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7/31/2019 Lecture 02 ConstitutiveModel Lecture Handout
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For f inite element analysis
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Q: What are const itutive models and why are theyneeded?*
A: They form a central component of most, if not all,
the predic tive models we develop of ground
response. From the simplest conceptual models, tothe most sophisticated mathematical model, we
need to idealise the behaviour of small elementsof soil.
A const itutive model can be defined as a set ofmathematical relationships between for example,
components of stress and components of strain.
2*Carter (2006)
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Soil compression
Odemeter Test
3
p
s
h
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Shear ing of soil - Contractant
triaxial
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31
31
q:stressDeviator
3
)2(p':stressMean
3
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Shearing of soil - Dilatant
5
V
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Strain
StressWork hardening
Peak
Work softening
Yield point
Perfectly plastic
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Strain
Stress
Strain
Stress
Strain
Stress
Strain
Stress
(a) Elastic
(d) Elastic-plastic softening
(b) Rigid plastic
(c) Elastic-Plastic
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Elastic
Ther e is a one- t o- one
r elat ionship bet ween st r essand st r ain.
i.e. = E
I t can be linear or non linear
Af t er loading and unloading,mat er ial r et urn t o it s or iginal
condit ion.
Soil f low (def or mat ion)mechanism depends on t hest ress increment .
Strain
Stress
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Strain
Stress
Elastic-Plastic
The st r ess-st r ain r elat ionshipis not unique.
I t is non linear .
Af t er loading and unloading,mat er ial doesnt r et ur n t o i t sor iginal condit ion.
Soil f low (def or mat ion)
mechanism depends on t hest r ess level.
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Conditions of application of elastic model:
For problems where loads are at the working loadlimit (i.e. much less than the ultimate load limit)
For initial displacement estimation of a structureunder loading.
Implications of adopting elastic model:
No-one can pretend that soil behaves as an elasticmater ial except under str ictest conditions.
Choice of elastic modulus and Poissons ratio arecr itical !!
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Yield surface is a boundary in soil elementstress field.
It defines the state of stress at which soilresponse changes from elastic to plastic.
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allowedNot0)f(
behaviourinelasticofOnset0)f(behaviourElastic0)f(
ij
ij
ij
Strain
StressYield stress- y
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xyyz
yzxz
zz
xz
xy
12
x
z
y
xx
xy
xzxx
yz
xz
zz
xy
zz
yz
zz
zzyzxz
yzxyyxy
xzxyxx
ij:tensorStress
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h
13
q
hv
h
q:stressDeviator
(p':stressMean v
3
)2
p
Yield Surface
Yield point x
z
y
hh
h
v
v
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Plastic strain increment arise if:1) The stress state is located on the yield surface, AND
2) The stress state remains on the yield surface after astress increment
Yield function f() tells us whether plastic strain isoccurring or not, however, we would like to know directionand magnitude of plastic strainFor that we need flow rule
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q
p
Yield Surface [f()]
Potential Surface [g()]
Critical State Line (CSL)
vp
qp
qp
vp
g
p
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Associated flow rule f() =g()Non associated flow rule f() g()
It would be great advantage that f() =g() , only 1 function is to be
generated to describe plastic responseAdvantages:
1) The solutions of the equations that emerge in the analyses is faster2) The validity of the numerical predictions can be more easily
guaranteed
For metals f() =g() For soil f() g()
The assumption of normality of plastic strain vectors to the yield locus
would result in much greater plastic volumetric dilation than actuallyobserved.
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st r ain or displacement )
(st r ess) Real soil response
Idealised soil model MC model
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First order approximation
Basic Law: i = ie + ip
ie = r ever sible (elast ic) st r ain
ip = ir r ever sible (plast ic) st r ain
ip
= 0 f or f < 0f = f (xx, yy, zz, xy, xz, yz)
Yield f unct ion
ip ie
i
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z
x
xz
xz
c
s r
Failure criterion: f = c + at an f
Or r c cos + s sins : cent r e of Mohr s st r ess circler : r adius of Mohr s st r ess cir cle
s sin
c cos
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Hexagonal shape
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strainsplasticofdirectionthedetermines
strainsplasticofmagnitudethedeterminesthatmultipliera
etc,:meansThis
:strainsplasticforruleFlow
py
px
pi
i
yx
i
g
gg
g
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g = 0
f = 0
n
nsp
vp
The assumption of normality of plastic strain increment vectors to
the yield laws would result in much greater plastic volumetricdilation than actually observed.
f g
M-C model: f r s sin c cosg r s sin - c cos : dilatancy angle
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Dense sand
Loose sand
vDense sand
Loose sand
v = x + y + z = 2x + y
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Dense packing of grains
V
Interlocking saw blades
i
Sliding takes place NOT on horizontal planes, but on planes inclined at anangle to the horizontal.
i (strength = dilatancy + friction)
The apparent externally mobilised angle of friction on the horizontal
planes () is larger than the angle of friction resisting sliding of the inclinedplanes (i).
i
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yyxy
xy 1
yy
xy
xy
G
Only plastic strainsf = 0
f < 0
tan
)1(2
xy
yy
EG
xy
yy
tan
xyyy
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y
x= z
z
|y - x|
y
E = 2G(1+)
y
v
v = x + y + z = 2x + y
sin1
sin2tan
11-2
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E0 E50y
2
y
Select E0 when soil behaveslinear elast ic f or a lar ger ange.
Select E50 f or gener al soils.
Select Eur f or unloadingproblems (t unnelling,
excavat ion)
However, E is relat ed t o
conf ining pr essur e,
loading/ unloading/ r eloading
Eur
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For one-dimensional compression:
1
0
v
hK
Select to match K0.
Under loading: = 0.3 ~ 0.4
Under unloading: = 0.15 ~ 0.25
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For undrained clay analysis, c is theundrained shear strength of the soil.
For drained sand analysis, c is normally zero.
f = c + tan
However, in FE analysis, a small valuec > 0.2 kPa to avoid numerical dif ficulties.
In advanced MC model, c can increase withsoil depth.
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= 0 can be used f or undrained analysis.
A high value is f or dense sand and can make
analysis t ake long t ime. A not t oo high value is r ecommended f or init ial
preliminar y analysis.
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c
u
Ef f ect ive st r ess = Tot al st r ess por e pr essur e = u
c = 0
Undrained ClaySand
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Except for very over-consolidated clays, wecan use = 0 for clay layers.
The dilatancy of sand depends on the sanddensity and its fr ict ion angle .
For quar tz sand, 30.
For sand with < 30, = 0.
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