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    An introduction to theFinite Element Method

    Inelastic constitutive models

    F. Auricchio

    [email protected]

    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.