introduction to computational methods for inelastic ... · introduction to computational methods...

Modelli e metodi computazionali per leghe a memoria di formaNapoli, Facolta di Ingegneria, Luglio 2008

Introduction to

computational methods for

inelastic material behaviors

Ferdinando Auricchio 1 2 3

1Dipartimento di Meccanica Strutturale, Universita di Pavia, Italy

2European School for Advanced Studies in Reduction of Seismic Risk, Pavia, Italy

3 Istituto di Matematica Applicata e Tecnologie Informatiche, CNR, Italy

July 18, 2008

F.Auricchio (DMS - ROSE - IMATI) Computational inelasticity July 18, 2008 1 / 74

Motivations

Uni-axial stress σ versus uni-axial strain ε

Non-linear kinematic hardening Generalized plast. with no hardening

ε

σ

σ y,0u

2σy,0u

2σy,0u

− σy,0u

Hkin nlHu u

Hkin nlHu u

A

B

ε

σ

σy,0u

βu

A

B

C

D

Smooth transition to horizontal asymptote

If unloaded from plastic range and reloadedbefore reverse plasticity (A → B → A),plasticity renewed exactly at the same stresswhere unloading began (A)

Smooth transition between elastic andplastic behavior also under cyclic load


If unloaded from plastic range and reloadedbefore reverse plasticity(A → B → D),plasticity renewed (C) before the stress atwhich unloading began (A)



Introduction I

Elastic material: a material for which stress is function only ofstrain, or, equivalently in the case of no internal constraint, a ma-terial for which strain is function only of stress

εεε = εεε(σ) (1)

Elastic material ⇒ constitutive response independent, for example, from processfollowed in reaching actual state

Position 1 is clearly an approximation of material real behavior, very close in somecases, rough in other cases

Thinking of strain as dependent variable, strain may depend not only on stress atcurrent time, but also on stress path followed to reach current state


Introduction II

Whenever path dependence is significant enough to be taken into account, thematerial must be regarded as inelastic

Inelastic material: material for which strain is function not only ofstress, but also of other quantities, in general, indicated as internal

variables, ξ:εεε = εεε(σ, ξ) (2)

Internal variables ξ can be a set of scalars, tensors, or whatever needed to properlydescribe the material response

Since internal variables ξ represent a new and extra set of variables, necessary tointroduce an extra set of equations, internal variable constitutive equations,represented, for example, as rate equations

ξ = g(σ, ξ) (3)

that is, through differential equations that determine the rates in time (timederivatives) of such new variables


Introduction III

Within small deformations, strain additively decomposed as

εεε = εεεe + εεε

i (4)

where:

I εεεe : elastic strain, i.e. part of strain related to stressthrough elastic relation or part of strain function onlyof stress, i.e. εεεe = εεεe(σ)

I εεεi : inelastic strain, i.e. difference between total andelastic strain, i.e. εεεi = εεε− εεεe

Inelastic strain, εεεi , may be assumed as internal variable, hence as an element of ξ:

ξ = εεεi , β ⇒εεε

i = g1(σ,εεεi , β)

β = g2(σ,εεεi , β)


Notation

For future convenience:

Split stress, σ, and strain, εεε, in deviatoric and volumetric components:

σ = s + pI

εεε = e +1

3θI

with

tr(s) = s : I = 0 and p =1

3tr(σ) =

1

3σ : I

tr(e) = e : I = 0 and θ = tr(εεε) = εεε : I

where tr(•) : trace operator; I: second rank identity tensor

Equip 2nd-rank tensor linear vector space with natural (Euclidean) inner product(trace of product):

‖a‖ = [a : a]12 = [ tr(a · a) ]

12

with a any second rank tensor


Viscoelasticity I

GOAL: develop a model showing inelastic strain evolution and rate-dependency(standard viscoelastic model)

Consider a simple case – isotropic linear viscoelastic material – detailing modeltime-continuous and time-discrete version

Due to model linearity, no special algorithm needed for time-discrete approach

Tangent tensor consistent with discrete model also addressed


Linear viscoelasticity: time-continuous model I

Isotropic material response ⇒ elastic volumetric and deviatoric responses decoupleI Moreover, linear elasticity

ee =s

2G, θe =

p

K

No inelastic volumetric response i.e. θi = 0 ⇒ volumetric response purely elastic

Hence, additive elasto-plastic strain split εεε = εεεe + εεεi can be rewritten as

e = ee + ei , θ = θe


Linear viscoelasticity: time-continuous model II

Assume ei as only internal variables, i.e. set ξ = ei

As internal variable rate equation, assume simplest linear relation, i.e.

ei =1

2ηs − 1

λei

withI η: material viscosity parameterI λ: internal characteristic time

Accordingly, obtain the so-called linear isotropic viscoelastic model

p = Kθ

s = 2G(e − ei )

ei =1

2ηs − 1

λei

First equation is trivial (function evaluation)


Linear viscoelasticity: time-discrete counterpart and algorithmic solution I

Present now model time discrete counterpart for a possible computer implementation

Since model is linear, no special algorithm should be used to obtain the solution

Tangent tensor consistent with discrete model is also addressed

Let [0, T ] ⊂ R be a time interval and consider two time values within it, tn andtn+1 > tn, such that tn+1 is the first time value of interest after tn

To minimize subscript appearance (to make equations more readable):

an = a(tn) , a = a(tn+1)

where a is any generic quantity.

Accordingly, in the discrete time setting subscript n indicates a quantity evaluated attime tn, while no subscript indicates a quantity evaluated at time tn+1

GOAL: develop a time marching procedure


Linear viscoelasticity: time-discrete counterpart and algorithmic solution II

IDEA: compute stress history from strain history (strain driven problem)I Thus, given strain εεε at time tn+1 and solution at time tn

(

σn,εεεn, ein

)

, computesolution at time tn+1

Using backward Euler integration formula on rate equation

ei =1

2ηs − 1

λei ⇒ ei − ei

n

∆t=

1

2ηs − 1

λei

with ∆t = tn+1 − tn

Evaluating remaining equation at tn+1, time-discrete model is

p = Kθ

s = 2G(e − ei )

ei = ein +

[

1

2ηs − 1

λei

]

∆t

(5)


Linear viscoelasticity: time-discrete counterpart and algorithmic solution III

Compute inelastic strain ei update as function of previous solution and current strain(RECALL: strain-driven process), combining 52 & 53

ei =

[

1 +

(

2G

2η+

1

λ

)

∆t

]

−1

ein +

(

2G

2η∆t

)

e

or in more compact form:

ei = α[

ein + βe

]

(6)

where:

α =

[

1 +

(

2G

2η+

1

λ

)

∆t

]

−1

, β =2G

2η∆t

Compute stress update using 51 & 52 as well as updated inelastic strain


Linear viscoelasticity: time-discrete counterpart and algorithmic solution IV

Compute now fourth-order tangent tensor Ctgdisc, consistent with time-discrete model

dσ = Ctgdiscdεεε (7)

Necessary in general to solve equilibrium problems

I Point-wise (local) equilibrium. Assume to assign a stress history σext(t). Tocompute strain history which produces the assigned stress history, we need to solveequation

R(εεε) = σ(εεε) − σext = 0

where all quantities are assumed to be evaluated at time tn+1

I Boundary-value (global) equilibrium. Assume to assign a standard boundary valueproblem where body load and boundary conditions may wary in time. Then at anytime tn+1 the following problem has to be solved:

R = Lint −Lext = 0 (8)

where Lint and Lext are respectively internal and external work with

Lint(εεε) =

∫

Ωδεεε : σ(εεε)dV (9)

and all field quantities are assumed to be evaluated at time tn+1


Linear viscoelasticity: time-discrete counterpart and algorithmic solution V

I To solve nonlinear residual equation R(εεε) = 0 often adopted an iterative Newtonmethod

F Start with a tentative solution εεεk

F Verify if tentative solution εεεk is an effective solution, i.e. if R(εεεk ) = 0F If tentative solution εεεk is NOT an effective solution, i.e. if R(εεεk ) 6= 0, then UPDATE

tentative solution, requiring

R(εεεk+1) ≈ 0

i.e.

R(εεεk+1) ≈ R(εεεk ) +∂R

∂εεε

∣

∣

∣

∣

εεε=εεεk

(εεεk+1 − εεεk ) = 0

hence

εεεk+1 = εεε

k−

[

∂R

∂εεε

]

−1

εεεk

R(εεεk )

F In both cases (eq. 3 or 8) we need ∂σ/∂εεε


Linear viscoelasticity: time-discrete counterpart and algorithmic solution VI

Linearize deviatoric elastic relation and inelastic strain evolution equation

ds = 2G(de − dei)

dei = αβde

Combine previous two relations

ds =[

2G(1 − αβ)II]

de

Recalling stress volumetric-deviatoric split, we get:

Ctgdisc = [K(I ⊗ I) + 2G(1 − αβ)Idev] (10)


Linear viscoelasticity: time-discrete counterpart and algorithmic solution VII

Overall time-discrete solution algorithm

Given σn,εεεn, ein and εεε

Compute en, θn and eUpdate ei [ Eq. 6 ]Update s and p

Compute σ

Compute algorithmic tangent [ Eq. 10 ]Exit


Viscoplasticity I

Simplest – but quite general – inelastic material model based on internal variables

εεε = εεεe(σ) + εεε

i

εεεi = g(σ,εεεi )

(11)

If g is linear/non-linear, then viscoelasticity

If g is strongly non-linear, then viscoplasticityI Strongly non-linear, i.e. elastic in a certain stress range ( elastic range )

history-dependent outside that range ( inelastic range )I Formally, assume existence of a continuous function F (σ, ξ) ( yield function ) s.t.

g(σ, ξ) = 0 when F ≤ 0

g(σ, ξ) 6= 0 when F > 0(12)

I F (σ, ξ) = 0 defines yield surface in stress spaceI F (σ, ξ) ≤ 0 defines elastic region enclosed by yield surface

If g is strongly non-linear & very slow process, then plasticityI In the limit, plasticity is rate-independent, i.e. results are independent of loading rate


Linear viscoplasticity: time-continuous model I

Bingham model: simple viscoplasticity model, originally proposed for 1D shear,later generalized to 3D regimes

Purely elastic volumetric response (θi = 0)

e = ee + ei , θ = θe

Isotropic elastic strain-stress relations

ee =s

2G, θe =

p

K

Only 1 tensor-value internal variable

ξ = ei


Linear viscoplasticity: time-continuous model II

Internal variable rate equation

ei =1

2η〈‖s‖ − σy〉 n (13)

whereI η: viscosity parameterI σy : parameter representing yield surface radiusI ‖s‖: norm of deviatoric stress, i.e. ||s|| =

√s : s =

√sij : sij

I < • >: Macauley bracket or positive part function defined as

< x > =

x if x ≥ 0

0 if x < 0

I n: direction of stress tensor

n =s

‖s‖lue

ei = 0 for ‖s‖ ≤ σy , so that F = ‖s‖ − σy = 0 is the yield surface equation

Model equations are piecewise linear ⇒ explicitly integratable in some cases


Linear viscoplasticity: time-continuous model III

The model reproduces two main features of the large class of viscoplastic models

For fast loading conditions, i.e. loading rates approaching infinity (˙||s|| → ∞), the

model response is elastic

This is a consequence that for fast loading the material does not have the time todevelop rate-dependent effects

For slow loading conditions, i.e. loading rates approaching zero (˙‖s‖ → 0+), the

stress norm remains constant at the yield value

This is a classical feature of models with rate equations in the form ξ = g(σ, ξ) andwith a well-defined yield function, depending on stress only.In the case of yield functions, for slow loading conditions the problem solution tendsto satisfy the yield function F = 0.


Linear viscoplasticity: time-discrete model/ algorithmic solution I

Being the model non-linear (due the positive part function), a special algorithmshould be used to solve the time-discrete model

Compute stress history from strain history (strain driven problem), i.e. given εεε and(

σn,εεεn, ein

)

, compute solution at tn+1

Using a backward Euler integration formula on rate equation and evaluatingremaining equations at time tn+1, time-discrete model can be expressed as:

p = Kθ

s = 2G(e − ei )

ei = ein +

∆t

2η〈‖s‖ − σy 〉 n

(14)

with ∆t = tn+1 − tn.

Combining last two equations, express inelastic strain ei in terms of previous solutionand current strain, according to the definition of strain-driven process; however, thisequation is strongly non-linear and we cannot easily solve it.


Linear viscoplasticity: time-discrete model/ algorithmic solution II

To solve non-linear algebraic problem consider a return map algorithm

Elastic-predictor plastic-corrector algorithm, based on a two-step procedure:

STEP 1 compute a purely elastic trial state

STEP 2 check if trial state non-admissable, then applyinelastic correction computed using trial state as initial condition

Rewrite deviatoric part of time-discrete problem as follows

s = 2G(e − ei)

ei = ein + λn

λ =∆t

2η〈‖s‖ − σy 〉

(15)


Linear viscoplasticity: time-discrete model/ algorithmic solution III

Trial state: assume that in [tn, tn+1] no inelastic deformation occurs (i.e. ei = ein,

λ = 0), leading to definition of elastic trial state:

λTR = 0

ei,TR = ein

sTR = 2G[

e − ein

]

(16)

Check trial state: if admissible elastic trial state (i.e. ‖sTR‖ ≤ σy ), then trial statedoes not violate model, representing solution at tn+1 and algorithm next part isskipped

Inelastic correction: if non-admissible trial state (‖sTR‖ > σy ), then trial state doesviolate model, hence a correction should be performed


Linear viscoplasticity: time-discrete model/ algorithmic solution IV

Inelastic correction: if trial step is non admissible, step is necessarily inelastic.Using Equations 15 and 16:

s = sTR − β (‖s‖ − σy ) n

with:

sTR = 2G(e − ein) , β =

2G∆t

2η

Now, setting: s = ‖s‖n , sTR = ‖sTR‖nTR we may observe that:

n = nTR (17)

which allows us to derive the following scalar relation (radial return):

‖s‖ = ‖sTR‖ − β (‖s‖ − σy ) (18)


Linear viscoplasticity: time-discrete model/ algorithmic solution V

Solving Equation 18 with respect to ‖s‖, we get:

‖s‖ =1

1 + β

(

‖sTR‖ + βσy

)

(19)

which can be substituted in the definition of λ – Equation 153 – resulting in:

λ =∆t

2η

1

1 + β

[

‖sTR‖ − σy

]

(20)

Using Equations 151 and 152, update inelastic strain and stress


Linear viscoplasticity: time-discrete model/ algorithmic solution VI

Compute now fourth-order tangent tensor Ctgdisc consistent with time-discrete model

dσ = Ctgdiscdεεε (21)

Linearize deviatoric elastic and inelastic strain evolution equations

s = 2G(e − ei )

ei = ein + λnTR

⇒

ds = 2G(de − dei )

dei = nTRdλ + λdnTR

Linearize expression for λ to compute dλ

dλ = AdiscnTR : de

with:

Adisc =2G∆t

2η + 2G∆t


Linear viscoplasticity: time-discrete model/ algorithmic solution VII

Keeping in mind that:

nTR =sTR

‖sTR‖ =sTR

[

(sTR : sTR)12

]

we can compute the linearization:

dnTR =1

‖sTR‖[

I − (nTR ⊗ nTR)]

dsTR =N

‖sTR‖dsTR

where:

dsTR = 2Gde

with fourth-order tensor N orthogonal projection on plane with unit normal nTR = n:

NnTR = 0 and NN = N


Linear viscoplasticity: time-discrete model/ algorithmic solution VIII

Combining all the previous relations we find:

ds =[

2G(1 − C)II + 2G (C − Adisc)(n ⊗ n)]

e

with:

C =2Gλ

‖sTR‖

Recalling volumetric-deviatoric stress split, get algorithmic tangent tensor:

Ctgdisc = K (I ⊗ I) + 2G (1 − C) Idev + 2G (C − Adisc)(n ⊗ n) (22)


Linear viscoplasticity: time-discrete model/ algorithmic solution IX

Given σn,εεεn, ein and εεε

Compute en, θn and eCompute trial state [ Eq. 16 ]Check trial stateIf ( ‖sTR‖ > σy ) then

Plastic stepCompute λ [ Eq. 20 ]Update ei [ Eq. 15 ]Update s and p [ Eq. 15 ]Compute algorithmic tangent [ Eq. 22 ]

ElseElastic step

Update s and p with trial state [ Eq. 16 ]Compute elastic tangent

End ifCompute stress σ

Exit


J2 classical plasticity: time-continuous model I

Model assumptions:

yield function depends only on deviatoric stress norm ‖s‖ =√

2J2 (von Mises or J2materials). Accordingly, we have:

θp = 0 , εεεp = ep

neglect hardening mechanisms

Model equations

p = K θ (23)

s = 2Gee = 2G [e − ep] (24)

F = ‖s‖ − σy (25)

ep = γ n (26)

γ ≥ 0 , F ≤ 0 , γF = 0 (27)


J2 classical plasticity: time-continuous model II

where:

Equation 23 is linear elastic relation between pressure p and volumetric strain θ,with K the bulk modulus

Equation 24 is linear elastic relation between deviatoric stress s and deviatoric elasticstrain ee , with G the shear modulus

Equation 25 is the von Mises yield function

Equation 26 is the constitutive equation for the deviatoric plastic strain in theframework of associative plasticity with n = s/‖s‖Equations 27 are the Kuhn-Tucker conditions, which reduce the plastic problem to aconstrained optimization problem


Continuous plastic rate parameter I

Use consistency condition F = 0 to compute plastic rate parameter as

γ = Acont[n : e] (28)

Using chain rule and rate form of equation 24, we have:

F = n : s = 2G(n : e) − 2G γ = 0

which has the solution:

γ = ALPcont[n : e]

where:

ALPcont = 1


Continuous elasto-plastic tangent tensor I

Starting from rate form of linear elastic relation and recalling consistency parameterexpression, we obtain a rate relation between total stress σ and total strain εεε:

σ = Dcontεεε

where

Dcont = K (I ⊗ I) + 2G [Idev − Acont(n ⊗ n)] (29)

with

Idev = I − 1

3(I ⊗ I)

and I the fourth-order identity tensor

Dcont consistent with continuous model is always symmetric


J2 classical plasticity: time-discrete couterpart and integration algorithm I

Discrete time counterpart of the model, with particular attention to possibleimplementation in finite element code

Use return map algorithm to integrate the constitutive model

Address elasto-plastic tangent tensor consistent with discrete model

Time interval: [0, T ]. Consider two time values within it, tn and tn+1 > tn, such thattn+1 is the first time value of interest after tn.

To minimize subscript appearance:

an = a(tn) , a = a(tn+1)

with a any generic quantity

Accordingly, in discrete time setting subscript n indicates a quantity evaluated attime tn, while no subscript indicates a quantity evaluated at time tn+1.


J2 classical plasticity: time-discrete couterpart and integration algorithm II

Compute stress history from strain history by an integration technique (strain drivenproblem), i.e. given strain e and solution at tn, (sn, en), compute solution at tn+1

Using a backward Euler integration formula, discrete plastic strain becomes:

ep = epn + λ n with λ =

∫ tn+1

tn

γ dt (30)

Substitution into elastic constitutive relation yields:

s = 2G [e − epn] − 2G λ n (31)

In the above, λ is unknown quantity to be computed by a return map algorithm

Return map: elastic-predictor plastic-corrector algorithm, hence a two stepalgorithm

I In the first step, a purely elastic trial state is computed;I in the second, if trial state violates material model constitutive equation, a plastic

correction is computed using trial state as initial condition


J2 classical plasticity: time-discrete couterpart and integration algorithm III

Trial state: assume that in [tn, tn+1] no plastic deformation occurs (i.e. ep = epn ,

λ = 0):

λTR = 0

ep,TR = epn

sTR = 2G [e − epn]

Check trial state: if elastic trial state is admissible, i.e. it does not violate limitequation F , then it represents the new solution at tn+1 and next part of thealgorithm is skipped. If elastic trial state is not admissible, a correction is performed.

Plastic correction: enforce limit equation satisfaction to compute discrete plasticrate parameter λ


Discrete plastic parameter and plastic rate parameter I

Express elastic relation and plastic evolution in terms of trial state and λ:

ep = ep,TR + λn

s = sTR − 2G λ n

Last relation implies n = nTR hence a scalar relation may be derived:

‖s‖ = ‖sTR‖ − 2Gλ

Discrete form of the limit equation can be enforced:

F =[

‖sTR‖ − 2Gλ]

− σy = 0

and solved for λ:

λLP =‖sTR‖ − σy

2G

Accordingly, discrete plastic rate parameter λ is computed solving a scalar equation

Closest point projection of trial state onto limit surface F = 0 reduces to radialprojection


Discrete elasto-plastic tangent tensor I

Simple and efficient approach for constructing elasto-plastic tangent tensor,consistent with discrete model

Use of a consistent tangent tensor preserves quadratic convergence of a Newtonmethod, for example within incremental solution in a finite element scheme

Recall linear elastic relation between s and e:

s = 2G [e − epn] − 2Gλn (32)

through linearization it becomes:

ds = 2Gde − 2Gdλ n − 2Gλ dn (33)


Discrete elasto-plastic tangent tensor II

Keeping in mind that:

n =s

‖s‖ =s

[

(s : s)12

]

we can compute the linearization:

dn =1

‖s‖ [I − (n ⊗ n)] ds =N

‖s‖ds

with N the orthogonal projection operator on the plane with unit normal n

Nn = 0 and NN = N

Accordingly, Equation 33 can be recast as follows:

[I + aN] ds = 2Gde − 2Gndλ (34)

where:

a =2Gλ

‖s‖


Discrete elasto-plastic tangent tensor III

Assume that from linearization of discrete limit equation

dλ = Adisc[n : de]

with Adisc a scalar quantity

Accordingly, Equation 34 simplifies to:

[I + aN] ds = 2G [(1 + Adisc)I − AdiscN] de

Inverting the coefficient matrix, we may solve for ds:

ds = [I + α1N] 2G [(1 + Adisc)I − AdiscN] de

where:

α1 = − a

1 + a


Discrete elasto-plastic tangent tensor IV

Combining this equation with incremental version of elastic volumetric relation, weobtain

dσ = Ddiscdεεε

where the algorithmic elasto-plastic tangent tensor is given by:

Ddisc = K (I ⊗ I) + 2G (1 − C) Idev + 2G (C − Adisc)(n ⊗ n) (35)

where Adisc comes from the linearized limit equation and

C =2Gλ

‖sTR‖

Ddisc consistent with discrete model is symmetric

Ddisc = Dcont when C LP = 0, or when load step reduces to zero

Linearization of discrete limit equation and use of Equations 33 and 34 return

dF = n : ds = 2G [n : de] − [2G ]dλ = 0

which can be solved for dλ, obtaining:

dλ = ALPdisc[n : de] with A

LPdisc = 1 = A

LPcont


A general plasticity framework: basics I

Inelastic body: strain determined by stress & additional (internal or hidden)variables

Inelastic behavior treated within “generalized plasticity” framework, i.e. assumeexistence of two continuous real-valued functions

I Limit function F : to distinguish between admissible and non-admissible states

F ≤ 0 ⇒ admissible stateF > 0 ⇒ non-admissible state

I Yield function f : to distinguish between elastic and inelastic states

f < 0 ⇒ elastic state ⇒ no inelastic effectsf ≥ 0 ⇒ inelastic state ⇒ inelastic effects may or may not occur

depending on loading or unloading

Limit function F and yield function f do not necessarily coincide

f > 0, F > 0

f < 0, F <= 0

F = 0

f = 0

f > 0, F <= 0

ε

σ

f = 0

f < 0

f > 0

σ.

>0, F<0, =0γ.

σ.

<0, F=0, =0γ.

σ.

>0, F=0, >0γ.

σ.

<0, F=0, =0γ.

Multi-axial case Uni-axial case


A general plasticity framework: simplifying assumptions I

Dependence on time t is not explicitly stated

Small deformation regime

Strain εεε additively decomposed into elastic and plastic part, εεεe and εεεp

εεε = εεεe + εεε

p

Linear elastic isotropic response:

ee =1

2Gs , θe =

1

Kp

with G and K shear and bulk modulus

Combining previous relations

σ = p1 + s = Kθe1 + 2Gee = [K (I ⊗ I) + 2G Idev]εεεe


A general plasticity framework: simplifying assumptions II

Include linear isotropic hardening and non-linear kinematic hardening

Internal variables:I plastic strain εεεp

I back stress α: yield surface center, to model kinematic hardening mechanismI accumulated plastic strain ep: to model isotropic hardening mechanism

Associated flow rule

εεεp = γ

∂f

∂σ

According to the existence of an elastic region, we require:

γ = 0 when f < 0γ ≥ 0 when f ≥ 0

Von Mises or J2 materials: yield and limit functions depend only on deviatoric stressnorm (‖s‖ =

√2J2, )

θp = 0 , εεεp = ep


A general plasticity framework: time-continuous model I

Linear elastic relationp = K θ

s = 2Gee = 2G [e − ep]

I p: pressure, θ: volumetric strain, K : bulk modulusI s: deviatoric stress, ee : elastic deviatoric strain G : shear modulus; e, ep : deviatoric

total and plastic strain

Overstress or relative stress Σ Σ = s − α

I in terms of deviatoric stress s and deviatoric back stress α

Yield function f ; limit function F f = f (Σ, ep) = ‖Σ‖ − σy (ep)

F = F (Σ, ep , γ)

I f in terms of relative stress norm (von Mises) and yield surface radius σyI F dependent explicitly on γI variable yield function radius, due to simplest possible form isotropic hardening

σy = σy,0 + Hiso ep

with σy,0 initial yield stress and Hiso a material constant. Non-linear isotropicmechanisms may needed to increase accuracy under complex loading


A general plasticity framework: time-continuous model II

Plastic strain evolution ep = γ∂f

∂σ= γ

∂f

∂Σ= γ n

I Due to the specific form of f , we have n = Σ/‖Σ‖ with ‖n‖ = 1

Accumulated plastic strain evolution ˙ep = ‖ep‖ = γ

Back-stress evolution α = Hkinep − Hnl ˙e

pα

I Hkin and Hnl material constantsI Originally proposed by Armstrong and Frederick, it can be rewritten as

α = Hkinγn − Hnl γα

Kuhn-Tucker conditions γ ≥ 0 , F ≤ 0 , γF = 0

I reduce plastic problem to a constrained optimization problem


A general plasticity framework: time-continuous model III

Three specific material models fit within the above general framework

They differ in terms of limit functions and kinematic hardening rules:

Material model Limit function Kinematic mech.Linear plasticity (LP) F = f = ‖Σ‖ − σy α = Hkinγn , Hnl = 0Non-linear kin.hard.(NLK) F = f = ‖Σ‖ − σy α = Hkinγn − Hnl γαGeneralized plasticity (GP) F = h(f ) [n : σ] − γ α = Hkinγn , Hnl = 0

For the generalized plasticity model h(f ) is a non-linear function


A general plasticity framework: time-continuous model IV


Non-linear kinematic hardening Generalized plast. with no hardening

ε

σ

σ y,0u

2σy,0u

2σy,0u

− σy,0u

Hkin nlHu u

Hkin nlHu u

A

B

ε

σ

σy,0u

βu

A

B

C

D


If unloaded from plastic range and reloadedbefore reverse plasticity (A → B → A),plasticity renewed exactly at the same stresswhere unloading began (A)


If unloaded from plastic range and reloadedbefore reverse plasticity(A → B → D),plasticity renewed (C) before the stress atwhich unloading began (A)


A general plasticity framework: time-continuous model V


Generalized plast. with hardening Cyclic Generalized plast. with hardening

ε

σ

σy,0u

β

β

u

u

ε

σ

σy,0u

β

β

2σy,0u

− σy,0u

u

u

Possible non horizontal asymptote

Update scheme for speed parameter

The model retains its capacity of smoothtransition between elastic and plasticbehavior also under cyclic load.


A general plasticity framework: continuous plastic rate I

Enforcing limit equation F = 0 or consistency condition F = 0, plastic rateparameter expressed as

γ = Acont[n : e] (36)

with Acont a scalar quantity depending on the particular material model

Elasto-plastic tangent tensor for continuous model:I Start from rate form of linear elastic relation and evolution equation for back stress

[

I 0 2Gn0 I − [Hkinn − Hnlα]

]

sαγ

=

2G e0

with I fourth-order identity tensor.I Recalling plastic rate parameter, we obtain:

[

I 00 I

]

sα

=

2G [I − Acont (n ⊗ n)] eAcont [Hkin (n ⊗ n) − Hnl (α⊗ n)] e

(37)


A general plasticity framework: continuous plastic rate II

I Combining first row with elastic rate ⇒ rate relation between total stress σ and totalstrain εεε:

σ = Dcontεεε

where:

Dcont = K (I ⊗ I) + 2G [Idev − Acont(n ⊗ n)]

with

Idev = I − 1

3(I ⊗ I)

I Elasto-plastic tangent tensor Dcont consistent with the continuous model is alwayssymmetric, as a consequence of the decoupling of the two equations in 37.


A general plasticity framework: time-discrete and integration algorithm I

Time interval: [0, T ]; focus on two time values within it, tn and tn+1

To make equations more readable, minimize subscript appearance

an = a(tn) , a = a(tn+1)

where a is any generic quantity. Accordingly, subscript n indicates a quantityevaluated at tn, while no subscript indicates a quantity evaluated at tn+1.

Stress history computed from strain history by integration technique (strain drivenproblem). Thus, knowing strain e and solution at time tn:

(sn, en, epn , e

pn , αn)

we need to compute solution at tn+1


A general plasticity framework: time-discrete and integration algorithm II

Backward Euler integration formula on discrete plastic strain, accumulated plasticstrain and back stress evolutionary equations

ep = epn + λ n (38)

ep = e

pn + λ (39)

and:

Rλα = αn + Hkinλn or α = T

λαn + HkinT

λλn (40)

with

λ =

∫ tn+1

tn

γdt

discrete plastic rate parameter and:

Rλ = 1 + Hnlλ , T

λ =1

1 + Hnlλ

Superscript λ indicates a dependence on plastic rate parameter


A general plasticity framework: time-discrete and integration algorithm III

Substitution of plastic evolution into elastic relation yields:

s = 2G [e − epn] − 2G λ n (41)

and subtraction of Equation 40 gives:

Σ = s − α = 2G [e − epn] − T

λαn − U

λn (42)

where:

Uλ =

[

2G + HkinTλ]

λ

Unknown quantity: λ, computed by return map algorithm

Return map algorithm: efficient and robust integration scheme based on limitequation discrete enforcement. Elastic-predictor plastic-corrector algorithm, hence, atwo part algorithm.

I In the first part, a purely elastic trial state is computedI in the second, if the trial state violates the material model constitutive equation, a

plastic correction is computed using the trial state as an initial condition


A general plasticity framework: time-discrete and integration algorithm IV

I Trial state: assume no plastic deformation in [tn, tn+1] (ep = epn , i.e. λ = 0, α = αn )

λTR = 0

ep,TR = epn

ep,TR = epn

αTR = αn

sTR = 2G [e − epn]

ΣTR = sTR − αTR = sTR − αn

I Check: If elastic trial state is admissible, i.e. it does not violate the limit equation F ,then it represents the new solution at tn+1 and the second part of the algorithm isskipped. If the elastic trial state is not admissible, a correction is performed.

I Plastic correction: enforcing the satisfaction of the limit equation, the discrete plasticrate parameter λ may be computed, as shown for various material model later.Equations 38, 39, 402, 41 and 42 can be now rewritten in terms of the trial state and λ:

ep = ep,TR + λn

ep = ep,TR + λ

α = TλαTR + HkinTλλn

s = sTR − 2G λ n

Σ = sTR − TλαTR − Uλn


A general plasticity framework: time-discrete and integration algorithm V

For all models considered, closest point projection of trial state onto limit surfaceF = 0 reduces to a radial return projection

Consequently, discrete plastic rate parameter λ computed solving only one scalarequation

I First note that relative stress Σ can be decomposed as

Σ = ΣλA − Uλn (43)

where:

ΣλA =

[

sTR − TλαTR]

I Secondly, observe that Σ = ‖Σ‖ n since

n =∂f

∂σ=

∂f

∂Σ

∂Σ

∂σ=

∂f

∂ΣIdev

=∂‖Σ‖∂Σ

Idev =Σ

‖Σ‖Idev =

Σ

‖Σ‖I Hence, Σλ

A must be also in the direction of n, i.e. ΣλA = ‖Σλ

A‖ n. Consequently

‖Σ‖ = ‖ΣλA‖ − Uλ (44)

I This situation differs from standard return map algorithm (i.e., linear plasticity) for

which direction n is determined by ΣTR , which is independent from λ.I In this more general setting, n is also function of λ; this makes the algorithm slightly

more complicated, but it is still possible to compute λ solving only one scalar equation.


A general plasticity framework: time-discrete and integration algorithm VI

The discrete plastic rate parameter λ is computed requiring satisfaction of discretelimit equation.

Assume that from linearization of discrete limit equation we get

dλ = Adisc[n : de]

where Adisc is a scalar quantity.


A general plasticity framework: time-discrete and integration algorithm VII

Linear elastic relation and discrete evolution equation for α are

s = 2G [e − epn] − 2Gλn

Rλα = αn + Hkinλn

Linearizing:

ds = 2Gde − 2Gdλ n − 2Gλ dn (45)

Rλdα + Hnldλ α = Hkindλ n + Hkinλ dn (46)

Recalling that n = Σ/‖Σ‖ = Σ/[

(Σ : Σ)12

]

we get:

dn =1

‖Σ‖ [I − (n⊗ n)] dΣ =N

‖Σ‖dΣ

with N orthogonal projection on plane normal to n, s.t. Nn = 0 , NN = N


A general plasticity framework: time-discrete and integration algorithm VIII

In matrix form in terms of unknowns ds , dα, dλ:

[

I + aN −aN 2Gn

−bN RλI + bN −Hkinn + Hnlα

]

dsdα

dλ

=

2Gde0

where:

a =2Gλ

‖Σ‖ , b =Hkinλ

‖Σ‖Using dλ = Adisc [n : de], previous system simplifies to

[

I + aN −aN

−bN RλI + bN

]

dsdα

=

2G [(1 + Adisc)I − AdiscN] deAdisc [Hkin(n ⊗ n) − Hnl (α ⊗ n)] de


A general plasticity framework: time-discrete and integration algorithm IX

Inverting coefficient matrix, we may solve for ds and dα:

dsdα

=

I + α1N −α3N

−α2N1

RλI + α2N

2G [(1 + Adisc)I − AdiscN] deAdisc [Hkin(n ⊗ n) − Hnl (α ⊗ n)] de

where:

α1 = − aRλ

b + Rλ + aRλ

α2 = − b

b + Rλ + aRλ

α3 = − a

b + Rλ + aRλ=

α1

Rλ


A general plasticity framework: time-discrete and integration algorithm X

Combining first row with incremental version of volumetric elastic relation

dσ = Ddiscdεεε

where algorithmic elasto-plastic tangent tensor is given by:

Ddisc = K (I ⊗ I) + 2G (1 − C) Idev

+ [2G (C − Adisc) + B(n : α)] (n ⊗ n) − B(α ⊗ n)(47)

with

B = AdiscTλCHnl , C =

2Gλ

‖ΣλA‖

while Adisc comes from linearized limit equation for the specific material modelI We note that in general the elasto-plastic tangent tensor Ddisc consistent with the

discrete model is non-symmetric, due to the presence of the non-linear kinematichardening term Hnl .


Non-linear kinematic hardening model I

For the NLK model continuous constitutive equations specialize to:

F = f = ‖Σ‖ − σy

α = Hkinγn − Hnl γα

I First equation states that limit and yield functions coincideI Second equation is non-linear kinematic hardening rule

Continuous consistency condition yields:

F = 2G [n : e] − 2G + Hiso + Hkin − Hnl [n : α] γ = 0

which can be solved in terms of γ, obtaining:

γ = ANLKcont[n : e] with A

NLKcont =

2G

2G1 − Hnl [n : α]

Tangent consistent with continuous NLK model

DNLKcont = K (I ⊗ I) + 2G

[

Idev − ANLKcont(n ⊗ n)

]


Non-linear kinematic hardening model II

Return map algorithm based on a scalar equation may be performed also for NLKI Discrete limit equation can be written as:

‖ΣλA‖ − Uλ − σy = 0

where:

‖ΣλA‖ =

[

(ΣλA : Σλ

A)] 1

2=

[

Sss − 2SsαTλ + Sαα(Tλ)2] 1

2

with:

Sss =(

sTR : sTR)

, Ssα =(

sTR : αTR)

, Sαα =(

αTR : αTR)

I Limit equation becomes a quartic equation in λ

g(λ) = C1λ4 + C2λ

3 + C3λ2 + C4λ + C5 = 0

where:

C1 = 4Hnl2G 2

0

C2 = 4Hnl2G0σy,n + 8G0G1Hnl

C3 = Hnl2[

(σy,n)2 − Sss

]

+ 4G 21 + 4Hnlσy,n [G0 + G1]

C4 = 2Hnl

[

(σy,n)2 + Ssα − Sss

]

+ 4σy,nG1

C5 = (σy,n)2 − Sαα + 2Ssα − Sss

with:2G0 = 2G + Hiso


Non-linear kinematic hardening model III

Solution search, i.e. search for minimum positive root, is not an easy task, due topolynomial order and to coefficient dependence on trial state

I Iterative algorithm of Newton type may be implementedF Adopted starting value of λ0 = 0F Unfortunately, this approach does not guarantee convergence to a positive rootF Different starting values of λ (such as λ0 = λn) have been explored but generate same

pathology

I Once Newton algorithm has failed, only robust approach is synthetic division of quarticpolynomial and compute in closed form roots of resulting cubic

I roots should then be corrected using a Newton algorithm, since synthetic division issensitive to roundoff


Non-linear kinematic hardening model IV

Discrete consistency condition can be formulated as

dF = 2G [n : de] −

2G + Hiso + TλHkin − T

λHnl [n : α]

dλ = 0

Solving for dλ, we obtain:dλ = A

NLKdisc [n : de]

where:

ANLKdisc =

2G

2G0 + TλHkin − TλHnl [n : α]

Elasto-plastic tangent tensor consistent with discrete NLK model is:

DNLKdisc = K (I ⊗ I) + 2G (1 − C) Idev

+[

2G (C − ANLKdisc ) + B(n : α)

]

(n ⊗ n) − B(α ⊗ n)

where:

B = ANLKdisc T

λCHnl

C =2Gλ

‖ΣλA‖


Non-linear kinematic hardening model V

NLK model represents an improvement with respect to LP modelI property of smoothly reaching an asymptotic stress valueI asymptote can only be horizontalI if unloaded from plastic range and reloaded before occurrence of reverse plasticity,

model renews plasticity exactly at the same stress where unloading began ⇒ behavioris identical to LP and in contrast with some experimental results

NLK model has been used to simulate behavior of metals under cyclic loading,achieving an high degree of accuracy.

Moreover, model extensions to include strain range memory, visco-plastic recoveryproperties, ratcheting effects presented in the literature

However, difficulties arise to implement the model in a return map framework, alldirectly related to non-linear kinematic hardening term.

Tangent tensor consistent with discrete model is non-symmetric


Generalized plasticity model I

Consider a simple generalized plasticity model (GP)

Limit equation and evolutionary equation for α are

F = h(f ) [n : σ] − γ

α = Hkinγn , Hnl = 0

where:

h(f ) =f

δ(β − f ) + Hβ, n =

∂f

∂σ

I β and δ: two positive constants with dimensions of stressI β: scalar measure of the distance between asymptotic and current radius of the yield

function σy

I δ: measures the speed of the model in approaching the asymptotic behaviorI H = Hiso + Hkin


GP: time-continuous model I

Requiring satisfaction of continuous limit function F = 0, we get:

F = h [n : s] − γ

= h [2G(n : e) − 2G γ] − γ = 0

Solving for γ, we obtain:

γ = AGPcont[n : e] , A

GPcont =

2G

2G +1

h(f )

I In the limiting case f → β, h(f ) → (1/H) ⇒ regain linear plasticity modelI Stress-strain curve continuous slope at transition between elastic and plastic behavior

Computed AGPcont, elasto-plastic tangent tensor consistent with continuous model

DGPcont = K (I ⊗ I) + 2G

[

Idev − AGPcont(n ⊗ n)

]


GP: time-discrete model I

We start by noting that:

n : σ = n : s = n : Σ + n : α = n : Σ + γHkin

= n :d

dt(‖Σ‖n) + γHkin =

d

dt‖Σ‖ + γHkin

since:

n : n = 1 ⇒ n : n = n : n = 0

As a result, limit equation can be rewritten as

F = h

[

d

dt‖Σ‖ + γHkin

]

− γ = 0

Integrated to yield discrete limit condition

λ − h [‖Σ‖ − ‖Σn‖ + λ Hkin] = 0


GP: time-discrete model II

If we set:

A1 = ‖ΣTR‖ − σy,n

A2 = ‖ΣTR‖ − ‖Σn‖A3 = δ − 2G

A4 = (δ + H) β

we obtain a quadratic equation:

aλ2 + bλ + c = 0 (48)

where:

a = 2G1A3

b = A4 − A1A3 + 2G1A2

c = −A1A2

Physically correct solution corresponds to smallest positive root


GP: time-discrete model III

For β = 0 the model reduces to linear plasticity, as expectedI In fact, β = 0 implies A4 = 0, while ‖Σn‖ ≤ σy,n implies A2 ≥ A1.I Accordingly, Equation 48 simplifies to:

(2G1λ − A1) (A3λ + A2) = 0

which has the roots: λ1 = −A2/A3 , λ2 = A1/(2G1).I First root never represents the correct one: in fact for δ > 2G it is negative; for

0 ≤ δ < 2G , we have λ1 > λ2.I Second root is always physically correct and coincides with the root of the linear

plasticity model.I If δ and the hardening are both zero, Equation 48 reduces to:

(2Gλ − A1) (−2Gλ + A2) = 0

which has roots: λ1 = A2/(2G) and λ2 = A1/(2G).I Here λ1 ≥ λ2 and again λ2 coincides with the linear plasticity root; hence, it is the

correct solution.


GP: time-discrete model IVUpon clearing fractions, the discrete limit equation becomes:

[δ(β − f ) + Hβ] λ = f [‖Σ‖ − ‖Σn‖ + Hkinλ]

Linearizing we get:

dλ = AGPdisc [n : de] with A

GPdisc =

2G(B1 + B2)

2G1B1 + (2G − δ)B2 + B3

and:

B1 = ‖Σ‖ − ‖Σn‖ + (Hkin + δ)λ

B2 = ‖Σ‖ − (σy,n + Hisoλ) = f

B3 = (δ + H) β

Elasto-plastic tangent tensor consistent with discrete GP model is:

DGPdisc = K (I ⊗ I) + 2G

(

1 − CGP

)

Idev +[

2G(

CGP − A

GPdisc

)]

(n ⊗ n)

where:

CGP =

2Gλ

‖ΣTR‖I D

GPdisc is symmetric

I DGPdisc returns linear plasticity tangent for β = 0 (which implies B2 = B3 = 0)


GP: remarks I

From constitutive behavior, main features of generalized plasticity are:I smooth transition between elastic response and asymptotic plastic limit due to plastic

strain non-linear evolution (h(f ) term of the limit function F )I with hardening non-horizontal asymptoteI elasto-plastic stress-strain curve continuous with first derivative at transition point

between elastic and plastic behaviorI speed in reaching asymptote controlled by δI for loading, unloading and reloading before occurrence of plasticity in reverse direction,

model renews plasticity before attainment of stress where unloading began (inaccordance with real material responses)

ε

σ

σy,0u

βu

A

B

C

D

ε

σ

σy,0u

β

β

u

u

ε

σ

σy,0u

β

β

2σy,0u

− σy,0u

u

u


GP: remarks II

From algorithmic point of view, GP model has a simple implementation:I discrete consistency condition generates a quadratic equation, which can be solved in

closed form and for which it is possible to delineate root propertiesI symmetry of continuous elasto-plastic tangent tensor is retained in discrete settingI only a few lines of extra code need to be added to convert an already implemented LP

routine (mainly a specific root for a quadratic equation must be computed).

GP seems to be a versatile and flexible model, involving parameters with clearphysical meaning, and at the same time it is a model with a simple andstraightforward algorithmic implementation.

Therefore, it seems that, within the simple versions of the models considered here,the GP model presents several advantages if compared to the NLK model.


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