particle-continuum dualit

for analysis we need to treat soil as a continuum

but its properties emerge from its particulate na


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particle-continuumduality

Leighton Buzz(picture width

Leda clay (pictu


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photoelastic discs

force chains

fabric

Drescher and de Josselin de Jong (1972)

particle-cduality


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particle-contin

Discrete Elemeof particle inte

homogenisation procedure required to produce contin

stress strain


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Laboratory element testin

we may calibrate models using triaxial test data but so

the ground and in numerical modelling will certainly baxial symmetry

beware of unintended responses in uncalibrated region

general stress state has 6 deg


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Laboratory element testing

axial symmetry

triaxial apparatus

centreline of circular loaded area

widely available source of data of stress:strain re


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plane strain occurs mor

no strain in long directi

plane strain withfixedprincipal axes

fairly easy to apply in laboratory


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true triaxial apparatus

3 degrees of freedom

no rotation of principal axes


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development of shear stress on ends of sample?

questions of uniformity of stress state/strain state

Simple shear apparatus

plane strain with rotation

no horizontal strain


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direct shear boxinhomogeneous deformati

strength information

pedagogic illustration of d

use for subsequent modell


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torsional hollow cy

4 degrees of freedo

butradial non-unif

average stress quan

with equal internal/external pressures

b = (2-3)/(1-3) = sin2

constraint on stress space exploration


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Stiffness

secant stiffness Gs = / tells us where we are now bindication of how we got there

tangent stiffness Gt=/ tells us about current inc

response

beware: use of term stiffness does not imply elasti


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Stiffness

progressive yielding of steel cantilever as analytica

of distinction between tangent and secant stiffness

elastic-plastic

plastic region penet

edges towards centr


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Stiffness: steel cantilever

tangent stiffness falls as yielding spreads

tangent stiffness zero when full plastic hinge has fosecant stiffness remains positive and non-zero

tip displacement


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load line limited by collapse load of cantilever

deflection line unlimited (ignore geometry changes)many to one mapping we can always map from deto load but not always from load to deflection

load line

deflection line

Stiffness: steel cantilever

stiffness good com


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stiffness


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Stiffness

variation of stiffness is due to plasticity

hence need for constitutive models

Quiou sand (LoPresti et al., 1997)

secant stiff


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Summary

modelling is all around us

care in selection of strain increment and str

pore pressure parameter as a variable

particle-continuum duality laboratory element testing not just axial s

but more general stress states not easily acc

tangent and secant stiffness

stiffness/compliance formulation?


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2.Elasticmod

(SBCSSM2,

G


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2. ElasticmodellingDavidMuirWood

[email protected]


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Elasticity

Hooke's law

elastic behaviour in triaxial compression:

drained/undrainedmeasurement of elastic properties with differen

elastic anisotropy

elastic nonlinearity hyperelasticity

worked example


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IfIhaveseenfurtheritisonlybystandingontheshouldersofgiants

HookewasNewtonspredecessorattheRoyalSocietyofLondon.

Hookewasaverysmallman.


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Tofill

the

vacancy

of

the

ensuing

page,

Ihave

here

added

adecimate

theInventionsIintendtopublish,thoughpossiblynotinthesameor

getopportunityandleasure;mostofwhich,Ihope,willbeasuseful

areyetunknownandnew.

2. ThetrueMathematicalandMechanichalformofallmannerofAr

withthetruebutmentnecessarytoeachofthem. AProblemwhich

Writerhatheveryetattempted,muchlessperformed.

abcccddeeeeefggiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuux.

3. ThetrueTheoryofElasticityorSpringiness,andaparticularExplic

severalSubjects

in

which

it

is

to

be

found:

And

the

way

of

computing

Bodiesmovedbythem.

ceiiinosssttuu

Hookes lawA description of helioscopes and some other instruments. Lon


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Tofill

the

vacancy

of

the

ensuing

page,

Ihave

here

added

adecimate




2. ThetrueMathematicalandMechanichalformofallmannerofAr

withthetruebutmentnecessarytoeachofthem. AProblemwhich

Writerhatheveryetattempted,muchlessperformed.

abcccddeeeeefggiiiiiiiillmmmmnnnnnooprrsssttttttuuuuuuuux.

Utpendetcontinuumflexile,sicstabitcontiguumrigiduminversum.

(Ashangstheflexiblechain,so,inverted,standstherigidarch.)

Hookes law


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Utpendetconti

stabitcontiguum

(Ashangsthefle

inverted,stands

StPeters:Po


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Tofill

the

vacancy

of

the

ensuing

page,

Ihave

here

added

adecimate




3. ThetrueTheoryofElasticityorSpringiness,andaparticularExplic

severalSubjectsinwhichitistobefound:Andthewayofcomputing

Bodiesmovedbythem.

ceiiinosssttuu

Uttensiosicvis. (Astheextensionsotheforce.)

Hookes law

force

extens

stress

strai


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Hookes law

principal stresses and strain increments:

( )( )

+=

=

z

y

x

z

y

x

z

y

x

z

y

x

1

1

1

211

E

'

'

'

'

'

'

1

1

1

E

1


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Hookes law

uniaxial tension

Youngs modulus E = (P/A)/(l

/l

)Poissons ratio = (d/d)/(l/l)

direct observation of elastic constants


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Hookes law

axial symmetry:

( )( )

+=

=

r

a

r

a

r

a

r

a

1

21

211

E

'

'

'

'

1

21

E

1

compliance/stiffness matrices not symmetric be

and strain variables not work conjugate


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Hookes law

separate change of size and change of shape

work conjugate stress and strain variables

=

=

q

p

G30

0K

q

'p

q

'p

G3/10

0K/1

( )=

213

EK

( )+=

12

EG

bulk modulus: shear modulus:

change of size and change of shape uncoupled

(off-diagonal zeros)


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Hookes law

only 2 independent elastic properties for isotrop

2

3

K3G

KG9E =

+

=


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Stiffness in conventional drained triaxial

total stress path in conventional compression

(constant cell pressure):

( )

3p

qthen0if

q3/2p

r

rara

=

=

=+=

=

=

q

p

q

p

G300K

q'p

q'p

G3/100K/1


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Stiffness in drained triaxial compre

=

=

=

=

21

K

G

Eq

G3q

a

p

q

p

a

q


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Stiffness in undrained triaxial comp

no coupling between change of size and change of shacompression and distortion for isotropic elastic soil

hence, in undrained test pure distortion p'=0 and

pressure parameter a = 0 for isotropic elastic soil

conventional triaxial compression q/p = 3 and u =

pore pressure from total mean stress alone

effective stress path independent of total stress path


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qhenceq

aq ==in undrained test

external view of undrained stiffness in terms of total s

=

q

p

G3/10

0K/1

u

u

q

p

distortional stress q = ar= 'a'r and hence


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=

qp

G3/100K/1

u

u

q

p

p = 0 (undrained)

p arbitrary (external change of total stress)

hence Ku =

( ) 21

213

EK u

u

uu ==

=


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=

qp

G3/100K/1

u

u

q

p

u = 1/2

( ) ( )G3E

12

EG

12

EG u

u

uu =+

==+

=

drained and undrained elastic propertiescannot be chosen independently


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Elastic stiffness from different de

oedometer

one dimensional co

( )( )( )

K211

1E

'E

z

zoed +=+

=

=

and Ko = /(1-)


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pressuremeter

direct measurement of shear modulus G


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plate loading test

( )=

1

G4

R

composite stiffnesses do not reveal individual elas


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elasticity

simple models assume isotropic linear elastic

two parameters: E, or G, K

convenient but not necessary

explore possibility of anisotropy

and nonlinearity


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anisotropic elasticity

many soils deposited over areas of large latera

implied symmetry: all horizontal directions eqB, C, D, E

cross anisotropy transverse isotropy

(anisotropy of real soils, after real stress/strainwill be more complex)


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different mo

c

(

+

=

vh

vh

vvvhvvh

vvhhhhh

vvhhhhh

xy

zx

yz

zz

yy

xx

1200000

G/10000

0G/1000

00E/1E/E/00E/E/1E/

00E/E/E/1

5 elastic properties

triaxial can only find 3: Ev, vh, Eh/(1-hh)

hh? Gvh? could use bender elements


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( )( )

+

+

=

22

22

xy

zx

yz

zz

yy

xx

00000

/*120000

0/*12000001/*/*

00/*/1/*

00/*/*/1

*E1

Graham & Houlsby 3 parameter cross anisotropy

axisymmetric triaxial conditions

=

J*G*K3

12

q

p

( )( )

( )( )

( )( )*21*13**1

*EJ

*21*16

*4*22*E*G

*21*19

2*4*1*E*K

2

2

2

++

=

++

=

+++

=

coupling of co

and disto


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( )( )

+

+

=

22

22

xy

zx

yz

zz

yy

xx

00000

/*120000

0/*12000001/*/*

00/*/1/*

00/*/*/1

*E1

Graham & Houlsby 3 parameter crossanisotropy

axisymmetric triaxial conditions

forcing particular link between cross

anisotropic elastic parameters

=

3J

*K

q

'p

coupling of

and dis


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effective stress path in undrainedtriaxial compression

2 = 1.52 = Eh/Ev

Winnipeg clay (Graham & Houlsby)


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experimental techniques

laboratory geophysics

bender elementsshear wave velocities: time, distance

Vvh = Vhv ? (elastic, symmetry)


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alternatively:

small undrained

unload-reload cyclesduring drained test

slope of effective

stress path indicateselastic anisotropy

vertical Eh = Ev, = 1

anisotropy evolves

with stress ratio

Eh 0 at critical state

Hostun sand (Gajo)


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Gault clay

evidence of stianisotropy fromelements

Gault clay and

evolution of elaanisotropy with


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cross anisotropy axis of symmetry of anisotropy with vertical axis of sample

more general anisotropy misalignment of axes distressed boundary measurements may be unrep

it is tempting to assume that things that you chooseobserve do not exist


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empirical description of elastic evolution of anisotropy

hypoelasticity

not necessarily conservative eproduction/dissipation in closed

for example, both G and K func

hyperelasticity

thermodynamically consistent

define strain energy density function U

or complementary energy V

hence stiffnessesi

i

i

i

V;

U

=

=


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complementary energy function (Boyce, 1980)

constant volume paths

constant distortional strain paths

road sub-base material: n = 0.2, = 0.3 (K1/G1 =

( )

( )( )

+=

+

+=

+

q

'p

G3

1

G3

1

G3

1

G6

n2n1

K

n

'p

'p

q

G6

1

K1n

1'pV

11

1

2

111n

q

p

2

11

1n

( )n

2

1

1o

G6

Kn11

'p

'p

=

1n

o

o 'p

'p

q

q

=


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Boyce complementary energy function

contours of constant volumetric strain andconstant distortional strain


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Cam clay has K dependent on p'

constant is thermodynamically unacceptable

Houlsby strain energy function

constant distortional strain paths: = constant

constant volume paths

asymptotic to

[ ]

=

+=

q

p

q

2

q23/

r

1

'pq

'p

e'pU p

( )ii2 'p'p'p6q =

2/3=


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Houlsby strain energy function

contours of constant volumetric strain andconstant distortional strain


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art in choosing energy functions to give desirable when differentiated twice (stiffness or compliance

varying slope of undrained path may come from norevolving anisotropy (or both)

subtle tests to distinguish

even more difficult to discover hyperelastic energ

to describe evolving elastic anisotropynonlinear elasticity or plasticity? (irrecoverable


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Summary

generalised Hookes law one-to-one link

stress and strain

elastic matrices symmetric with work-con

variables two independent elastic properties for isot

material

deduction of elastic properties from stand

evolving anisotropy - nonlinearity


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Example: Plane strain isotropic elastic elemen

under general stress path

Takey direction as plane strain direction:

y = 0 = (y - x - z)/E

hence y = (x + z) z z

xx

z z

y=0y


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y = (x + z)

General plane strain stress path: x = z

then y

= (1 + ) z z

x = z

z

y=0y


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General plane strain stress path: x = z

strain inz direction z = (z - x - y)

z

= (1 +)(1 - [1 + ]) z/E

x = (1 +)([1 -] -) z/Ez z

x = zx

z zy=0


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z = (1 +)(1 - [1 + ]) z/E

x = (1 +)([1 -] -) z/E

One-dimensional compression (oedometer):

= / [1 -]

stiffness: Eoed = z/ z = E(1 -)/ (1 +) (1

For example: = 0.25, Ko = / [1 -] = 1/3;Eoed /E = 1.2


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Exercise: Plane strain isotropic elastic elemen


Assume = 0.25;

1. Plot relationship betweenz stiffness z/

2. Plot relationship betweenx stiffness z/


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3.Themostwidely

soilmodel:

Mohr

Cou(G


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3. Themostwidelyusedsoilmodel:MoDavidMuirWood

[email protected]


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Mohr-Coulomb familiar as a strength/failure mod

limiting stress ratio characterised by frictional str

convert to stress ratio M

( 2'

'3

'p

qM

si1

si1

'

'

'

'

a

a

3

1

r

a

+

==

+=

=

Elastic-perfectly plastic Mohr-Coulom

combine with isotropic elastic stiffness model

aim: to produce elastic-plastic stiffness matrix

strain increment stress increment


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Mohr-Coulomb failure; elastic s

failure locus divides streselastic and inaccessible re

plasticity only occurs for on failure locus: q/p'=M

stress increments implyinpossible

assume strain increment divided into elastic and pla

= e + p

elastic strain accompanies any change in stress

=

=

q

'p

G3/10

0K/1or

G30

0K

q

'peq

ep

eq

ep

stiffness compliance


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What are the deformations at yield

assume that yielding mobilises a plastic mechanism

mechanism defines ratio (not magnitude) of plastic s

=

=

1*Mor*M

pq

pp

pq

pp

where is an arbitrary scalar multiplier

work conjugatestrain increments

plotted on local axes

parallel to

corresponding stress

axes


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Stiffness matrix: consistency con

=

e

q

e

p

G

K

q

p

30

0'

for plastic strainchange lies alon

Mp' + q = 0

combine and rearrange

( )

( )

=

1

*M

G30

0K1M

G30

0K

1Mq

p

=

=

q

p

e

q

e

p

G

K

G

K

q

p

30

0

30

0'

=

=

30

0

30

0'

G

K

G

K

q

p

q

p

p

qq

p

pp


99/356

Stiffness matrix which stiffness? ela

+

=

2

2

G9MGK3

G*M3K*MM

G3*KMM

1

G30

0K

q

'p

stiffness formulation: calculate stress incrementsfrom total strain increments

elastic predictorandplastic corrector

implementation: if elastic predictionproduces a stres

in inaccessible region beyond failure locus,plastic crequired to bring stress state back to failure locus

if plastic:

+=

q

p

*MMM

*M1

G3*KMM

GK3

q

'p

we do not initially know whether a strain increment

elastic orplastic


100/356

Plastic dissipation? Associated f

+=

q

p

*MMM

*M1

G3*KMM

GK3

q

'p

if plastic:

confirm that q = Mp'

asymmetric unless M = M* (associated flow)

plastic work Wp = p'pp + qq

p =(M M*)p'q

plastic work Wp = 0 for M = M* (associated flow

physically unreasonable

(but basis of bearing capacity calculations etc)


101/356

Mohr-Coulomb model: typical re

typical response in constant p' test

dilation/compression depends on sign of M*


102/356

standard elasplastic Mohr

non-associate

M* M

simplicity

sharp stiffness changes

tangent stiffness eitherelastic or zero

continuing volume change


103/356

Mohr-Coulomb model: fitting to

standard model availabgeotechnical finite elem

subjectivity in choice o

stiffness G, K;

strength M;

dilatancy M*

exercise


104/356

Mohr-Coulomb model: system/elemen

system response for example, footing shows pr

yielding even though model for individual soil elemelastic-perfectly plastic

exercise response of two element box model


105/356

Summary

elastic-perfectly plastic Mohr-Coulomb m

widely available constitutive model for so

associated plastic flow is not physically re

choice of soil parameters requires engineejudgement (exercise)


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z

x

x > z

z > x

passive:

active: x = Ka z

x = z

Mohr-Coulomb failure


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z

x

x = Kp

x = Ka z

x = z

initial stress state: Ka < Ko < Kp (z congeneral stress path: x = z

stress path: x = z


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z

x

x = Kp

x = Ka z

x = z

stress path: x = z

active failure when Ko zo + z = Ka (zo

z /zo = (Ko Ka)/(Ka -) (for > Ka

zo

Ko zo

zo + z


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z

x

x = Kp

x = Ka z

x = z

stress path: x = z

passive failure when Ko zo + z = Kp (z

z /zo = (Kp Ko)/( - Kp) (for > K

zo

Ko zo


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z

x

x = Kp

x = Ka z

x = z

stress path: x = z

strains to failure: z = (1 +)(1 - [1 + ])

x = (1 +)([1 -] -)

zo

Ko zo


111/356

Exercise: Plane strain isotropic elastic Mohr-C

element under general stress path

Assume = 0.25, = 30

1. Plot relationship between normalised strain

passive) failure Ez/

zoand initial state K

oand

2. Plot relationship between normalised strain t

passive) failure Ex/zo and initial state Ko and

What happens if Kp > > Ka ?


112/356


113/356

4.Applicati

MohrCoulomb

m

(exercise)(G


114/356

Soil modelling: South East Asia: October-November 2010

4. Application of Mohr-Coulomb model (exercise)David Muir Wood

[email protected]


115/356

Mohr-Coulomb model: system/element response

system response for example, footing shows progressive

yielding even though model for individual soil elements is

elastic-perfectly plastic

exercise response of two element box model


116/356

numerical modelling

for example, finite element, finite difference

equilibrium compatibility of deformations

stresses strains

stress:strain relationship

constitutive model

lectures primarily concerned with introduction to various aspects

and possibilities of constitutive modelling


117/356

A P

v b

h ha.

h

h

vv

two element box modelas analogue of footing

single

equilibrium

kinematic compatibility

stress:strain response


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horizontal stress h

vertical

stress v

h=

v

= 1 = 0

passive failure h> v

initial stress h= K

active failure

v> h

stress path characterised by=h/v

single e


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= 0

vertical

stress v

= /(1) = 1

horizontal stress h

= /(1),

h

= 1

h =

v

= 0, h

strain inc0

a. b.

stress paths elastic strains: effe

single e


120/356

horizontal stress h

verticalstress v

Mohr-Coulomb

failure

Mohr-Coulomb

failure

P

A

A

h

v

hA= -hP

=

a.

kinematic compatibilityequal & opposite horizontal strains in elemen

two-elem


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kinematic compatibility

equal & opposite horizontal strains in elemen

element P:v = 0; equivalent to compressioloaded horizontallyP = 0

for element P,h/h=E/[(1 +)(1 )]

we can deduce the stress path direction A fA

Av/h= E/[(1 +)(1 )] =AE/(1 +)[A(1 ) ]

A=/[2(1 )](which is half the elastic Ko


122/356

horizontal stress h

verticalstress v

Mohr-Coulomb

failure

Mohr-Coulomb

failure

P

A

A

h

v

hA= -hP

=

a.

equal & opposite horizontal strains in elemenwhich element reaches failure first?


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P

A

v v

vh

elemen

yield

elem

yi

element P reaches failure firsth remains constant

A for element A changes from A=/2(1A= 0


124/356

P

Av v

vh

elemen

yield

element A

yields

element A reaches failure first

A=h/v now changes and remains conA=Ka

stressv can continue to increase until elemreaches failure


125/356

P

Av v

vh

elemen

yield

element A

yields

element A reaches failure first

compatibility of horizontal strains: compressielement P matches extension of element A

hA=ehA+

phA=

ehP

vA =evA+

pvA =

evA

phA


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Use= 0.25,= 30,E/b= 200

CalculateKa,Kp and slope of stress pathA=/[2(1 )]

Plot initial normalised stress statesv/b = 1h/b =Ko forKo= (1 +Ka)/2,1and(1 +K

Plot stress paths for elements A and P for eastress state. For each value of Ko discover welement reaches failure first and calculate thcorresponding vertical strain in element A (nwithE/b)

Plot the stress paths followed as the second heads for failure

Calculate the normalised vertical strain in elewhich the second element reaches failure

Plot the normalised stress:strain response ofooting for each initial state


127/356


4. Exercise: Two element box modelDavid Muir Wood (University of Bristol/Dundee)

Introduction

A two element box model (Fig 1a) provides a simple analogue of a foundation. We can use this simplemodel to demonstrate the three constraints that have to be satisfied in any boundary value problem:equilibrium, kinematic compatibility, and constitutive response. Equilibrium is straightforward: forcesmust balance. Kinematic compatibility means that gaps should not open up in our model. Constitutiveresponse indicates the relationship between stress changes and strain changes for the material: we willassume an isotropic elastic response up to Mohr-Coulomb failure. But first we will analyse the responseof a single element - like a laboratory test - and present the response graphically.

All stresses are effective stresses.

1. Single element: elastic response: plane strain

Plane strain testing in the laboratory is not particularly common but this is what we need for our singleelement test (Fig 1b). The nearest equivalent to the standard triaxial test will be a plane strain test inwhich the lateral stress is kept constant but we will look more generally at tests in which the verticalstress and lateral stress are increased in constant proportion: h=v (Fig 2).

Hookes Law for the plane strain direction of zero strain tells us that:

y = 0 = 1

E

[y (v+ h)] (1)

so that:y =(v+ h) =(1 + )v (2)

Hookes Law for the vertical direction tells us that:

v = 1

E[v (y+ h)] =

1

E(1 + )(1 )v (3)

and for the horizontal direction:

h=

1

E[h (y+ v)] =

1

E(1 + )( )v (4)

Then for any value of not only can we plot the stress paths in the h:v effective stress plane, but wecan also plot the horizontal and vertical strain increments for the several paths (Fig 3b). Evidently = 0corresponds to a test with constant lateral stress; = 1 is a sort of isotropic plane strain compressiontest; = /(1 ) corresponds to one-dimensional compression h = 0.

Mohr-Coulomb failure implies a limiting ratio of vertical and horizontal effective stresses. We know thevalues of this limiting ratio as the active and passive earth pressure coefficients - depending on whetherthe vertical or the horizontal stress is the greater.

v > h : v

h=Kp=

1 + sin

1sin (5)


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A P

v b

h

v

v

hh

a.

b.h

h

vv

Figure 1: (a) Two element box model as analogue of footing; (b) single plane strain element

horizontal stress h

vertical

stress v

h= v

= 1 = 0

passive failure h> v

initial stress h= Kov

active failure

v> h

Figure 2: Single element: stress paths

v < h : v

h=Ka=

1sin

1 + sin (6)

These limiting ratios can be plotted on the effective stress plane (Fig 2).

The elastic response is (assumed to be) independent of the initial stress state. We can characterise theinitial stress state by the coefficient of earth pressure at restKo (whereKa Ko Kp and hence deducethe dependency of the vertical (or horizontal) strain at failure on the value ofKo and on the directionof the stress path .

2. Two element box model: elastic response, Mohr-Coulomb failure

The two element model (Fig 1a) provides a simple analogue of the bearing capacity problem. There is aninitial surcharge stress b on the surface of both elements. The initial horizontal stress in each element(which must from equilibrium be identical) can be chosen as h=Kob where Ka Ko Kp.

The right-hand passive element P (Fig 1a) is just a single element being loaded horizontally with P = 0,so we can plot (we have already plotted) the stress path for this element (but with vertical and horizontalstresses interchanged) and indicate the strains in the horizontal direction of passive loading (which wasthe vertical direction in the previous section) (Fig 4a, b).

The left-hand active element A (Fig 1a) is being loaded vertically by the footing. Equilibrium requiresthat the horizontal stress in element A should always be the same as the horizontal stress in element P.We can indicate this constraint in the stress plane (Fig 4a). Consideration of the nature of the problem


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= 0

vertical

stress v

= /(1) = 1

horizontal stress h

= /(1),

h

= 1,

h = v

v

= 0, v

= /(1),

v

= 0, h

strain increment0

a. b.

Figure 3: Single element: (a) stress paths and (b) stress:strain response

shows that the horizontal stress must increase as the footing load is increased so that A>0.

Kinematic compatibility requires that no gap should open up between the two elements. The horizontalcompressive strain in element P resulting from the increase of horizontal stress must be equal to thehorizontaltensilestrain in element A resulting from the increase of vertical stress. These two constraintstogether allow us to deduce the relevant value ofAfor the loading of element A and hence the value of thevertical strain in element A and the vertical stress: vertical strain response of our analogue foundation.The logic is illustrated in Fig 4b. The value of A for the active element is unknown but it can becalculated from (3) and (4), recalling that element P is being loaded horizontally. Equating appropriatestrains and noting that h/v =A we find that:

A = 2(1 )

(7)

which is exactly half the value for one-dimensional compression: for h/v = 0,h/v =/(1 ).

We are free to choose our initial value ofKo: its value will control which element reaches failure first.

Suppose the right-hand passive element reaches failure first so that h=Kpb(Fig 5a). The horizontalstress cannot increase any further so that from this moment onwards the left-hand active element isbeing loaded with A = 0 - and we can again use the stress path and strain information from theprevious section to deduce the overall footing response. The horizontal strain will continue to increase inboth elements. In the passive element, which has reached failure, there will be no further elastic strainsand the vertical and horizontal strains will be linked by the angle of dilation . Let us, for simplicity,assume that the failure occurs at constant volume. Then vertical and horizontal strains will be equaland opposite: the horizontal squeezing of the element will translate directly into a vertical stretching.The three sections of the vertical stress:vertical strain response for element A are shown schematicallyin Fig 6a, b.

Or suppose that the left-hand active element reaches failure first (Fig 5b). The ratio of stresses on thiselement must then remain constant but the magnitudeof the vertical stress will continue to increaseuntil the passive element also reaches failure. The stress path followed by the passive element does notchange direction so the continuing loading of the footing can be followed by applying the equilibrium

constraint of equal horizontal stresses in the two elements (Fig 5b). The active element will experienceboth elastic strains (along a path with A =Ka) and plastic strains sufficient to give a total horizontalstrain matching the strain in the passive element. Compatibility of horizontal strains requires that


130/356

horizontal stress h

vertical

stress v

Mohr-Coulomb

failure

Mohr-Coulomb

failure

P

P

A

A

h

v

hA= -hP hP

= = 0

a.

b.

Figure 4: Box model: (a) stress paths and (b) relationship between vertical stress and horizontal strain betweenthe two elements

P

P

A

A

= 0 after

yield at P

= Kaafter

yield at A

vv

hh

a. b.

Figure 5: Box model: stress paths with (a) passive element P, and (b) active element A reaching failure first


131/356

P

P

A

Av v

v v

v

vh

h

a.

c. d.

b.

element P

yields

element P

yields

element A

yields

element A

yields

Figure 6: Box model: (a)(c) stress paths and (b)(d) stress:strain response for (a) (b) element P reaching yieldfirst; (c) (d) element A reaching failure first

elastic compression of element P should be equal and opposite to combined elastic and plastic extensionof element A:

hA=ehA+

phA =

ehP (8)

so that phA

=ehP e

hA. Then the constant volume condition at failure implies that:

vA = evA+

pvA =

evA

phA=

evA +

ehP+

ehA (9)

We can discover the dependence of the strain to yield (when the first element reaches the Mohr-Coulombfailure condition) and the strain to eventual perfectly plastic failure (when the second element has alsoreached the Mohr-Coulomb failure condition) on the initial stress state Ko. The three sections of thevertical stress:vertical strain response for element A are shown schematically in Fig 6c, d.

3. Graphical and calculation exercise

1. Choose a normalised Youngs modulusE/b = 200, Poissons ratio = 0.25 and angle of friction= 30.


132/356

2. CalculateKa, Kp and slope of stress path A = /[2(1 )].

3. Plot initial normalised stress states v/b = 1, h/b = Ko for values ofKo = (1 +Ka)/2, 1 and(1 + Kp)/2.

4. Plot the stress paths for elements A and P for these three initial stress states. For each value ofKo discover which element reaches failure first and calculate the corresponding vertical strain inelement A (normalised with E/b).

5. Plot the stress paths followed as the second element heads for failure.

6. Calculate the normalised vertical strain in element A at which the second element reaches failure.

7. Plot the normalised stress:strain response of the footing for each initial state. (Shown schemati-cally in Fig 6.)

8. ChallengeWhat is the effect of anisotropy of the elastic response on these several plotted relation-ships?

9. ChallengeWhat is the effect of a non-zero angle of dilation >0 on the strains that occur before

and during full failure of the footing?

References

Muir Wood, D (2004) Geotechnical modelling, Spon Press (section 7.5.1)

Nordal, S (1983) Elasto-plastic behaviour of soils analysed by the finite element method. Doctor ofEngineering thesis, Norges Tekniske Hogskole, Trondheim (Chapter 3)


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5.Cam

(SBCSSM5,

G


134/356


5. Cam clayDavid Muir Wood

[email protected]


135/356

extension l

load

P

a.

b.

P

l l1 l2

Py2

Py1

loading and unloading of annealed copper w

linear (elastic?) response when load less thamaximum load

nonlinear response when load exceeds prevmaximum load (yield load)

cycle of loading and unloading that exceeds maximum load leaves permanent (plastic) exwire - but also enlarged elastic (stiff) region


136/356

extension l

load

P

a.

b.

P

l l1 l2

Py2

Py1

penalty for increased elastic region (Py2 Py

permanent increase in length of wire (2hardening law

elastic(recoverable) deformation occurs wheload changes

change in load causing yield produces both(recoverable) andplastic(irrecoverable) defo


137/356

l1 l2

Py2

Py1

unloading-reloading lines

plastic deformation = separation ofunloading-reloading lines

increase in yield load

increase in permanent extension

elastic properties do not change when yieldi(assumption)

concept ofunloading-reloading linesfor load

current yield load

plastic deformation is equal to separation ofunloading-reloading paths at any load


138/356


139/356

q

p'

p'

v

effective stress plane

compression plane

unloading-reloading line

Cam clay

volumetric chanimportant: deve

model in paralleeffective stress (p, q)and compplane(p, v)

bulk modulusK

shear modulusisotropic nonlineelasticity

v=v lnp: u

reloading line pression planestresses withinyield locus


140/356

q

p' p'o

assume thatshapeof yield locus is not affecloading history

current yield locus characterised by sizepo aintersection withp axis

second question: what is the penalty for incr(hardening law)


141/356

q

p' p'oA p'oB

p'

v

iso-ncl

url A

A

url B

B

penalty for increpo? (hardening

volumetric hard

model: change linked only with in volume

pp=

vp

o

po


142/356

q

p' p'oA p'oB

p'

v

iso-ncl

url A

A

url B

B

isotropic norma

compression linplastic volumetr

pp=

vp

o

po

=

elastic volumetr

ep= vp

p

v= vp=

v(ep+pp) =

isotropic normpression line v=N lnp


143/356

q

p' p'oBp'oA

= M

ylA

ylB

geometry of yield locus fixed: soil propertyM

yield locus expands to accommodate the chastress: sizepo

po not affected by stress increment causing


144/356

q

p'

v v

p' lnp'p'=1

vv

NN

slope

slope

= 0

= 0

= 0

= M

= M

= M

iso ncl

url

url

iso ncl

yieldalwathrou

geomsimilinterspointpathsq/p

cons


145/356

q

p'

v v

p' lnp'p'=1

vv

NN

slope

slope

= 0

= 0

= 0

= M

= M

= M

iso ncl

url

url

iso ncl

anisoconslinesin serithm

compplane

unloareloalines

slopelnp,


146/356

Cam clay state boundary surface

in p, q, v space

each yield locus is associated with

an unloading-reloading line

state boundary surface encloses all

permissible states


147/356

p'o p'

q

Mp

p

qp

p

third question: what ismechanismof plasticdeformation? (relative amounts of plastic dis

and plastic volumetric strain:

p

q/

p

p)assumenormalityof plastic strain incrementlocus at current stress

= 0,pq/p

p = 0; =M,p

q/p

p =


148/356

Cam clay

volumetric hardening model: penalty for incre

of elastic region is permanent change in volupacking

adopt a graphical approach to the explanatioway in which the model operates inp, qeffec

plane and inp

, v compression planebehaviour bounded in these two planes: strevolume cannot change indefinitely (failure, limdensity of packing)

compute stress:strain response: distortional unbounded and can increase indefinitely


149/356

q

p'

p'

v iso-ncl

A

A

B

B

graphical coof the Cam cstress:strain

increment ABdrained comtest q/p

stress incremdefines interwith new yieand p

o

project dounloading-reline in complane


150/356

p'

v iso-ncl

A

B-vp

separation of urls produces irrecoverable volchange vp

plastic volumetric strain pp = vp/v


151/356

q

p'

A

B D

1

normality to yield locus givesD = pp/p

qa

plastic distortional strain


152/356

q

q

3G

A

B

qe qp

q

elastic distortional strain increment is eq

=

elastic and plastic components summed to gdistortional strain

then repeat for next increment


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q

p'

= M

as stress ratio =q/p increases, so ratio of

strain incrementspp/p

q reduces

yield locus can only expand if plastic volumeoccurs

as M, soil tends to state of perfect plast

critical state: continuing shearing with no furchange in stresses or volume


154/356

Cam clay drained compression test (1)

B-C-D-E-F: slope of yield locus becomes flatter

ratio distortional/volumetric strain becomes larger

F: no plastic volumetric strain (normality)

q: end of test!


155/356


how did we reach B? normal compression to A?

behaviour beyond B not affected by previous history


156/356


how did we reach B? overconsolidation to A?

behaviour beyond B not affected by previous history


157/356

Cam clay drained compression test (4)heavily overconsolidated KP: yielding at Q with p

p < 0

yield locus shrinks: softening stress:strain response

elastic: q > 0

elastic: p' > 0

q

Q-R-S-T: slope of yield locus becomes flatter:

ratio distortional/volumetric strain becomes larger:

T: no plastic volumetric strain (normality)

q: end of test!


158/356

Cam clay ambiguity of response

softening regime: yielding with > M

same stress change associated either withplastic (QR) or elastic

(QR') strain

bifurcation of response

numerical ambiguity


159/356

Taylor, 1948

plastic softening confined to thin localised region

elastic unloading

Physical consequence of uncertainty of response


160/356

Cam clay hardening and softening response


161/356

q

p'

p'

v

iso-ncl

P

P

Q

Qurl A

yl A

url B

yl B

stress increme

inside yield locpurely elastic

ep=p/vp


162/356

q

p'

p'

v

iso-ncl

P

P

R

R

url A

yl A

url B

yl B

stress increme

p = 0 ep=

expands yield lfrom yA to yBpo=p

oB p

oA

plastic volumepp= ( )po


163/356

q

p'

p'

v

iso-ncl

P

P

Q

Q

R

R

S

S

url A

yl A

url B

yl B

stress increme

pPS=p

PQan

poPS=poPR

project down tounloading-reloalines in compreplane

v= 0: undrain


164/356

q

p'

p'

v

iso-ncl

P

P

Q

Q

R

R

S

S

url A

yl A

url B

yl B

stress incremev= 0: undrain

elastic + plasticvolumetric strai

ep+pp= 0

p

vp+ ( ) pvp


165/356

q

p'

p'

v

iso-ncl

P

P

Q

Q

R

R

S

S

url A

yl A

url B

yl B

stress incremev= 0: undrain

elastic + plasticvolumetric strai

plastic comprespo>0

elastic expansio


166/356


167/356

p'

v iso-ncl

A

B

-vp

url A

url B

separation of urs produces irrecoverable vochangevp

plastic volumetric strainpp= vp/v

elastic volumetric strainep= pp


168/356


169/356

q

q

3G

A

B

qe qp

q

elastic distortional strain increment iseq=

elastic and plastic components summed to gdistortional strain


170/356


171/356


172/356

v

Cam clay undrained compression

step through undrained test on

normally compressed soil

conventional compression q = 3p


173/356


step through undrained test on lightly

overconsolidated soil

conventional compression q = 3p


174/356


175/356

Weald clay undrained response division of pore pressure


176/356


177/356

6.Camclaygrap

calculations(exe

(SBCSS


178/356


6. Cam clay: triaxial testsGraphical construction and direct calculation

David Muir Wood

University of Dundee

Graphical construction: drained test

Figure 1 can be used to follow the operation of the Cam clay model using essentially entirely graphicalconstruction. The test is constructed in four linked plots. Figures 1a, b are the two finite plots: theeffective stress plane (a) and the compression plane (b). Each yield locus in the effective stress plane islinked with a specific unloading-reloading line in the compression plane, which intersects the isotropicnormal compression line (iso-ncl) at the corresponding value of isotropic preconsolidation pressure po,

which is the mean stress where the yield locus intersects the p

axis. With the known constraints ofa particular test (constant cell pressure, drained, for example) the path of the test can be definitivelytracked in these two plots.

Figure 2 shows a detail of one step of a conventional drained triaxial compression test. The effectivestress path is constrained to climb at gradient q/p = 3 in the effective stress plane. This path thusfixes the points A and B at which it crosses two successive yield loci. Direct vertical projection down tothe compression plane at the same values of mean effective stress then fixes the points A and B on thevolumetric path from the intersections with the corresponding unloading-reloading lines.

The stress-strain response produces relationships which are unbounded: there is no limit, in principle, to

the shear strain that can be imposed. First we calculate the plastic volumetric strain. The vertical sepa-ration, at constant mean stress, of the unloading-reloading lines through A and B gives the irrecoverablechange in specific volume, vp (Fig 3). The plastic volumetric strain is then:

pp= vp

v (1)

The plastic distortional strain is then calculated from the condition of normality to the yield locus appliedat the current stress point in the effective stress plane (Fig 4). The direction of the dotted arrow, drawnperpendicular to the yield locus at point A indicates the ratio of plastic distortional strain increment toplastic volumetric strain increment, 1/D:

1

D =

pqpp

(2)

We can calculate the elastic distortional strain from the change in distortional, deviator stress, q(Fig5):

eq =q

3G (3)

We can sum the plastic and elastic components to find the total distortional strain increment:


179/356

0

50

100

150

200

0

50

100

150

200

100

5

00

2

.1

2

.2

2

.3

2

.4

2

.5

q

100

50

0q

p'

p'

v

2.

1

2.

2

2.

3

2.

4

2.

5

v

q

q

Camc

lay:dr

ainedtest

graphicalconstruction

a.

b.

c.

d.

iso-nc

l

Figure 1: Cam clay: drained test


180/356

q

p'

p'

v iso-ncl

A

A

B

B

Figure 2: Drained test: stress increment

p'

v iso-ncl

A

B-vp

Figure 3: Plastic volumetric strain increment


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q

p'

A

B D

1

Figure 4: Normality to yield locus

q

q

3G

A

B

qe qp

q

Figure 5: Elastic and plastic distortional strain increments

q = eq+ pq (4)

and we can project across to the strain diagrams, Figs 1c, d, to plot the corresponding points on thedistortional stress:distortional strain plot (Fig 1c) and the volume:strain plot (Fig 1d).

(In fact, the inclusion of the elastic distortional strain is equivalent to imposing a skew or shear on thedistortional stress:distortional strain plot, since the elastic distortional strain is directly proportional toq. So, for the purposes of this exercise, we can simply work in terms of plastic strain and recall that forcompleteness we would need to add the elastic strains. Alternatively, we need to assume a value of shearmodulus: for example,G = 2MPa? All the other constitutive parameters for Cam clay are implicit inthe plots of Figs 1.)

Graphical construction: undrained test

Figure 6 can be used to follow the operation of the Cam clay model for an undrained test, again usingessentially entirely graphical construction. The test is constructed in four linked plots. Figures 6a, b arethe two finite plots: the effective stress plane (a) and the compression plane (b). Each yield locus in theeffective stress plane is linked with a specific unloading-reloading line in the compression plane. Withthe known constraints of a particular test (constant cell pressure, undrained, for example) the path ofthe test can be definitively tracked in these two plots.

Figure 7 shows a detail of one step of a conventional undrained triaxial compression test. The path isconstrained to remain at constant volume in the compression plane (Fig 7b). This path thus fixes the


182/356


183/356


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0 50 100 150 2002.1

2.2

2.3

2.4

2.5

0 50 100 150 200

100

50

0

q

p'

p'

v

iso-ncl

a.

b.

A

A

B

B

Figure 7: Undrained test: stress increment

50 100 150

2.2

2.3

2.4

p'

v iso-ncl

A

B

-vp

Figure 8: Plastic volumetric strain increment


185/356


186/356

ln p'

v iso-ncl

A

B-vp

Figure 11: Plastic change in specific volume: semilogarithmic compression plane

q

p= 3 (12)

and we can start the calculation process by choosing a value for the increment qand hence the newvalue ofqB.

4. From (12) we can calculate the corresponding increment p and hence the new value ofpB.

5. From (10) we can calculate the size of yield locus poB required to accommodate this new stress state.

6. From (11) we can calculate the new specific volumevB. Through steps 4-6 we have established theposition of the new point in the effective stress and compression planes (Figs 1a, b) as was demonstratedgraphically in Fig 2.

7. Now we have to calculate the strain increments. We can divide the change in specific volume into twoparts: one due to the change in po, the plasticpart; and one due to the change in p

, the elastic part.The plastic part, vp is the volume separation of the unloading-reloading lines at the start and end ofthe increment: the geometry of the compression plane shows that this is (Figs 3, 11):

vp = ( ) l n (poB/p

oA) (13)

The plastic volumetric strain increment is then:

pp= vp

v (14)

8. The plastic strain increment is normal to the yield locus at the current stress, so the plastic dilatancy,D, is given by:

1

D=

pqpp

= 2

M2 2 (15)

where stress ratio = q/p. Hence we can calculate the plastic distortional strain increment, pq.


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188/356


189/356


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7. Cam clay: compliance relationshipsDavid Muir Wood

[email protected]


191/356

p', q p'o pp q

p

p, q

p', q pe, q

e

elasticproperties

yield

locus

hardeningrule

flow

rule

o

opp

'p

'p

v

2

M 22

pq

pp

22

2

o M

M

'p

'p

Cam clay calculation process

q

'p

G3

10

0'vp

eq

ep


192/356

Cam clay: convert graphics into equations

q

p' p'o

v v

p' lnp'p'=1

NN

slope

slope

= 0

= 0

= 0

= M

= M

= M

iso ncl

url

url

iso ncl

pp

qp

p


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q

p' p'oA

= M

ylA

elliptical yield locus

q2

M2 p(po p

) = 0 or p

po= M

2

M2 +

soil parameterMcurrent size (history) po


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q

p' p'oA p'oB

p'

v

iso-ncl

url A

A

url B

B

volumetric hardening

pp=

v

p

p

=

v(M2 +2)

(M2


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decomposition of strain increment into elastic an

components=e +pelastic compliance

ep

eq = vp 0

0 13G

p

q plastic compliance

pp

p

q=

vp

(M2

+2

)M2 2 2

2

42

M22

normality (flow rule) symmetric plastic compliarelationship


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22

2i

2

i

22

o

op

M

M

'p

'p

M

2

'p

'p

0'vp

'p

'vp

'p

undrained stress path cannotbe the same as the yield locus

balance of elastic expansion/plastic compression: < M

balance of elastic compression/plastic expansion: > M

> M, p' > 0 < M, p' < 0


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0'vp'p

'vp'p

o

op

initial stress A inside current yield locus

elastic response: p'o = 0

hence: p' = 0: constant p'

until yield at B

(isotropic elastic model)


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reminder: pore pressure has external and internal components

u = p p' = p + aq (pore pressure parameter a)

a is not a soil constant depends on history

222 2M

2

a


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General elastic-perfectly plastic

define boundary of elastic region: yield surfacef

divide strain increment into elastic and plastic pa=e +p

inside yield surface response iselastic: = D

stress

elasticyield = failure

inaccessible


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define boundary of elastic region: yield surfacef

when the soil is yielding the stress change must

the yield surface (consistency): f() = 0and

f= f

T

= 0

stress

elasticyield = failure

inaccessible


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describe mechanism of plastic deformation via apotentialg()such that:

p =g

whereis a scalar multiplier which will define thmagnitudes of the plastic strain increments

= De = D( p) = D D

combine with consistency condition to deduce:

=f

TD

f

TD

g


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example: elastic perfectly plastic Mohr-Coulomb

M

Melastic

inaccessible

p' p'

q q

qp

pp

p

isotropic elastic D=

K 0

0 3G


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M

M

elastic

inaccessible

p' p'

q q

qp

pp

p

plastic potential g() =q Mp +k

wherekfits the plastic potential to the current str


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M

M

elastic

inaccessible

p' p'

q q

qp

pp

p

plastic potential g() =q Mp +k pp

pq

=

g/p

g/q

=

M

1

plastic mechanism pp

pq= M


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elastic

plastic

initial

stress

p'

qq = Mp'

consistency requires M p +q= 0: stress cha

lie along failure locus for plastic strains p

q

=

K 0

0 3G

epeq

=

K 0

0 3G

p

q

M

1


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final result p

q

=

K 0

0 3G

1

KM M

+ 3G M M

K2 3MGK

3M GK 9G2

elastic predictorandplastic correctorwhich is as

unlessM=M (associated flow)


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if plastic strains occurring

p

q

=

3GK

KM M + 3G

1 M

M M M

check thatq=M p

plastic workWp =ppp+qpq= (M M)p

pq

plastic workWp = 0forM=M - physically un- in practiceM < M


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q

p'

q

p

q

q

M* > 0

M* < 0

a. b.

c.compression

dilation

Mohr-Coulomb: typical response in tests withp

dilation/compression depends on sign ofM


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stress

hardening yield surface f(, ) = 0

plastic potential g() = 0

p

elastic

elastic-hardening plastic model

natural extension of perfectly plastic model

yield surface now function not only of stresses bhardening parameter(s): f(, ) = 0


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divide strain into elastic and plastic parts =

elastic model= De

plastic mechanism defined by plastic potentialg(

p =g

= De = D( p) = D D

combine consistency and flow rule

f= f

T

+f

p

T g

= 0


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example: elastic-hardening plastic Cam clay mod

p'o p'

q

Mpp

qp

p

yield function and plastic potential assumed iden

g() =f(, po) = q2

M2 p(po p

) = 0

single hardening parameterpo


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yield function and plastic potential assumed iden

g() =f(, po) = q2

M2 p(po p

) = 0

plastic strain increments given by: pp

pq

=

gp

gq

=

2p po

2qM2


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compute hardening function functionH

H= f

po

popp

g

p =

p

vpo

2p

and full elastic-plastic stiffness relationship (K=

p

q=

K 0

0 3G

K2 (2p po)

2 6GKq(2ppo)

M2

6GKq(2ppo)

M236G2q2

M4

K(2p po)2 + 12Gq2M

4 + vpp

o(2p

elastic predictor(is the resulting stress state withlocus?) andplastic corrector


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0 0.05 0.10

1

2increasing overconsolidation ratio

normally consolidated

q

p

q

increasing

overconsolidation ratio

normally consolidated

0 0.05 0.1

-0.05

0

0.05

a.

b.

M=1.2

Cam clay: drained compression withp = 0


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0 50 1000

50

100

p': kPa

increasing overconsolidation ratioa.

0 0.05 0.0

50

100

increasing

overconsolidationratio

b.

0 0.05 0.

-40

-20

0

20

40

increasing overconsolidation ratio

c.

q: kPa q: kPa

u: kPa

q

q

Cam clay: undrained compression


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Cam clay

stiffness formulation always works: strain inc

stress increment

elastic prediction - followed by plastic correctpredicted elastic stress state violates the yie

however, compliance formulation may be les

cumbersome: stress increment strain incrbut note that this breaks down if soil wants toand is unsure whether to unload plastically o


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example: add prefailure nonlinearity to Mohr-Coumodel: elastic-hardening plastic model

qp

p'

q plastic potentialsM

p

elastic + plastic

inaccesssibleq

p'

failure: q- pp'= 0

elastic

yield: q- yp'= 0

frictional yield function

f(, ) =f(p, q , y) =q yp = 0

single hardening parametery


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Dilatancy

Ottawa sand (Taylor, 1948)

shear box

direct observation of dilatancy

link mobilised friction Q/P anddilatancy y/x


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example: add prefailure nonlinearity to Mohr-Coumodel: elastic-hardening plastic model

qp

p'

q plastic potentialsM

p

elastic + plastic

inaccesssibleq

p'

failure: q- pp'= 0

elastic

yield: q- yp'= 0

nonassociated flow - normality to yield function gexcessive dilation (andWp = 0)

g() =q M p lnprp

plastic mechanism pp

pq

=

gp

gq

=

M

1


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hyperbolic distortional hardening rule

00

0.1 0.2

1

mobilised friction

shear strain

yp

= pqa+pq

or incrementally

y =(p y)

2

appq


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define a hardening function H

H= f

p

T g

general stiffness relationship is

=

D D

g

f

TD

f

TD

g +H

=Dep

elastic predictorandplastic corrector


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isotropic elastic properties:

D= K 0

0 3G partial differentials required for construction of stmatrix:

yp =

y/

p

py/

pq

=

0(p y)

2 /ap

f

= f/p

fq

= y

1

f

y= p


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drained and undrained compression tests

0

0

0

0

0.1

0.05

0.2

0.2

1

M= 0.8, p= 1.0

M= 1.0, p= 1.0

M= 1.2, p= 1.0

mobilised friction

shear strain

volumetric strain

compression

expansion

00

1

p'/p

q/p'i

M= 1.2, p= 1

M= 1, p= 1

M= 0.8,

but no softeningcontinuing volumetric deformation (unlessM=


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compare elastic-perfectly plastic Mohr-Coulomb

3GK

3G + KM M 1 M

M M M and elastic-hardening plastic Mohr-Coulomb mo

3GK + KH 3GK(M y)

3GKy

3GKy(M

y) + 3GH3G Ky(M y) + H

hardening function H= p(p y)

2

(ap)

H 0; M y; M (M y

expanding yield locus; variable dilatanc


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9.Selectionofsoilparam

(exe(G


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9. Practicalexercise:choiceofsoilpaDavidMuirWood

[email protected]


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triaxial compression of Hostun sand

use spreadsheet to fin

of parameters for [ela

plastic Mohr-Coulom

and elastic-hardeningCoulomb models

there should be many there are many waymatches of similar qu

it would be more real

more challenging to

several sets of data si

0

0.5

1

1.5

0 0.05 0.1 0.15 0.2

shear strain eq

stressratioq/p

observation

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.

Paper Geotechical Modeling Short Couse by Wood

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