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
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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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stress and strain variables (
( ) 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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stress and strain variables (
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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stress and strain variables (
( ) 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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stress and strain variables (
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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stress and strain variables (
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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Laboratory element testing
plane strain occurs mor
no strain in long directi
plane strain withfixedprincipal axes
fairly easy to apply in laboratory
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Laboratory element testing
true triaxial apparatus
3 degrees of freedom
no rotation of principal axes
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Laboratory element testing
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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Laboratory element testing
direct shear boxinhomogeneous deformati
strength information
pedagogic illustration of d
use for subsequent modell
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Laboratory element testing
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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Soilmodelling:SouthEastAsia:OctoberNovem
2. ElasticmodellingDavidMuirWood
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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
theInventionsIintendtopublish,thoughpossiblynotinthesameor
getopportunityandleasure;mostofwhich,Ihope,willbeasuseful
areyetunknownandnew.
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
theInventionsIintendtopublish,thoughpossiblynotinthesameor
getopportunityandleasure;mostofwhich,Ihope,willbeasuseful
areyetunknownandnew.
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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Stiffness in undrained triaxial comp
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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Stiffness in undrained triaxial comp
=
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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Stiffness in undrained triaxial comp
=
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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Elastic stiffness from different de
pressuremeter
direct measurement of shear modulus G
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Elastic stiffness from different de
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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Example: Plane strain isotropic elastic elemen
under general stress path
y = (x + z)
General plane strain stress path: x = z
then y
= (1 + ) z z
x = z
z
y=0y
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Example: Plane strain isotropic elastic elemen
under general stress path
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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Example: Plane strain isotropic elastic elemen
under general stress path
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
under general stress path
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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Soilmodelling:SouthEastAsia:OctoberNovem
3. Themostwidelyusedsoilmodel:MoDavidMuirWood
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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
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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
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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)
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Mohr-Coulomb model: typical re
typical response in constant p' test
dilation/compression depends on sign of M*
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standard elasplastic Mohr
non-associate
M* M
simplicity
sharp stiffness changes
tangent stiffness eitherelastic or zero
continuing volume change
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Mohr-Coulomb model: fitting to
standard model availabgeotechnical finite elem
subjectivity in choice o
stiffness G, K;
strength M;
dilatancy M*
exercise
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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
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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
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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 ?
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4.Applicati
MohrCoulomb
m
(exercise)(G
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Soil modelling: South East Asia: October-November 2010
4. Application of Mohr-Coulomb model (exercise)David Muir Wood
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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
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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
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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
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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
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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
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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
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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
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Soil modelling: South East Asia: October-November 2010
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
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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
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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.
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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
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Soil modelling: South East Asia: October-November 2010
5. Cam clayDavid Muir Wood
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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,
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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
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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 =
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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
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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
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p'
v iso-ncl
A
B-vp
separation of urls produces irrecoverable volchange vp
plastic volumetric strain pp = vp/v
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q
p'
A
B D
1
normality to yield locus givesD = pp/p
qa
plastic distortional strain
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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
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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!
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Cam clay drained compression test (2)
how did we reach B? normal compression to A?
behaviour beyond B not affected by previous history
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Cam clay drained compression test (3)
how did we reach B? overconsolidation to A?
behaviour beyond B not affected by previous history
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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!
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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
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Taylor, 1948
plastic softening confined to thin localised region
elastic unloading
Physical consequence of uncertainty of response
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Cam clay hardening and softening response
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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
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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
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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
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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
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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
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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
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q
q
3G
A
B
qe qp
q
elastic distortional strain increment iseq=
elastic and plastic components summed to gdistortional strain
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v
Cam clay undrained compression
step through undrained test on
normally compressed soil
conventional compression q = 3p
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Cam clay undrained compression
step through undrained test on lightly
overconsolidated soil
conventional compression q = 3p
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Weald clay undrained response division of pore pressure
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6.Camclaygrap
calculations(exe
(SBCSS
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Soil modelling: South East Asia: October-November 2010
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:
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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
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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
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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
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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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Soil modelling: South East Asia: October-November 2010
7. Cam clay: compliance relationshipsDavid Muir Wood
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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
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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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Cam clay undrained compression
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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Cam clay undrained compression
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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Cam clay undrained compression
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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Soilmodelling:SouthEastAsia:OctoberNovem
9. Practicalexercise:choiceofsoilpaDavidMuirWood
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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.