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An introduction to theFinite Element Method
Inelastic constitutive models
F. Auricchio
http://www.unipv.it/dms/auricchio
Universita degli Studi di PaviaDipartimento di Meccanica Strutturale
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Inelastic constitutive models
Elastic material material for which strain de-
pends only on stress
Inelastic material material for which strain de-pends on stress and possibly on other quantities
Limit our consideration to 1D problems(scalar eqns.)
Elastic material =f()
Inelastic material =f(, )
For an inelastic material the quantityis an extra prob-lem variable (internal variable), hence it requires anew equation (often in evolutive form)
Elastic material =f()
Inelastic material
=f(, )
=g(, )
Internal variable in the sense that it is a variablewhich cannot be measured externally
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Inelastic constitutive models
Standard assumption: strain additive decomposi-tion in elastic and inelastic contribution
=e +i
withe =e() , i = e()
The latter equation is the inelastic strain definition
In the case of linear response for the elastic component
e = E
withEconstant parameter characterizing the materialelastic response
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Standard visco-elastic model
Develop a model which show inelastic strain evolution and
rate-dependency
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Standard visco-elastic model
Internal-variable evolutionary equation:
i =1
1
i (1)
with
: viscosity parameter: internal characteristic time
Full model
=e +i
e =
Ei =
1
1
i
which can be condensed as follows
=E
i
i =1
1
i
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Standard visco-elastic model
We may consider two types of loading histories
Strain control:[input] given=(t)[output] compute(t) andi(t)
Stress control:[input] given=(t)[output] compute(t) andi(t)
For analytical approaches (closed form solutions), wemay consider either strain or stress controlled loadinghistories. Clearly, one of the other can be easier to solve.
For numerical approaches (approximated solutions), ingeneral we consider strain as indipendent variable.
Consider time interval of interest [0, T] and subdi-vide it in sub-intervals
Indicated the generic time sub-interval as [tn, tn+1]
Consider known all the quantities attnas well as theindependent variable at time tn+1. The dependentvariable at timetn+1 should be computed.
For simplicity in a time-discrete numerical framequantities at tn are indicated with the subscript n,while quantities at tn+1 are indicated with no sub-script
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Standard visco-elastic model
Numerical approach (approximated solution)
Strain control: given n, in, n as well as [ =(tn+1)], compute [output] and
i
Stress control: given n,in,n as well as
ext [ext =ext(tn+1)], compute[output] and
i such that
R() =() ext = 0 (2)
Equation 2 can be interpreted as an equilibrium con-dition, solved in general with an iterative Newtonmethod, using the update formula
do while R(k) not equal 0
k+1 =k dR()d
k1
R(k)
k =k+1
end do
dR()/dis the algorithmically consistent tan-gent operatorfundamental to obtain a quadratic
convergenceNote that from Equation 2
dR()
d =
d()
d =ET = tangent modulus
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Standard visco-elastic model
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Visco-elast.model: analytical sols
Given a stress loading history=(t) given
it is possible to integrate Equation 1
i(t) =
t
exp
t
()
d (3)
where for convenience the initial time is set equal tot=for which we assume i = 0
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Visco-elast.model: analytical sols
Consider now some specific stress loading histories:
Viscosity test orcreep test
Constant load starting at t= 0(t) = 0 for t 0
Possible now to compute the -relation
=e +i con
e
=
1
E
i =
1 exp
t
from which
=
E
+
1 exp t
=
1
E+
exp
t
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Standard visco-elastic model
We can now consider the resulting strain historyFor small time values [t0+], then
i 0
e = 1
E
e
For large time values [t+], then
i 1
e = 1
E
e +i =
1E
+
Att= 0 elastic response with modulus E
Att= +elastic response with relaxed modulus
E =1
1
E+
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Visco-elastic model
Maxwell model:elastic spring and viscous damper in series
Kelvin model:elastic spring and viscous damper in parallel
Standard visco-elastic model:
elastic spring and viscous damper in parallel + elasticspring in series (analyzed model)
More complex spring-dashpot combinations
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Standard visco-elastic model
Numerical treatment
Time-continuous model
=E( i)
i =1
1
i
Time-discrete model Compute stress history from strain history by an
integration technique (strain driven problem)
Thus, knowing strain at tn+1 and the solution attn[i.e. (n, n,
in)], we need to compute the solution
at timetn+1 [i.e. (, i)]
Discrete solution
Time-integration+
solution algorithm
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Standard visco-elastic model
Time-integrationUse implicit Backward Euler formula
=E( i)
i int
=1
1
i
(4)
where t=tn+1
tn
Solution algorithmSince model is linear, no special algorithm is required.
Combining Equation 4, inelastic strain i results as afunction of previous solution and current strain (consis-tent with definition of strain-driven process):
i =
in+
(5)
where:
=
1 +
E
+
1
t
1, =
E
t
Once updated the inelastic strain, possible to use Equa-tion 41 to update the stress.
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Standard visco-elastic model
Tangent modulusd=Ed Edi
anddi =d
Accordingly
ET=E(1 )
Exercise. Develop a matlab code in which: 1) assign a strain history[0, max] in a time interval [0, Tmax]; 2) subdivide time interval [0, Tmax] in
subintervals of length t; 3) for each time istant solve constitutive problem4) plot stress and inelastic strain versus time.
Exercise. Develop a matlab code in which: 1) assign a stress history[0,
max] in a time interval [0, T
max]; 2) subdivide time interval [0, T
max] in
subintervals of length t; 3) for each time istant solve constitutive problem
such to satisfy Equation 2 4) plot stress and inelastic strain versus time.
Exercise. Test the visco-elastic model for slow and fast loading histories.
Exercise. Extend the first two exercises to piecewise linear loading histo-ries.
Exercise. Test the visco-elastic model for loading histories with changingrates.
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Standard visco-plastic model
Develop a model which is rate-dependent and inelastic only
for high stress values (threshold)
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Standard visco-plastic model
Internal-variable evolutionary equation:i =
1
y (6)
with
: viscosity parametery: yielding stress
MacAuley bracket
< x >=
x if x >0
0 if x 0 !!
Full model
=e +i
e =
E
i =1
y
which can be condensed as follows
=E
i
i =1
y
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Standard visco-plastic model
Fundamental difference wrt visco-elastic model is the pres-ence of ayield (limit) functionFdefined as
F = y
Note:
IfF 0, i.e. > y then i = 0
INELASTIC CASE
Fis a function classically defined in stress space
Generalization to remove limitation >0
i =1
|| y
y| y|
for any
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Standard visco-plastic model
One-dimensional case
Extension to three-dimensional case
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Standard visco-plastic model
Numerical treatment
Time-continuous model
=E( i)
i =1
y
Time-discrete model Compute stress history from strain history by an
integration technique (strain driven problem)
Thus, knowing strain at tn+1 and the solution attn[i.e. (n, n,
in)], we need to compute the solution
at timetn+1 [i.e. (, i)]
Discrete solution
Time-integration+
solution algorithm
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Standard visco-plastic model
Time-integration.Use implicit Backward Euler formula
=E( i)
i int
=1
y
(7)
where t=tn+1
tn
Solution algorithm.Now the model is non-linear (difference!!), need to de-velop a special algorithm (predictor-corrector)
1. Assume elastic step (i.e. < y)
2. Update variables3. Verify position 1
4. If test 3 satisfied solution is elastic,otherwise assume plastic step
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Standard visco-plastic model
Consider only step 4:
=E( i)
i int
=1
( y)
(8)
Combining Equation 8, inelastic strain i results as a
function of previous solution and current strain (consis-tent with definition of strain-driven process):
i =
in++
(9)
where:
=
1 +
Et
1
, =
Et
, =
yt
Once updated the inelastic strain, possible to use Equa-tion 82 to update the stress.
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Standard visco-plastic model
Tangent modulusElastic response
d=Ed
AccordinglyET=E
Visco-plastic responsed=Ed Edi
anddi =d
Accordingly
ET=E(1 )
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Standard visco-plastic model
Exercise. Develop a matlab code in which: 1) assign a strain history
[0, max] in a time interval [0, Tmax]; 2) subdivide time interval [0, Tmax] insubintervals of length t; 3) for each time istant solve constitutive problem4) plot stress and inelastic strain versus time.
Exercise. Develop a matlab code in which: 1) assign a stress history[0, max] in a time interval [0, Tmax]; 2) subdivide time interval [0, Tmax] insubintervals of length t; 3) for each time istant solve constitutive problemsuch to satisfy Equation 2 4) plot stress and inelastic strain versus time.
Exercise. Test the visco-plastic model for slow and fast loading histories.
Exercise. Extend the first two exercises to piecewise linear loading histo-ries.
Exercise. Test the visco-plastic model for loading histories with changingrates.
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Plastic model
Develop a model which is rate-independent and inelastic
only for high stress values
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Plasticity model
Full model
=e +i
e =
E
i =
||
0
F()0F = 0
F = 0
The model can be condensed as follows
=E ii =
||
keeping in mind the two sets of requirements0 , F()0
F = 0 , F = 0
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Plasticity model
Fundamental difference wrt visco-elastic model is thepresence of ayield (limit) function Fdefined as
F =|| y
Fundamental difference wrt visco-plastic model is thatF >0 is non admissable
Possible situations
IfF
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Plasticity model
One-dimensional case
Extension to three-dimensional case
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Plasticity model
Numerical treatment
Time-continuous model
=E( i)
i =
||
0 , F()0F = 0 , F = 0
Time-discrete model
Compute stress history from strain history by anintegration technique (strain driven problem)
Thus, knowing strain at tn+1 and the solution attn[i.e. (n, n,
in)], we need to compute the solution
at timetn+1 [i.e. (, i)]
Discrete solution
Time-integration+
solution algorithm
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Plasticity model
Time-integrationUse implicit Backward Euler formula
=E( i)
i in=
||(11)
where
= t
tn+1
tn
dt
t=tn+1 tn
and
0 , F()0
F = 0 , (F Fn) = 0
Solution algorithm.Now the model is non-linear (difference!!), need to de-velop a special algorithm (predictor-corrector)
1. Assume elastic step
2. Compute elastic trial state
3. Verify position 1
4. If test 3 satisfied solution is elastic,otherwise assume plastic step
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Plasticity model
1. Assume elastic step = 0
2. Compute trial stateT R =E( in)
i,TR =in(12)
3. Verify position 1
if|T R|< y thenstep elasticupdate variablesexit
if|T R
|> y thenstep plasticprocede
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Plasticity model
4. Plastic stepObserve parallelism
=T R E
||
||=
T R
|T R|
hence
||=|T R
| ECompute consistency parameter
F = 0 =|T R| y
E
Update plastic strain
i =in+T R
|T R|
Update the stress
=E( i)
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Plasticity model
Tangent modulusElastic response
d=Ed
AccordinglyET=E
Plastic responsed=Ed Edi
and
di =dT R
|T R|
The linearization of the discrete consistency para-meter computed linearing the yield condition:
dF = 0 Ed EdT R
|T R|= 0
AccordinglyET= 0
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Plasticity model
Exercise. Develop a matlab code in which: 1) assign a strain history
[0, max] in a time interval [0, Tmax]; 2) subdivide time interval [0, Tmax] insubintervals of length t; 3) for each time istant solve constitutive problem4) plot stress and inelastic strain versus time.
Exercise. Develop a matlab code in which: 1) assign a stress history[0, max] in a time interval [0, Tmax]; 2) subdivide time interval [0, Tmax] insubintervals of length t; 3) for each time istant solve constitutive problemsuch to satisfy Equation 2 4) plot stress and inelastic strain versus time.
Exercise. Test the plastic model for slow and fast loading histories.
Exercise. Extend the first two exercises to piecewise linear loading histo-ries.
Exercise. Test the plastic model for loading histories with changing rates.
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