Computational Solid Mechanicsagelet.rmee.upc.edu/master/Chapter 4. J2 Plasticity Algorithms...

Computational Solid Mechanics Computational Plasticity

C. Agelet de Saracibar ETS Ingenieros de Caminos, Canales y Puertos, Universidad Politécnica de Cataluña (UPC), Barcelona, Spain

International Center for Numerical Methods in Engineering (CIMNE), Barcelona, Spain

Chapter 4. J2 Plasticity Algorithms

Contents 1. Introduction 2. J2 Rate independent plasticity algorithms

1. Return mapping algorithm 2. Consistent elastoplastic tangent operator 3. Step by step algorithm 4. Nonlinear isotropic hardening

3. J2 Rate dependent plasticity algorithms 1. Return mapping algorithm 2. Consistent elastoplastic tangent operator 3. Step by step algorithm 4. Nonlinear isotropic hardening

4. J2 Computational plasticity assignment

J2 Plasticity Algorithms > Contents

Contents

April 20, 2015 Carlos Agelet de Saracibar 2






Contents


J2 Plasticity Algorithms > Introduction

Time integration algorithm


pnE 1

pn+E

1n+E

Time integration algorithm






Contents


1. Additive split of strains

2. Constitutive equations

3. Associative plastic flow rule

4. Yield function

5. Kuhn-Tucker loading/unloading conditions

J2 Plasticity Algorithms > J2 Rate Independent Plasticity Algorithms

J2 Rate independent plasticity model


( ) { } { }23, , , diag , ,e

e e q K Hψ= ∂ = = σ q 1E

S E CE S: , C :=

{ } { } { }: : ,0,0 , : , , , : , ,e p p p e eξ ξ= + = = = − −ε ε ξ ε ξE E E , E E E

( )p fγ= ∂

SE S

( ) ( ) ( )23

: , , : dev Yf f q qσ= = − − −σ q σ qS

( ) ( )0, 0, 0f fγ γ≥ ≤ =S S

Associative plastic flow rule: plastic strains at time n+1

Using a Backward-Euler (BE) time integration scheme yields,


Return mapping algorithm


( )p fγ= ∂

SE S

( )11 1 1n

p pn n n nfγ

++ + += + ∂SE E S

1 1 1

1 1

1 1 1

2 3

p pn n n n

n n n

n n n n

γ

ξ ξ γ

γ

+ + +

+ +

+ + +

= + = + = −

ε ε n

ξ ξ n

Constitutive equations: stress state at time n+1

The time-discrete constituve equation at time n+1 takes the form,

Substituting the plastic strains at time n+1 yields,




( ) ( ) { }23, diag , ,e

e e p K Hψ= ∂ = = − 1E

S E CE C E E C :=

( )1 1 1 1e p

n n n n+ + + += = −S CE C E E

( )( )( ) ( )

1

1

1 1 1 1

1 1 1

n

n

pn n n n n

pn n n n

f

f

γ

γ+

+

+ + + +

+ + +

= − − ∂

= − − ∂

S

S

S C E E S

C E E C S

Trial state at time n+1 The trial state at time n+1 is defined by freezing the plastic behaviour at the time step




( ) ( )( )

,1

, ,1 1 1

, ,1 1 1 1 1

1 1

:

:

:

:

p trial pn ne trial p trialn n n

trial e trial p trial pn n n n n n

trial trialn nf f

+

+ + +

+ + + + +

+ +

=

= −

= = − = −

=

E E

E E E

S CE C E E C E E

S

Return mapping algorithm The return mapping algorithm takes the form,




( )11 1 1 1n

trialn n n nfγ

++ + + += − ∂SS S C S

1 1 1 1 1 1 1

1 1 1

21 1 1 13

: 2

: 2 3

:

trial trialn n n n n n n

trialn n n

trialn n n n

q q K

H

γ γ µ

γ

γ

+ + + + + + +

+ + +

+ + + +

= − = − = − = +

σ σ n σ n

q q n

1. Additive split of strains at time n+1

2. Stresses at time n+1. Return mapping algorithm

3. Plastic internal variables at time n+1

4. Yield function at time n+1

5. Kuhn-Tucker loading/unloading conditions at time n+1




( ) ( )21 1 1 1 13: : devn n n n Y nf f qσ+ + + + += = − − −σ qS

1 1 1 10, 0, 0n n n nf fγ γ+ + + +≥ ≤ =

( )11 1 1 1n

trialn n n nfγ

++ + + += − ∂SS S C S

1 1 1: e pn n n+ + += +E E E

( )11 1 1n

p pn n n nfγ

++ + += + ∂SE E S

Theorem 1. Elastic step/plastic step If the yield function is convex and the constitutive matrix is definite-positive, the following condition holds,

and Kuhn-Tucker loading/unloading conditions can be decided just in terms of the trial state according to,




( ) ( )1 1trialn nf f+ +≥S S

( )( )

1 1

1 1

Elastic step

Plastic s

0 0

0 t0 ep

trialn n

trialn n

f

f

γ

γ

+ +

+ +

< ⇒ =

> ⇒ >

S

S

Theorem 2. Closest-point-projection The stress state at time n+1 is the closest-point-projection of the trial stress state at n+1 onto the space of admissible stresses, measured in the complementary energy norm, where the complementary energy is given by,




( )1 1arg min trialn n σ+ += Ξ − ∀ ∈S S S S E

( )( ) ( )

1

211 12

1 11 12

trial trialn n

trial trialn n

−+ +

−+ +

Ξ − = −

= − −C

S S S S

S S C S S





( )11 1 1 1n

trialn n n nfγ

++ + + += − ∂SS S C S

1 1 1 1 1 1 1

1 1 1

21 1 1 13

: 2

: 2 3

:


trialn n n

trialn n n n

q q K

H

γ γ µ

γ

γ

+ + + + + + +

+ + +

+ + + +

= − = − = − = +

σ σ n σ n

q q n

Return mapping algorithm Taking into account that the return mapping algorithm takes place on the deviatoric plane, it can be written as,




1 1 1 1 1 1 1

1 1 1

21 1 1 13

dev dev : dev 2

: 2 3

:


trialn n n

trialn n n n

q q K

H

γ γ µ

γ

γ

+ + + + + + +

+ + +

+ + + +

= − = − = − = +

σ σ n σ n

q q n

The solution for the return mapping algorithm yields,




( )21 1 1 1 1 13dev dev 2trial trial

n n n n n nHγ µ+ + + + + +− = − − +σ q σ q n

( )21 1 1 1 1 1 1 13dev dev 2trial trial trial

n n n n n n n nHγ µ+ + + + + + + +− = − − +σ q n σ q n n

( )( )21 1 1 1 1 1 13dev 2 dev trial trial trial

n n n n n n nHγ µ+ + + + + + +− + + = −σ q n σ q n

( )21 1 1 1 13

1 1

dev 2 dev trial trialn n n n n

trialn n

Hγ µ+ + + + +

+ +

− + + = −

=

σ q σ q

n n

For the non-trivial case (plastic loading), using the discrete Kuhn-Tucker loading/unloading conditions, the discrete plastic multiplier (or discrete plastic consistency parameter) reads,




( )1 1 1 1if 0 then , , 0n n n nf qγ + + + +> =σ q

( ) ( )

( ) ( )( ) ( )

21 1 1 1 1 13

2 2 21 1 1 13 3 3

2 21 1 1 1 3 3

, , dev

dev 2

, , 2 0

n n n n n Y n

trial trial trialn n n Y n

trial trial trialn n n n

f q q

K H q

f q K H

σ

γ µ σ

γ µ

+ + + + + +

+ + + +

+ + + +

= − − −

= − − + + − −

= − + + =

σ q σ q

σ q

σ q

( ) 12 21 13 32 trial

n nK H fγ µ−

+ += + +

( ) ( )1 1 1 1 1 1 1 10, , , 0, , , 0n n n n n n n nf q f qγ γ+ + + + + + + +≥ ≤ =σ q σ q

The return mapping algorithm takes the form,




( )( )( )

12 21 1 1 13 3

12 21 1 13 3

12 21 1 1 13 3

2 2

: 2 2 3

2: 23


trial trialn n n


K H f

q q K H f K

K H f H

µ µ

µ

µ

−

+ + + +

−

+ + +

−

+ + + +

= − + +

= − + + = + + +

σ σ n

q q n

1 1 1 1 1 1 1

1 1 1

21 1 1 13

: 2

: 2 3

:


trialn n n

trialn n n n

q q K

H

γ γ µ

γ

γ

+ + + + + + +

+ + +

+ + + +

= − = − = − = +

σ σ n σ n

q q n

Return mapping algorithm: Geometric interpretation




, 1nσ +

1dev n+σ

1n+q

( )1 1

23n nYR qσ+ += −

1n+n

1dev trialn+σ

dev nσ

,nσnq

( )23n nYR qσ= −

1trialn+n

The update of the plastic internal variables takes the form,




1 1 1

1 1

1 1 1

2 3

p pn n n n

n n n

n n n n

γ

ξ ξ γ

γ

+ + +

+ +

+ + +

= + = + = −

ε ε n

ξ ξ n

( )( )( )

12 21 1 13 3

12 21 13 3

12 21 1 13 3

2

2 2 3

2

p p trial trialn n n n

trialn n n

trial trialn n n n

K H f

K H f

K H f

µ

ξ ξ µ

µ

−

+ + +

−

+ +

−

+ + +

= + + + = + + + = − + +

ε ε n

ξ ξ n

Consistent elastoplastic tangent operator The consistent elastoplastic tangent operator is computed taking the variation of the stress tensor at time n+1, yielding, where the variations of the trial stress tensor, trial yield function, and trial unit normal to the yield surface at time n+1, have to be computed.


Consistent elastoplastic tangent operator


( )( )

( )

12 21 1 1 13 3

12 21 1 1 13 3

12 21 13 3

2 2

2 2

2 2



trial trialn n

K H f

d d K H df

K H f d

µ µ

µ µ

µ µ

−

+ + + +

−

+ + + +

−

+ +

= − + +

= − + +

− + +

σ σ n

σ σ n

n

The variation of the trial stress tensor at time n+1 takes the form,




( ) ( )( ) ( ) ( ) ( )( ) ( )( )

, , , ,1 1 1 1 1

1 1 1 1

, , , ,1 1 1 1 1

1

tr 2 tr 2 dev

tr 2 tr 2 dev

tr 2 tr 2 dev

tr

trial e trial e trial e trial e trialn n n n n

p pn n n n n n


n

d d d

d

λ µ κ µ

λ µ κ µ

λ µ κ µ

λ

+ + + + +

+ + + +

+ + + + +

+

= + = +

= + − = + −

= + = +

=

σ ε 1 ε ε 1 ε

ε 1 ε ε ε 1 ε ε

σ ε 1 ε ε 1 ε

ε 1 ( )( ) ( ) ( )( )

1 1 1

11 1 13

2 tr 2 dev

: 2 2 :

n n n

n n n

d d d

d d d

µ κ µ

λ µ κ µ

+ + +

+ + +

+ = +

= ⊗ + = ⊗ + − ⊗

ε ε 1 ε

1 1 ε ε 1 1 1 1 ε

The variation of the trial yield function at time n+1 takes the form,




( )( )

21 1 1 13

21 3

1 1 1 1 1 1

1 1

dev

dev

dev : dev : 2 dev

: 2

trial trial trial trialn n n Y n

trialn n Y n

trial trial trial trial trialn n n n n n n

trialn n

f q

q

df d d d

d

σ

σ

µ

µ

+ + + +

+

+ + + + + +

+ +

= − − −

= − − −

= − = =

=

σ q

σ q

σ q n σ n ε

n ε

The variation of the unit normal to the yield surface at time n+1 takes the form,




( )( )

1

1

1 1 11

1 1 1

11 1 1 1dev

11 1 1dev

dev devdev dev

dev dev

: dev

trialn n

trialn n

trial trial trialtrial n n n nn trial trial trial

n n n n

trial trial trial trialn n n n n

trial trial trn n n

d d d

d+

+

+ + ++

+ + +

+ + + +−

+ + +−

− −= =

− −

= − −

= − ⊗

σ q

σ q

σ q σ qnσ q σ q

n σ n σ q

n n σ

( )1

1

11 1 1dev

11 1 1dev

: 2 dev

1 : 23

trialn n

trialn n

ial

trial trialn n n

trial trialn n n

d

d

µ

µ

+

+

+ + +−

+ + +−

= − ⊗

= − ⊗ − ⊗

σ q

σ q

n n ε

1 1 n n ε

Substituting the variations shown before, the following discrete tangent constitutive equation can be obtained, where the consistent elastoplastic tangent operator at time n+1 is given by

and the following parameters have been introduced




1 1 1:epn n nd d+ + +=σ ε

1 1 1 1 112 23

ep trial trialn n n n nκ µδ µδ+ + + + +

= ⊗ + − ⊗ − ⊗

1 1 I 1 1 n n

( )11 1 12 2

1 3 3

2 2: 1 , : 12dev

nn n ntrial

n n K Hµ γ µδ δ δ

µ+

+ + ++

= − = − −+ +−σ q

J2 Plasticity algorithm Step 1. Given the strain tensor at time n+1 (strain driven problem), and the stored plastic internal variables at time n (plastic internal variables database) Step 2. Compute the trial state at time n+1


J2 Plasticity algorithm


( ) ( )( )

,1

, ,1 1 1

, ,1 1 1 1 1

21 1 1 13

:

:

:

: dev



trial trial trial trialn n n Y nf qσ

+

+ + +

+ + + + +

+ + + +

=

= −

= = − = −

= − − −σ q

E E

E E E

S CE C E E C E E

Step 3. Check the trial yield function at time n+1 Step 4. Compute the discrete plastic multiplier at time n+1




( ) ( )1 11 1if 0 then set , and exittrialtrial ep

n nn nf + ++ +

≤ • = • =

( ) 12 21 13 32 trial

n nK H fγ µ−

+ += + +

Step 5. Return mapping algorithm (closest-point-projection)




( )1

1 1 1 1trialn

trial trialn n n nfγ

++ + + += − ∂

SS S C S

( )( )( )

12 21 1 1 13 3

12 21 1 13 3

12 21 1 1 13 3

2 2

: 2 2 3

2: 23


trial trialn n n


K H f

q q K H f K

K H f H

µ µ

µ

µ

−

+ + + +

−

+ + +

−

+ + + +

= − + +

= − + + = + + +

σ σ n

q q n

Step 6. Update plastic internal variables database at time n+1




( )1

1 1 1trialn

p p trialn n n nfγ

++ + += + ∂

SE E S

( )( )( )

12 21 1 13 3

12 21 13 3

12 21 1 13 3

2

2 2 3

2


trialn n n

trial trialn n n n

K H f

K H f

K H f

µ

ξ ξ µ

µ

−

+ + +

−

+ +

−

+ + +

= + + + = + + + = − + +

ε ε n

ξ ξ n

Step 7. Compute the consistent elastoplastic tangent operator




1 1 1 1 112 23


= ⊗ + − ⊗ − ⊗

1 1 I 1 1 n n

( )11 1 12 2

1 3 3

2 2: 1 , : 12dev

nn n ntrial

n n K Hµ γ µδ δ δ

µ+

+ + ++

= − = − −+ +−σ q

Nonlinear isotropic hardening Exponential saturation law + linear isotropic hardening


Nonlinear isotropic hardening


( ) ( ):q ξ ξψ ξ ξ′= −∂ = −∂ Π = −Π

( ) ( )( ): : 1 expYq Kξψ σ σ δξ ξ∞= −∂ = − − − − −

( ) ( ) ( )( )1 expY Kξ σ σ δξ ξ∞′Π = − − − +

Time discrete nonlinear isotropic hardening




( ) ( )( ) ( )

( ) ( )

1 1 1

1 1

1 1 1

: 2 3

:

: 2 3

n n n n

trial trialn n n n

trialn n n n n

q

q q

q q

ξ ξ γ

ξ ξ

ξ γ ξ

+ + +

+ +

+ + +

′ ′= −Π = −Π +

′ ′= −Π = −Π =

′ ′= −Π + +Π

Plastic loading: Yield function at time n+1 Nonlinear residual scalar equation on the plastic multiplier at time n+1




( ) ( ) ( )( )2 2 21 1 1 13 3 3

2 0trialn n n n n nf f Hγ µ ξ γ ξ+ + + +′ ′= − + − Π + −Π =

( )( ) ( ) ( )( )

1 1 1

2 2 21 1 1 13 3 3

: 0

: 2 0

n n n

trialn n n n n n

g g f

g f H

γ

γ µ ξ γ ξ

+ + +

+ + + +

= = =

′ ′= − + − Π + −Π =

Newton-Raphson iterative solution algorithm Step 1. Initialize iteration counter and plastic multiplier Step 2. Compute the residual g at time n+1, iteration k Step 3. While the absolute value of the current residual at time n+1, iteration k, is greater than a tolerance Step 4. Solve the linarized equation Step 5. Update the plastic multiplier at time n+1, iteration k+1 Step 6. Compute the residual g at time n+1, iteration k+1 Step 7. Increment iteration counter k=k+1 and go to Step 3




10, 0knk γ += =

1 1 1 0k k kn n ng Dg γ+ + ++ ∆ =

11 1 1

k k kn n nγ γ γ++ + += + ∆

Newton-Raphson iterative solution algorithm




( )( ) ( )( )

( ) ( )( )( )

21 1 1 3

2 213 3

2 2 21 1 1 1 13 3 3

2 221 13 33

11 1 1

: 2

: 2

: 2

:

k trial kn n n

kn n n

k k k k kn n n n n n

k kn n n

k k kn n n

g f H

Dg H

H

γ µ

ξ γ ξ

γ µ γ ξ γ γ

µ ξ γ γ

γ γ γ

+ + +

+

+ + + + +

+ +

++ + +

= − +

′ ′− Π + −Π

′′∆ = − + ∆ − Π + ∆

′′= − + Π + + ∆

∆ = −

1 1 1 0k k kn n ng Dg γ+ + ++ ∆ =

Consistent elastoplastic tangent operator The consistent elastoplastic tangent operator is computed taking the variation of the return mapping equation, yielding, where the variations of the trial stress tensor, plastic multiplier, and trial unit normal to the yield surface at time n+1, have to be computed.




1 1 1 1

1 1 1 1 1 1

2

2 2

trial trialn n n n

trial trial trialn n n n n nd d d d

γ µ

γ µ γ µ+ + + +

+ + + + + +

= −

= − −

σ σ nσ σ n n

The variation of the plastic multiplier at time n+1 is computed setting the variation of the residual equal to zero,




( ) ( ) ( )( )( ) ( )

( )( )( )( )

2 2 21 1 1 13 3 3

2 2 21 1 1 1 13 3 3

2 221 1 13 33

12 22

1 1 13 33

: 2

: 2

: 2 0

2

trialn n n n n n

trialn n n n n n

trialn n n n

trialn n n n

g f H

dg df d H d

df d H

d H df

γ µ ξ γ ξ

γ µ ξ γ γ

γ µ ξ γ

γ µ ξ γ

+ + + +

+ + + + +

+ + +

−

+ + +

′ ′= − + − Π + −Π

′′= − + − Π +

′′= − + Π + + =

′′= + Π + +






1 1 1:epn n nd d+ + +=σ ε

1 1 1 1 112 23


= ⊗ + − ⊗ − ⊗

1 1 I 1 1 n n

( ) ( )11 1 12 22

1 13 33

2 2: 1 , : 1dev 2

nn n ntrial

n n n n Hµ γ µδ δ δ

µ ξ γ+

+ + ++ +

= − = − −− ′′+ Π + +σ q


1. Return mapping algorithm 2. Consistent elastoplastic tangent modulus 3. Step by step algorithm 4. Nonlinear isotropic hardening

3. J2 Rate dependent plasticity algorithms 1. Return mapping algorithm 2. Consistent elastoplastic tangent modulus 3. Step by step algorithm 4. Nonlinear isotropic hardening



Contents


1. Additive split of strains

2. Constitutive equations

3. Associative plastic flow rule

4. Yield function

5. Plastic multiplier

J2 Plasticity Algorithms > J2 Rate Dependent Plasticity Algorithms

J2 Rate dependent plasticity model


( ) { } { }23, , , diag , ,e

e e q K Hψ= ∂ = = σ q 1E

S E CE S: , C :=

{ } { } { }: : ,0,0 , : , , , : , ,e p p p e eξ ξ= + = = = − −ε ε ξ ε ξE E E , E E E

( )p fγ= ∂

SE S

( ) ( ) ( )23

: , , : dev Yf f q qσ= = − − −σ q σ qS

( )1 0fηγ = ≥S

Associative plastic flow rule: plastic strains at time n+1

Using a Backward-Euler (BE) time integration scheme yields,




( )p fγ= ∂

SE S

( )11 1 1n

p pn n n nt fγ

++ + += + ∆ ∂SE E S

Constitutive equations: stress state at time n+1

The time-discrete constituve equation at time n+1 takes the form,

Substituting the plastic strains at time n+1 yields,




( ) ( ) { }23, diag , ,e

e e p K Hψ= ∂ = = − 1E

S E CE C E E C :=

( )1 1 1 1e p

n n n n+ + + += = −S CE C E E

( )( )( ) ( )

1

1

1 1 1 1

1 1 1

n

n

pn n n n n

pn n n n

t f

t f

γ

γ+

+

+ + + +

+ + +

= − − ∆ ∂

= − − ∆ ∂

S

S

S C E E S

C E E C S

Trial state at time n+1 The trial state at time n+1 is defined by freezing the plastic behaviour




( ) ( )( )

,1

, ,1 1 1

, ,1 1 1 1 1

1 1

:

:

:

:



trial trialn nf f

+

+ + +

+ + + + +

+ +

=

= −

= = − = −

=

E E

E E E

S CE C E E C E E

S





( )11 1 1 1n

trialn n n nt fγ

++ + + += − ∆ ∂SS S C S

1 1 1 1 1 1 1

1 1 1

21 1 1 13

: 2

: 2 3

:


trialn n n

trialn n n n

t t

q q t K

t H

γ γ µ

γ

γ

+ + + + + + +

+ + +

+ + + +

= − ∆ = − ∆ = − ∆ = + ∆

σ σ n σ n

q q n

1. Additive split of strains at time n+1

2. Stresses at time n+1. Return mapping algorithm

3. Plastic internal variables at time n+1

4. Yield function at time n+1

5. Plastic multiplier at time n+1




( ) ( )21 1 1 1 13: : devn n n n Y nf f qσ+ + + + += = − − −σ qS

( )11 1 1 1n

trialn n n nt fγ

++ + + += − ∆ ∂SS S C S

1 1 1: e pn n n+ + += +E E E

( )11 1 1n

p pn n n nt fγ

++ + += + ∆ ∂SE E S

( )11 1 0n nfηγ + += ≥S





( )11 1 1 1n

trialn n n nt fγ

++ + + += − ∆ ∂SS S C S

1 1 1 1 1 1 1

1 1 1

21 1 1 13

: 2

: 2 3

:


trialn n n

trialn n n n

t t

q q t K

t H

γ γ µ

γ

γ

+ + + + + + +

+ + +

+ + + +

= − ∆ = − ∆ = − ∆ = + ∆

σ σ n σ n

q q n

Return mapping algorithm Taking into account that the return mapping algorithm takes place on the deviatoric plane, it can be written as,




1 1 1 1 1 1 1

1 1 1

21 1 1 13

dev dev : dev 2

: 2 3

:


trialn n n

trialn n n n

t t

q q t K

t H

γ γ µ

γ

γ

+ + + + + + +

+ + +

+ + + +

= − ∆ = − ∆ = − ∆ = + ∆

σ σ n σ n

q q n

The solution for the return mapping algorithm yields,




( )21 1 1 1 1 13dev dev 2trial trial

n n n n n nt Hγ µ+ + + + + +− = − − ∆ +σ q σ q n

( )21 1 1 1 1 1 1 13dev dev 2trial trial trial

n n n n n n n nt Hγ µ+ + + + + + + +− = − − ∆ +σ q n σ q n n

( )( )21 1 1 1 1 1 13dev 2 dev trial trial trial

n n n n n n nt Hγ µ+ + + + + + +− + ∆ + = −σ q n σ q n

( )21 1 1 1 13

1 1

dev 2 dev trial trialn n n n n

trialn n

t Hγ µ+ + + + +

+ +

− + ∆ + = −

=

σ q σ q

n n

For the non-trivial case (plastic loading), the discrete plastic multiplier reads,




( )1 1 1 1 1if 0 then , ,n n n n nf qγ γ η+ + + + +> =σ q

( ) ( )

( ) ( )( ) ( )

21 1 1 1 1 13

2 2 21 1 1 13 3 3

2 21 1 1 1 13 3

, , dev

dev 2

, , 2

n n n n n Y n

trial trial trialn n n Y n

trial trial trialn n n n n

f q q

t K H q

f q t K H

σ

γ µ σ

γ µ γ η

+ + + + + +

+ + + +

+ + + + +

= − − −

= − − ∆ + + − −

= − ∆ + + =

σ q σ q

σ q

σ q

12 2

1 13 32 trialn nt K H f

tηγ µ

−

+ + ∆ = + + + ∆

The return mapping algorithm takes the form,




12 2

1 1 1 13 3

12 2

1 1 13 3

12 2

1 1 1 13 3

2 2

: 2 2 3

2: 23


trial trialn n n


K H ft

q q K H f Kt

K H f Ht

ηµ µ

ηµ

ηµ

−

+ + + +

−

+ + +

−

+ + + +

= − + + + ∆ = − + + + ∆ = + + + + ∆

σ σ n

q q n

1 1 1 1 1 1 1

1 1 1

21 1 1 13

: 2

: 2 3

:


trialn n n

trialn n n n

t t

q q t K

t H

γ γ µ

γ

γ

+ + + + + + +

+ + +

+ + + +

= − ∆ = − ∆ = − ∆ = + ∆

σ σ n σ n

q q n

The update of the plastic internal variables takes the form,




1 1 1

1 1

1 1 1

2 3

p pn n n n

n n n

n n n n

t

t

t

γ

ξ ξ γ

γ

+ + +

+ +

+ + +

= + ∆ = + ∆ = − ∆

ε ε n

ξ ξ n1

2 21 1 13 3

12 2

1 13 3

12 2

1 1 13 3

2

2 2 3

2


trialn n n

trial trialn n n n

K H ft

K H ft

K H ft

ηµ

ηξ ξ µ

ηµ

−

+ + +

−

+ +

−

+ + +

= + + + + ∆ = + + + + ∆ = − + + + ∆

ε ε n

ξ ξ n

Consistent elastoplastic tangent operator The consistent elastoplastic tangent operator is computed taking the variation of the stress tensor at time n+1, yielding,

where the variations of the trial stress tensor, trial yield function, and trial unit normal to the yield surface at time n+1, have to be computed.




12 2

1 1 1 13 3

12 2

1 1 1 13 3

12 2

1 13 3

2 2

2 2

2 2



trial trialn n

K H ft

d d K H dft

K H f dt

ηµ µ

ηµ µ

ηµ µ

−

+ + + +

−

+ + + +

−

+ +

= − + + + ∆

= − + + + ∆

− + + + ∆

σ σ n

σ σ n

n

The variation of the trial stress tensor at time n+1 takes the form,




( ) ( )( ) ( ) ( ) ( )( ) ( )( )

, , , ,1 1 1 1 1

1 1 1 1

, , , ,1 1 1 1 1

1

tr 2 tr 2 dev

tr 2 tr 2 dev

tr 2 tr 2 dev

tr


p pn n n n n n


n

d d d

d

λ µ κ µ

λ µ κ µ

λ µ κ µ

λ

+ + + + +

+ + + +

+ + + + +

+

= + = +

= + − = + −

= + = +

=

σ ε 1 ε ε 1 ε

ε 1 ε ε ε 1 ε ε

σ ε 1 ε ε 1 ε

ε 1 ( )( ) ( ) ( )( )

1 1 1

11 1 13

2 tr 2 dev

: 2 2 :

n n n

n n n

d d d

d d d

µ κ µ

λ µ κ µ

+ + +

+ + +

+ = +

= ⊗ + = ⊗ + − ⊗

ε ε 1 ε

1 1 ε ε 1 1 1 1 ε

The variation of the trial yield function at time n+1 takes the form,




( )( )

21 1 1 13

21 3

1 1 1 1 1 1

1 1

dev

dev

dev : dev : 2 dev

: 2

trial trial trial trialn n n Y n

trialn n Y n

trial trial trial trial trialn n n n n n n

trialn n

f q

q

df d d d

d

σ

σ

µ

µ

+ + + +

+

+ + + + + +

+ +

= − − −

= − − −

= − = =

=

σ q

σ q

σ q n σ n ε

n ε

The variation of the unit normal to the yield surface at time n+1 takes the form,




( )( )

1

1

1 1 11

1 1 1

11 1 1 1dev

11 1 1dev

dev devdev dev

dev dev

: dev

trialn n

trialn n

trial trial trialtrial n n n nn trial trial trial

n n n n

trial trial trial trialn n n n n

trial trial trn n n

d d d

d+

+

+ + ++

+ + +

+ + + +−

+ + +−

− −= =

− −

= − −

= − ⊗

σ q

σ q

σ q σ qnσ q σ q

n σ n σ q

n n σ

( )1

1

11 1 1dev

11 1 1dev

: 2 dev

1 : 23

trialn n

trialn n

ial

trial trialn n n

trial trialn n n

d

d

µ

µ

+

+

+ + +−

+ + +−

= − ⊗

= − ⊗ − ⊗

σ q

σ q

n n ε

1 1 n n ε






1 1 1:epn n nd d+ + +=σ ε

1 1 1 1 112 23


= ⊗ + − ⊗ − ⊗

1 1 I 1 1 n n

( )11 1 1

2 213 3

2 2: 1 , : 1dev 2

nn n ntrial

n n

t

K Ht

µ γ µδ δ δηµ

++ + +

+

∆= − = − −

− + + +∆

σ q

J2 Plasticity algorithm Step 1. Given the strain tensor at time n+1 (strain driven problem), and the stored plastic internal variables at time n (plastic internal variables database) Step 2. Compute the trial state at time n+1




( ) ( )( )

,1

, ,1 1 1

, ,1 1 1 1 1

21 1 1 13

:

:

:

: dev



trial trial trial trialn n n Y nf qσ

+

+ + +

+ + + + +

+ + + +

=

= −

= = − = −

= − − −σ q

E E

E E E

S CE C E E C E E

Step 3. Check the trial yield function at time n+1 Step 4. Compute the discrete plastic multiplier at time n+1




( ) ( )1 11 1if 0 then set , and exittrialtrial ep

n nn nf + ++ +

≤ • = • =

12 2

1 13 32 trialn nt K H f

tηγ µ

−

+ + ∆ = + + + ∆

Step 5. Return mapping algorithm (closest-point-projection)




( )1

1 1 1 1trialn

trial trialn n n nt fγ

++ + + += − ∆ ∂

SS S C S

12 2

1 1 1 13 3

12 2

1 1 13 3

12 2

1 1 1 13 3

2 2

: 2 2 3

2: 23


trial trialn n n


K H ft

q q K H f Kt

K H f Ht

ηµ µ

ηµ

ηµ

−

+ + + +

−

+ + +

−

+ + + +

= − + + + ∆ = − + + + ∆ = + + + + ∆

σ σ n

q q n

Step 6. Update plastic internal variables database at time n+1




( )1

1 1 1trialn

p p trialn n n nt fγ

++ + += + ∆ ∂

SE E S

12 2

1 1 13 3

12 2

1 13 3

12 2

1 1 13 3

2

2 2 3

2


trialn n n

trial trialn n n n

K H ft

K H ft

K H ft

ηµ

ηξ ξ µ

ηµ

−

+ + +

−

+ +

−

+ + +

= + + + + ∆ = + + + + ∆ = − + + + ∆

ε ε n

ξ ξ n

Step 7. Compute the consistent elastoplastic tangent operator




1 1 1 1 112 23


= ⊗ + − ⊗ − ⊗

1 1 I 1 1 n n

( )11 1 1

2 213 3

2 2: 1 , : 1dev 2

nn n ntrial

n n

t

K Ht

µ γ µδ δ δηµ

++ + +

+

∆= − = − −

− + + +∆

σ q

Nonlinear isotropic hardening Exponential saturation law + linear isotropic hardening




( ) ( ):q ξ ξψ ξ ξ′= −∂ = −∂ Π = −Π

( ) ( )( ): : 1 expYq Kξψ σ σ δξ ξ∞= −∂ = − − − − −

( ) ( ) ( )( )1 expY Kξ σ σ δξ ξ∞′Π = − − − +

Time discrete nonlinear isotropic hardening




( ) ( )( ) ( )

( ) ( )

1 1 1

1 1

1 1 1

: 2 3

:

: 2 3

n n n n

trial trialn n n n

trialn n n n n

q t

q q

q q t

ξ ξ γ

ξ ξ

ξ γ ξ

+ + +

+ +

+ + +

′ ′= −Π = −Π + ∆

′ ′= −Π = −Π =

′ ′= −Π + ∆ +Π

Plastic loading: Yield function at time n+1

Nonlinear residual scalar equation on the plastic multiplier at time n+1




( ) ( ) ( )( )2 2 21 1 1 13 3 3

1

2

0

trialn n n n n n

n

f f t H tγ µ ξ γ ξ

γ η

+ + + +

+

′ ′= − ∆ + − Π + ∆ −Π

= >

( )

( ) ( )( )1 1 1 1

2 2 21 1 1 13 3 3

: 0

: 2 0

n n n n

trialn n n n n n

g g f

g f t H tt

γ γ η

ηγ µ ξ γ ξ

+ + + +

+ + + +

= = − =

′ ′= − ∆ + + − Π + ∆ −Π = ∆

Newton-Raphson iterative solution algorithm Step 1. Initialize iteration counter and plastic multiplier Step 2. Compute the residual g at time n+1, iteration k Step 3. While the absolute value of the current residual at time n+1, iteration k, is greater than a tolerance Step 4. Solve the linarized equation Step 5. Update the plastic multiplier at time n+1, iteration k+1 Step 6. Compute the residual g at time n+1, iteration k+1 Step 7. Increment iteration counter k=k+1 and go to Step 3




10, 0knk γ += =

1 1 1 0k k kn n ng Dg γ+ + ++ ∆ =

11 1 1

k k kn n nγ γ γ++ + += + ∆

Newton-Raphson iterative solution algorithm




( ) ( )( )( )

( )

21 1 1 3

2 213 3

2 2 21 1 1 1 13 3 3

2 221 13 33

: 2

: 2

: 2

k trial kn n n

kn n n

k k k k kn n n n n n

k kn n n

g f t Ht

t

Dg H t t tt

t H tt

ηγ µ

ξ γ ξ

ηγ µ γ ξ γ γ

ηµ ξ γ γ

+ + +

+

+ + + + +

+ +

= − ∆ + + ∆

′ ′− Π + ∆ −Π

′′∆ = − + + ∆ ∆ − Π + ∆ ∆ ∆ ∆ ′′= − + Π + ∆ + + ∆ ∆ ∆

1 1 1 0k k kn n ng Dg γ+ + ++ ∆ =

Consistent elastoplastic tangent operator The consistent elastoplastic tangent operator is computed taking the variation of the return mapping equation, yielding, where the variations of the trial stress tensor, plastic multiplier, and trial unit normal to the yield surface at time n+1, have to be computed.




1 1 1 1

1 1 1 1 1 1

2

2 2

trial trialn n n n

trial trial trialn n n n n n

td d d t t d

γ µ

γ µ γ µ+ + + +

+ + + + + +

= − ∆

= − ∆ − ∆

σ σ nσ σ n n

The variation of the plastic multiplier at time n+1 is computed setting the variation of the residual at time n+1 equal to zero,




( ) ( )( )( )

( )

2 2 21 1 1 13 3 3

2 2 21 1 1 1 13 3 3

2 221 1 13 33

21 3

: 2 0

: 2

: 2 0

2

trialn n n n n n

trialn n n n n n

trialn n n n

n n

g f t H tt

dg df d t H t d tt

df d t t Ht

d t

ηγ µ ξ γ ξ

ηγ µ ξ γ γ

ηγ µ ξ γ

γ µ ξ

+ + + +

+ + + + +

+ + +

+

′ ′= − ∆ + + − Π + ∆ −Π = ∆ ′′= − ∆ + + − Π + ∆ ∆ ∆ ′′= − ∆ + Π + ∆ + + = ∆

′′∆ = + Π +( )1

221 133

trialn nt H df

tηγ

−

+ + ∆ + + ∆






1 1 1:epn n nd d+ + +=σ ε

1 1 1 1 112 23


= ⊗ + − ⊗ − ⊗

1 1 I 1 1 n n

( )( )1

1 1 12 221

13 33

2 2: 1 , : 1dev 2

nn n ntrial

n nn n

t

t Ht

µ γ µδ δ δηµ ξ γ

++ + +

++

∆= − = − −

− ′′+ Π + ∆ + +∆

σ q


1. Return mapping algorithm 2. Consistent elastoplastic tangent modulus 3. Step by step algorithm 4. Nonlinear isotropic hardening

3. J2 Rate dependent plasticity algorithms 1. Return mapping algorithm 2. Consistent elastoplastic tangent modulus 3. Step by step algorithm 4. Nonlinear isotropic hardening



Contents


Implement in MATLAB the BE time-stepping algorithm for J2 rate-independent/rate-dependent hardening plasticity models, including linear and nonlinear isotropic hardening, and linear kinematic hardening

Perform the numerical simulation of uniaxial cyclic plastic loading/elastic unloading examples for the following cases: o Rate-independent/rate-dependent perfect plasticity o Rate-independent/rate-dependent linear isotropic hardening plasticity o Rate-independent/rate-dependent nonlinear isotropic hardening

plasticity, considering an exponential saturation law o Rate-independent/rate-dependent linear kinematic hardening

plasticity o Rate-independent/rate-dependent nonlinear isotropic and linear

kinematic hardening plasticity

J2 Plasticity Algorithms > J2 Computational Plasticity Assignment

J2 Computational plasticity assignment


For the perfect plasticity models, plot the stress11-strain11 and the dev[stress11]-strain11 curves.

For the linear isotropic/linear kinematic hardening models, plot the stress11-strain11 and dev[stress11]-strain11 curves showing the influence of the isotropic/kinematic hardening parameters

For the nonlinear isotropic hardening model, plot the stress11-strain11 and dev[stress11]-strain11 curves showing the influence of the exponential coefficient of the exponential saturation law




For the rate-dependent plasticity models, plot the stress11-strain11, dev[stress11]-strain11, and the stress11-time and dev[stress11]-time curves showing the influence of the viscosity parameter and the loading rate.

Show that the rate-independent response can be recovered from the rate-dependent results using very small values for the viscosity or the loading rate




Write a comprehensive deliverable report (10 pages) providing the data of the cyclic loading and material properties considered, the stress-strain curves, and the stress-time curves for the rate-dependent plasticity examples. Add suitable comments on the results, comparing the influence of the different material parameters and loading conditions.

Add a printed copy of the subroutines as an Appendix




Computational Solid Mechanicsagelet.rmee.upc.edu/master/Chapter 4. J2 Plasticity Algorithms...

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