Computational Solid Mechanics

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 2. 1D Plasticity Algorithms

Contents 1. Introduction 2. 1D Rate independent plasticity algorithms

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

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

4. 1D Computational plasticity assignment

1D Plasticity Algorithms > Contents

Contents

April 14, 2015 Carlos Agelet de Saracibar 2






Contents


1D 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

1D Plasticity Algorithms > 1D Rate Independent Plasticity Algorithms

1D Rate independent plasticity model


( ) { } { }, , , diag , ,ee e q q E K Hψ σ= ∂ = =

ES E CE S: , C :=

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

( )p fγ= ∂

SE S

( ) Yf q qσ σ= − − +S

( ) ( )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 1 1

sgn

sgn

p pn n n n n

n n n

n n n n n

q

q

ε ε γ σ

ξ ξ γ

ξ ξ γ σ

+ + + +

+ +

+ + + +

= + − = + = − −

Constitutive equations: stress state at time n+1

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

Substituting the plastic strain variables at time n+1 yields,




( ) ( ) { }, diag , ,ee e p E K Hψ= ∂ = = −

ES 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, yielding




( ) ( )( )

,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 1 1 1

sgn

:

: sgn

trialn n n n n

trialn n n

trialn n n n n

E q

q q Kq q H q

σ σ γ σ

γ

γ σ

+ + + + +

+ + +

+ + + + +

= − − = − = + −

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




( )1 1 1 1 1:n n n n Y nf f q qσ σ+ + + + += = − − +S

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 yield function 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

Proof 1. Convexity of the yield function yields,

Using the return mapping equation, yields,

Definite-positiveness of the constitutive matrix yields,




( ) ( ) ( ) ( )11 1 1 1 1n

trial trialn n n n nf f f

++ + + + +− ≥ − ∂SS S S S S

( ) ( ) ( ) ( )1 11 1 1 1 1n n

trialn n n n nf f f fγ

+ ++ + + + +− ≥ ∂ ∂S SS S S C S

( )11 1 1 1n

trialn n n nfγ

++ + + += − ∂SS S C S

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

1 11 1 1 1 1

1 1

0n n

trialn n n n n

trialn n

f f f f

f f

γ+ ++ + + + +

+ +

− ≥ ∂ ∂ ≥

≥

S SS S S C S

S S

Proof 2. The trial yield function at time n+1 determines the discrete plastic loading/elastic unloading conditions according to,




( )( ) ( )

( )( )

( )( ) ( )

1

1

1 1

1 1 1

1

1 1 1 1

1 1 1 1

if 0 then

0

0 Elastic step

Plastic

0

else if 0 then

0, 0 0 ste

en if

p

d

n

trialn

trialn n

n n n

trialn

trialn n n n

n n n n

f

f f

f

f

f

f f

γ γ

γ

γ γ+

+

+ +

+ + +

+

+ + + +

+ + + +

<

≤ <

= ⇒ =

≥

= − ∂

> = ⇒ =

S

S

S S

S

S

S S C S

S S

Theorem 2. Closest-point-projection The stress state at time n+1 is the closest-point-projection (cpp) 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

Proof 1. The stress state at time n+1, the closest-point-projection of the trial stress state at n+1 onto the space of admissible stresses, is the solution of the following constrained minimization problem,

Using the Lagrange multipliers method, the constrained minimization problem can be transformed into an unconstrained minimization problem defined as,




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

( ) ( ) ( )1 1, :trial trialn n fγ γ+ += Ξ − +S S; S S SL

( )1 1 1arg min ,trialn n nγ+ + += ∀S S S; SL

The optimality conditions of the unconstrained minimization problem read, The return mapping algorithm arises as the solution of the unconstrained minimization problem yielding,

1D Plasticity Algorithms > 1D Rate Independent Plasticity Algorthms



( ) ( ) ( )

( ) ( )1 1 1

1

1 1 1 1 1 1 1

11 1 1 1

, :

: 0

;n n n

n

trial trialn n n n n n n

trialn n n n

f

f

γ γ

γ+ + +

+

+ + + + + + +

−+ + + +

∂ = ∂ Ξ − + ∂

= − − + ∂ =

S S S

S

S S S S S

C S S S

L

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

( )11 1 1 1n

trialn n n nfγ

++ + + += − ∂SS S C S





( )11 1 1 1n

trialn n n nfγ

++ + + += − ∂SS S C S

( )

( )

1 1 1 1 1

1 1 1

1 1 1 1 1

sgn

:

: sgn

trialn n n n n

trialn n n

trialn n n n n

E q

q q Kq q H q

σ σ γ σ

γ

γ σ

+ + + + +

+ + +

+ + + + +

= − − = − = + −

The solution for the return mapping algorithm yields,




( ) ( )1 1 1 1 1 1 1sgntrial trialn n n n n n nq q E H qσ σ γ σ+ + + + + + +− = − − + −

( ) ( )( ) ( )

1 1 1 1 1 1 1 1

1 1 1

sgn sgn

sgn

trial trial trial trialn n n n n n n n

n n n

q q q q

E H q

σ σ σ σ

γ σ+ + + + + + + +

+ + +

− − = − −

− + −

( )( ) ( )( )

1 1 1 1 1

1 1 1 1

sgn

sgnn n n n n

trial trial trial trialn n n n

q E H q

q q

σ γ σ

σ σ

+ + + + +

+ + + +

− + + − =

= − −

( )( ) ( )

1 1 1 1 1

1 1 1 1sgn sgn

trial trialn n n n n

trial trialn n n n

q E H q

q q

σ γ σ

σ σ

+ + + + +

+ + + +

− + + = −

− = −

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 q qγ σ+ + + +> =

( )( )

( ) ( )

1 1 1 1 1 1

1 1 1 1

1 1 1 1

, ,

, , 0

n n n n n Y n

trial trial trialn n n Y n

trial trial trialn n n n

f q q q q

q E K H q

f q q E K H

σ σ σ

σ γ σ

σ γ

+ + + + + +

+ + + +

+ + + +

= − − +

= − − + + − +

= − + + =

( ) ( )11 1 1 1, , 0trial trial trial

n n n nE K H f q qγ σ−+ + + += + + >

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

The return mapping algorithm takes the form,




( )

( )

1 1 1 1 1

1 1 1

1 1 1 1 1

sgn

:

: sgn

trialn n n n n

trialn n n

trialn n n n n

E q

q q Kq q H q

σ σ γ σ

γ

γ σ

+ + + + +

+ + +

+ + + + +

= − − = − = + −

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

11 1 1 1 1 1 1

11 1 1 1 1

11 1 1 1 1 1 1

, , sgn

: , ,

: , , sgn

trial trial trial trial trial trialn n n n n n n

trial trial trial trialn n n n n

trial trial trial trial trialn n n n n n n

E K H f q q E q

q q E K H f q q K

q q E K H f q q H q

σ σ σ σ

σ

σ σ

−+ + + + + + +

−+ + + + +

−+ + + + + + +

= − + + −

= − + +

= + + + −( )trial

The update of the plastic internal variables takes the form,




( )

( )

1 1 1 1

1 1

1 1 1 1

sgn

sgn

p pn n n n n

n n n

n n n n n

q

q

ε ε γ σ

ξ ξ γ

ξ ξ γ σ

+ + + +

+ +

+ + + +

= + − = + = − −

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

11 1 1 1 1 1

11 1 1 1

11 1 1 1 1 1

, , sgn

, ,

, , sgn

p p trial trial trial trial trialn n n n n n n

trial trial trialn n n n n

trial trial trial trial trialn n n n n n n

E K H f q q q

E K H f q q

E K H f q q q

ε ε σ σ

ξ ξ σ

ξ ξ σ σ

−+ + + + + +

−+ + + +

−+ + + + + +

= + + + − = + + +

= − + + −

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


Consistent elastoplastic tangent modulus


( ) ( )11 1 1 1 1sgntrial trial trial trial

n n n n nE K H f E qσ σ σ−+ + + + += − + + −


n n n n nd d E K H df E qσ σ σ−+ + + + += − + + −

( )( ) ( )

( ) ( )

,1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1

sgn

sgn sgn

trial e trial pn n n n n

trial trial trial trial trial trial trialn n n n n n n

trial trial trial trial trialn n n n n n

d Ed E d E d

df d q q d q

q d q E d

σ ε ε ε ε

σ σ σ

σ σ σ ε

+ + + +

+ + + + + + +

+ + + + + +

= = − =

= − = − −

= − = −

The consistent elastoplastic tangent modulus takes the form,




( ) ( )

( ) ( ) ( )( )( )

11 1 1 1 1

1 11

1 1 1 1 1

11 1

sgn

sgn sgn

1


n n


n n

d d E K H df E q

d E d

E K H q E d E q

d E E E K H d

σ σ σ

σ ε

σ ε σ

σ ε

−+ + + + +

+ +

−+ + + + +

−+ +

= − + + −

=

− + + − −

= − + +

( )( )11 1, : 1ep ep

n nd E d E E E E K Hσ ε −+ += = − + +

1D Plasticity algorithm Step 1. Given the strain 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


1D Plasticity algorithm


( ) ( )

,1

, ,1 1 1

, ,1 1 1 1 1

1 1 1 1

:

:

:

:



trial trial trial trialn n n Y nf q qσ σ

+

+ + +

+ + + + +

+ + + +

=

= −

= = − = −

= − − +

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 discrete plastic multiplier at time n+1




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

n n nf E E+ + +

≤ • = • =

( ) 11 1

trialn nE K H fγ −+ += + +

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




( ) ( )( )( ) ( )

11 1 1 1 1

11 1 1

11 1 1 1 1

sgn

:

: sgn


trial trialn n n


E K H f E q

q q E K H f K

q q E K H f H q

σ σ σ

σ

−+ + + + +

−+ + +

−+ + + + +

= − + + − = − + +

= + + + −

( )1

1 1 1 1trialn

trial trialn n n nfγ

++ + + += − ∂

SS S C S

Step 6. Update plastic internal variables database at time n+1 Step 7. Compute the consistent elastoplastic tangent modulus




( ) ( )( )( ) ( )

11 1 1 1

11 1

11 1 1 1

sgn

sgn

p p trial trial trialn n n n n

trialn n n


E K H f q

E K H f

E K H f q

ε ε σ

ξ ξ

ξ ξ σ

−+ + + +

−+ +

−+ + + +

= + + + − = + + +

= − + + −

( )( )1: 1epE E E E K H −= − + +

( )1

1 1 1trialn

p p trialn n n nfγ

++ + += + ∂

SE E S

Nonlinear isotropic hardening Exponential saturation law + linear 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

:

:

:

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




( ) ( ) ( )( )1 1 1 1 0trialn n n n n nf f E Hγ ξ γ ξ+ + + +′ ′= − + − Π + −Π =

( )( ) ( ) ( )( )

1 1 1

1 1 1 1

: 0

: 0n n n

trialn n n n n n

g g f

g f E H

γ

γ 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 given 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




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

( )( )

1 1 1 1

1 1 1 1 1

1 1

11 1 1

:

:

:

:

k trial k kn n n n n n

k k k k kn n n n n n

k kn n n

k k kn n n

g f E H

Dg E H

E H

γ ξ γ ξ

γ γ ξ γ γ

ξ γ γ

γ γ γ

+ + + +

+ + + + +

+ +

++ + +

′ ′= − + − Π + −Π

′′∆ = − + ∆ −Π + ∆

′′= − +Π + + ∆

∆ = −

1 1 1 0k k kn n ng Dg γ+ + ++ ∆ =

Consistent elastoplastic tangent modulus The consistent elastoplastic tangent modulus is computed taking the variation of the stress at time n+1, yielding, where the variations of the trial stress tensor, and plastic multiplier, at time n+1, have to be computed.




( )( )

1 1 1 1 1

1 1 1 1 1

sgn

sgn



E q

d d d E q

σ σ γ σ

σ σ γ σ

+ + + + +

+ + + + +

= − −

= − −

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,




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

( )( )( )( )

1 1 1 1

1 1 1 1 1

1 1 1

1

1 1 1

: 0

:

: 0

trialn n n n n n

trialn n n n n n

trialn n n n

trialn n n n

g f E H

dg df d E H d

df d E H

d E H df

γ ξ γ ξ

γ ξ γ γ

γ ξ γ

γ ξ γ

+ + + +

+ + + + +

+ + +

−

+ + +

′ ′= − + − Π + −Π =

′′= − + −Π +

′′= − +Π + + =

′′= +Π + +

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




1 1 1ep

n n nd E dσ ε+ + +=

( )( )( )1

1 11epn n nE E E E Hξ γ

−

+ +′′= − +Π + +






Contents


1. Additive split of strains

2. Constitutive equations

3. Associative plastic flow rule

4. Yield function

5. Plastic multiplier

1D Plasticity Algorithms > 1D Rate Dependent Plasticity Algorithms

1D Rate dependent plasticity model


( ) { } { }, , , diag , ,ee e q q E K Hψ σ= ∂ = =

ES E CE S: , C :=

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

( )p fγ= ∂

SE S

( ) Yf q qσ σ= − − +S

( )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,




( ) ( ) { }, diag , ,ee e p E K Hψ= ∂ = = −

ES 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 at the time step, yielding




( ) ( )( )

,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 1 1 1

sgn

:

: sgn

trialn n n n n

trialn n n

trialn n n n n

t E q

q q t Kq q t H q

σ σ γ σ

γ

γ σ

+ + + + +

+ + +

+ + + + +

= − ∆ − = − ∆ = + ∆ −

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




( )1 1 1 1 1:n n n n Y nf f q qσ σ+ + + + += = − − +S

( )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 1 1 1

sgn

:

: sgn

trialn n n n n

trialn n n

trialn n n n n

t E q

q q t Kq q t H q

σ σ γ σ

γ

γ σ

+ + + + +

+ + +

+ + + + +

= − ∆ − = − ∆ = + ∆ −

The solution for the return mapping algorithm yields,




( ) ( )1 1 1 1 1 1 1sgntrial trialn n n n n n nq q t E H qσ σ γ σ+ + + + + + +− = − − ∆ + −

( ) ( )( ) ( )

1 1 1 1 1 1 1 1

1 1 1

sgn sgn

sgn

trial trial trial trialn n n n n n n n

n n n

q q q q

t E H q

σ σ σ σ

γ σ+ + + + + + + +

+ + +

− − = − −

− ∆ + −

( )( ) ( )( )

1 1 1 1 1

1 1 1 1

sgn

sgnn n n n n

trial trial trial trialn n n n

q t E H q

q q

σ γ σ

σ σ

+ + + + +

+ + + +

− + ∆ + − =

= − −

( )( ) ( )

1 1 1 1 1

1 1 1 1sgn sgn

trial trialn n n n n

trial trialn n n n

q t E H q

q q

σ γ σ

σ σ

+ + + + +

+ + + +

− + ∆ + = −

− = −

For the non-trivial case (plastic loading), the yield function and the discrete plastic multiplier are greater than zero, and the following expressions hold,




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

( )( )

( ) ( )

1 1 1 1 1 1

1 1 1 1

1 1 1 1 1

, ,

, ,

n n n n n Y n

trial trial trialn n n Y n


f q q q q

q t E K H q

f q q t E K H

σ σ σ

σ γ σ

σ γ γ η

+ + + + + +

+ + + +

+ + + + +

= − − +

= − − ∆ + + − +

= − ∆ + + =

( )1

1 1 1 1, , 0trial trial trialn n n nt E K H f q q

tηγ σ

−

+ + + + ∆ = + + + > ∆

The return mapping algorithm takes the form,




( )

( )

1 1 1 1 1

1 1 1

1 1 1 1 1

sgn

:

: sgn

trialn n n n n

trialn n n

trialn n n n n

t E q

q q t Kq q t H q

σ σ γ σ

γ

γ σ

+ + + + +

+ + +

+ + + + +

= − ∆ − = − ∆ = + ∆ −

( ) ( )( )( ) ( )

11 1 1 1 1

11 1 1

11 1 1 1 1

sgn

:

: sgn


trial trialn n n


E K H t f E q

q q E K H t f K

q q E K H t f H q

σ σ η σ

η

η σ

−+ + + + +

−+ + +

−+ + + + +

= − + + + ∆ − = − + + + ∆

= + + + + ∆ −

The update of the plastic internal variables takes the form,




( )

( )

1 1 1 1

1 1

1 1 1 1

sgn

:

: sgn

p pn n n n n

n n n

n n n n n

t q

tt q

ε ε γ σ

ξ ξ γ

ξ ξ γ σ

+ + + +

+ +

+ + + +

= + ∆ − = + ∆ = − ∆ −

( ) ( )( )( ) ( )

11 1 1 1

11 1

11 1 1 1

sgn

:

: sgn


trialn n n


E K H t f q

E K H t f

E K H t f q

ε ε η σ

ξ ξ η

ξ ξ η σ

−+ + + +

−+ +

−+ + + +

= + + + + ∆ − = + + + + ∆

= − + + + ∆ −

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





n n n n nE K H t f E qσ σ η σ−+ + + + += − + + + ∆ −


n n n n nd d E K H t df E qσ σ η σ−+ + + + += − + + + ∆ −

( )( ) ( )

( ) ( )

,1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1

sgn

sgn sgn

trial e trial pn n n n n

trial trial trial trial trial trial trialn n n n n n n

trial trial trial trial trialn n n n n n

d Ed E d E d

df d q q d q

q d q E d

σ ε ε ε ε

σ σ σ

σ σ σ ε

+ + + +

+ + + + + + +

+ + + + + +

= = − =

= − = − −

= − = −

The consistent elastoplastic tangent modulus takes the form,




( ) ( )

( ) ( )( )( )

11 1 1 1 1

1 11 2

1 1 1

11 1

sgn

sgn

1


n n

trial trialn n n

n n

d d E K H t df E q

d E d

E K H t q E E d

d E E E K H t d

σ σ η σ

σ ε

η σ ε

σ η ε

−+ + + + +

+ +

−+ + +

−+ +

= − + + + ∆ −

=

− + + + ∆ −

= − + + + ∆

( )( )11 1, : 1ep ep

n nd E d E E E E K H tσ ε η −+ += = − + + + ∆

1D Plasticity algorithm Step 1. Given the strain 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

1 1 1 1

:

:

:

:



trial trial trial trialn n n Y nf q qσ σ

+

+ + +

+ + + + +

+ + + +

=

= −

= = − = −

= − − +

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 1 1if 0 then set , and exittrialtrial ep

n n nf E E+ + +

≤ • = • =

( ) 11 1

trialn nt E K H t fγ η −+ +∆ = + + + ∆

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




( ) ( )( )( ) ( )

11 1 1 1 1

11 1 1

11 1 1 1 1

sgn

:

: sgn


trial trialn n n


E K H t f E q

q q E K H t f K

q q E K H t f H q

σ σ η σ

η

η σ

−+ + + + +

−+ + +

−+ + + + +

= − + + + ∆ − = − + + + ∆

= + + + + ∆ −

( )1

1 1 1 1trialn

trial trialn n n nt fγ

++ + + += − ∆ ∂

SS S C S

Step 6. Update plastic internal variables database at time n+1 Step 7. Compute the consistent elastoplastic tangent modulus




( ) ( )( )( ) ( )

11 1 1 1

11 1

11 1 1 1

sgn

sgn


trialn n n


E K H t f q

E K H t f

E K H t f q

ε ε η σ

ξ ξ η

ξ ξ η σ

−+ + + +

−+ +

−+ + + +

= + + + + ∆ − = + + + + ∆

= − + + + ∆ −

( )( )1: 1epE E E E K H tη −= − + + + ∆

( )1

1 1 1trialn

p p trialn n n nt fγ

++ + += + ∆ ∂

SE E S

Nonlinear isotropic hardening Exponential saturation law + linear 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

:

:

:

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




( ) ( ) ( )( )1 1 1 1 1 0trialn n n n n n nf f t E H tγ ξ γ ξ γ η+ + + + +′ ′= − ∆ + − Π + ∆ −Π = >

( )( ) ( )( )

1 1 1 1

1 1 1 1

: 0

: 0

n n n n

trialn n n n n n

g g f

g f t E H tt

γ γ η

ηγ 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




( ) ( )( )

( )

( )

1 1 1 1

1 1 1 1 1

1 1

11 1 1

:

:

:

:

k trial k kn n n n n n

k k k k kn n n n n n

k kn n n

k k kn n n

g f t E H tt

Dg E H t t tt

E t H tt

ηγ ξ γ ξ

ηγ γ ξ γ γ

ηξ γ γ

γ γ γ

+ + + +

+ + + + +

+ +

++ + +

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

∆ = −

1 1 1 0k k kn n ng Dg γ+ + ++ ∆ =

Consistent elastoplastic tangent modulus The consistent elastoplastic tangent modulus is computed taking the variation of the stress at time n+1, yielding, where the variations of the trial stress tensor, and plastic multiplier, at time n+1, have to be computed.




( )( )

1 1 1 1 1

1 1 1 1 1

sgn

sgn



t E q

d d d t E q

σ σ γ σ

σ σ γ σ

+ + + + +

+ + + + +

= − ∆ −

= − ∆ −

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,




( ) ( )( )

( )

( )

( )

1 1 1 1

1 1 1 1 1

1 1 1

1

1 1

: 0

:

: 0

trialn n n n n n

trialn n n n n n

trialn n n n

n n n n

g f t E H tt

dg df d t E H t d tt

df d t E t Ht

d t E t H dft

ηγ ξ γ ξ

ηγ ξ γ γ

ηγ ξ γ

ηγ ξ γ

+ + + +

+ + + + +

+ + +

−

+ + +

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

′′∆ = +Π + ∆ + + ∆ 1

trial

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




1 1 1ep

n n nd E dσ ε+ + +=

( )1

1 11epn n nE E E E t H

tηξ γ

−

+ +

′′= − +Π + ∆ + + ∆






Contents


Implement in MATLAB the BE time-stepping algorithm for 1D 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

1D Plasticity Algorithms > 1D Computational Plasticity Assignment

1D Computational plasticity assignment


For the perfect plasticity models, plot the stress-strain curves For the linear isotropic/linear kinematic hardening models,

plot the stress-strain curves showing the influence of the isotropic/kinematic hardening parameters

For the nonlinear isotropic hardening model, plot the stress-strain curves showing the influence of the exponential coefficient of the exponential saturation law on the stress-strain curves

For the rate-dependent plasticity models, plot the stress-strain, and the stress-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




Documents

Computational Solid Mechanics