6.7 Derivation of Consistent Elastoplastic Moduli

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For finite element calculations, we also need to determine the consistent elastoplastic moduli

at t = tn+1. These moduli are the derivatives of the second Piola–Kirchhoff stress with

FINITE STRAIN ELASTOPLASTICITY 295

respect to the Lagrange strain:

Bn+1 := ∂Sn+1

∂En+1

= 2

∂Sn+1

∂Cn+1

. (6.208)

Recall that Sn+1 takes the form (Equation (6.197)

Sn+1 = Jn+1U

 

(Jn+1)C

−1

n+1

+ μJ

−2/3

n+1 DEVn+1Cp−1

n

− 2_γn+1μn+1Nn+1. (6.209)

The first term on the right-hand side of Equation (6.209) is the volumetric part, the second

is the deviatoric trial stress and the third is the plastic correction.

6.7.1 The volumetric stress

Taking into account that

∂J

∂C

= J

2

C

−1 (6.210)

∂C

−1

∂C

= −I

C

−1 (6.211)

with I

C

−1 defined in Equation (6.82), one gets

2

∂

∂C

[JU

 

(J )C

−1] = JU

 

(J )C

−1 ⊗ C

−1 + J 2U

 

(J )C

−1 ⊗ C

−1 − 2JU

 

(J )I

C

−1

(6.212)

= J 2U

 

(J )C

−1 ⊗ C

−1 + Jp(C

−1 ⊗ C

−1 − 2I

C

−1 ). (6.213)

The index n + 1 was dropped for convenience. For U(J) defined in Equation (6.51),

Equation (6.213) takes the form

2

∂

∂C

[JU

 

(J )C

−1] = KJ2C

−1 ⊗ C

−1 − K(J2 − 1)I

C

−1 . (6.214)

6.7.2 Trial stress

Since

∂J

−2/3

∂_________C

= −13

J

−2/3C

−1 (6.215)

and

DEV(·) = (·) − 13

[(·) : C]C

−1 (6.216)

one gets

2

∂T trial

∂C

= −23

μJ

−2/3DEV(Cp−1

) ⊗ C

−1 − 23

μJ

−2/3C

−1 ⊗ (Cp−1 : II )

+ 23

μJ

−2/3(C : Cp−1

)IC

−1 (6.217)

296 FINITE STRAIN ELASTOPLASTICITY

= −23

μJ

−2/3 _Cp−1 ⊗ C

−1 − 13

(C : Cp−1

)C

−1 ⊗ C

−1_

− 23

μJ

−2/3C

−1 ⊗ Cp−1 + 23

μJ

−2/3(C : Cp−1

)I

C

−1 (6.218)

= 23

μJ

−2/3(C : Cp−1

) _I

C

−1 + 13

C

−1 ⊗ C

−1_

− 23

μJ

−2/3 _Cp−1 ⊗ C

−1 + C

−1 ⊗ Cp−1_ (6.219)

= 23

μJ

−2/3(C : Cp−1

) _I

C

−1 − 13

C

−1 ⊗ C

−1_

− 23

μJ

−2/3  _Cp−1 − 13

(C : Cp−1

)C

−1_ ⊗ C

+C ⊗ _Cp−1 − 13

(C : Cp−1

)C

−1_! (6.220)

= 23

μJ

−2/3(C : Cp−1

) _I

C

−1 − 13

C

−1 ⊗ C

−1_

− 23

μJ

−2/3 _DEV(Cp−1

) ⊗ C

−1 + C

−1 ⊗ DEV(Cp−1

)_ . (6.221)

Accordingly,

2

∂T trial

n+1

∂Cn+1

= 23

μJ

−2/3

n+1 (Cn+1 : Cp−1

n ) _I

C

−1

n+1

− 13

C

−1

n+1

⊗ C

−1

n+1

_

− 23

μJ

−2/3

n+1

_DEVCp−1

n

⊗ C

−1

n+1

+ C

−1

n+1

⊗ DEVCp−1

n

_ := Btrial

n+1. (6.222)

6.7.3 Plastic correction

This is the most difficult part. One obtains

2

∂

∂Cn+1

_−2_γn+1μn+1Nn+1_ = −4μn+1Nn+1 ⊗ ∂_γn+1

∂Cn+1

− 4_γn+1Nn+1 ⊗ ∂μn+1

∂C

− 4_γn+1μn+1

∂Nn+1

∂Cn+1

. (6.223)

Concentrating on the last term,

∂Nn+1

∂Cn+1

= ∂

∂Cn+1

_trial

n+1

_trial

n+1

 

= 1

_trial

n+1

 

_∂_trial

n+1

∂Cn+1

− Nn+1 ⊗

∂_trial

n+1

 

∂Cn+1

_

. (6.224)

FINITE STRAIN ELASTOPLASTICITY 297

In complete analogy to Equation (6.222), one finds

2

∂Atrial

n+1

∂Cn+1

= −23

J

−2/3

n+1 (Cn+1 : Q2,n) _I

C

−1

n+1

− 13

C

−1

n+1

⊗ C

−1

n+1

_

+ 23

J

−2/3

n+1

_DEV(Q2,n) ⊗ C

−1

n+1

+ C

−1

n+1

⊗ DEV(Q2,n)_ (6.225)

and accordingly,

2

∂_trial

n+1

∂Cn+1

= 2μn+1 _I

C

−1

n+1

− 13

C

−1

n+1

⊗ C

−1

n+1

_

− 23

__trial

n+1

⊗ C

−1

n+1

+ C

−1

n+1

⊗ _

−1

n+1

_ := Htrial

n+1. (6.226)

Furthermore,

∂_trial

n+1

 

∂Cn+1

= ∂

∂Cn+1

__IJ,trial

n+1 _KL,trial

n+1 CIJ,n+1CKL,n+1 (6.227)

= 12

Htrial

n+1 : Nn+1 + _trial

n+1

N2

n+1 (6.228)

and

Htrial

n+1 : Nn+1 = 2μn+1Nn+1 − 23

_trial

n+1

C

−1

n+1. (6.229)

Equations (6.224), (6.226), (6.227) and (6.229) yield

∂Nn+1

∂Cn+1

= 1

_trial

n+1

 

 Htrial

n+1

− Nn+1 ⊗ _μn+1Nn+1 + _trial

n+1

DEVn+1(N2

n+1)_! (6.230)

since

DEV(N2) = N2 − 13

(N · N : C)C

−1 = N2 − 13

C

−1. (6.231)

For the second term, one starts from the expression for μn+1:

μn+1 = 13

μJ

−2/3

n+1 Cp−1 : Cn+1 (6.232)

⇓

∂μn+1

∂Cn+1

= −13

μn+1C

−1

n+1

+ 13

μJ

−2/3

n+1 Cp−1

n (6.233)

and

μn+1 = μn+1 + 13

J

−2/3

n+1 (Q2,n : Cn+1) (6.234)

⇓

∂μn+1

∂Cn+1

= ∂μn+1

∂Cn+1

+ 13

J

−2/3

n+1 DEVn+1(Q2,n). (6.235)

298 FINITE STRAIN ELASTOPLASTICITY

Note the following interesting expression:

∂

∂Cn+1

_J

−2/3

n+1 TRn+1(·)_ = J

−2/3

n+1 DEVn+1(·). (6.236)

The first term is obtained by taking the derivative of Equation (6.193):

∂_trial

n+1

 

∂Cn+1

− 2

∂μn+1

∂Cn+1

_

1 + h

eq

2

3μ

_

_γn+1 − 2μn+1

_

1 + h

eq

2

3μ

_ ∂_γn+1

∂Cn+1

− 23

h

 

1

∂_γn+1

∂Cn+1

− 23

f

∂_γn+1

∂Cn+1

= 0 (6.237)

(h

eq

2 is assumed to be constant) from which

2μn+1

∂_γn+1

∂Cn+1

= 1

_1 + h

eq

2

3μ

+ h

 

1

3μn+1

+ f

 

3μn+1

_

·

·

_∂_trial

n+1

 

∂Cn+1

− 2_γn+1

_

1 + h

eq

2

3μ

_ ∂μn+1

∂Cn+1

_

. (6.238)

Collecting terms, the derivative of the plastic correction yields

2

∂

∂Cn+1

_−2_γn+1μn+1Nn+1_ =

−1

_1 + h

eq

2

3μ

+ h

 

1

3μn+1

+ f

 

3μn+1

_

·

·

_

Nn+1 ⊗ 2

∂_trial

n+1

 

∂Cn+1

− 2_γn+1

_

1 + h

eq

2

3μ

_

Nn+1 ⊗ 2

∂μn+1

∂Cn+1

_

− 2_γn+1Nn+1 ⊗ 2

∂μn+1

∂Cn+1

− 2_γn+1μn+1

_trial

n+1

 

_

Htrial

n+1

− Nn+1 ⊗ 2

∂_trial

n+1

 

∂Cn+1

_

(6.239)

= −_ 1

δ0

− f0_Nn+1 ⊗ 2

∂_trial

n+1

 

∂Cn+1

+ 2γn+1

_ 1

δ0

_

1 + h

eq

2

3μ

_

− 1

_

Nn+1 ⊗ 2

∂μn+1

∂Cn+1

− f0Htrial

n+1 (6.240)

where

f0 := 2μn+1_γn+1

_trial

n+1

 

(6.241)

δ0 := 1 + h

eq

2

3μ

+ h

 

1

3μn+1

+ f

 

3μn+1

. (6.242)

FINITE STRAIN ELASTOPLASTICITY 299

Substituting Equations (6.228), (6.229), (6.231) and (6.235) into Equation (6.240) yields

2

∂

∂Cn+1

_−2_γn+1μn+1Nn+1_

= −f1 _2μn+1Nn+1 ⊗ Nn+1 + 2_trial

n+1

Nn+1 ⊗ DEVn+1(N2

n+1)_

+ 2γn+1

_ 1

δ0

_

1 + h

eq

2

3μ

_

− 1

_ 23

_trial

n+1

Nn+1 ⊗ Nn+1 − f0Htrial

n+1 (6.243)

where

f1 := 1

δ0

− f0 (6.244)

or

2

∂

∂Cn+1

_−2_γn+1μn+1Nn+1_

= −δ1Nn+1 ⊗ Nn+1 − δ2Nn+1 ⊗ DEVn+1(N2

n+1) − f0Htrial

n+1 (6.245)

where

δ1 := f12μn+1 −

_ 1

δ0

_

1 + h

eq

2

3μ

_

− 1

_ 43

γn+1_trial

n+1

 (6.246)

and

δ2 := 2_trial

n+1

f1. (6.247)

Summarizing,

Bn+1 = KJ2

n+1C

−1

n+1

⊗ C

−1

n+1

− K(J2

n+1

− 1)I

C

−1

n+1

+ Btrial

n+1

− δ1Nn+1 ⊗ Nn+1 − δ2Nn+1 ⊗ DEVn+1(N2

n+1) − f0Htrial

n+1 (6.248)

where

Btrial

n+1

= 2μn+1 _I

C

−1

n+1

− 13

C

−1

n+1

⊗ C

−1

n+1

_ − 23

_T trial

n+1

⊗ C

−1

n+1

+ C

−1

n+1

⊗ T trial

n+1_ (6.249)

Htrial

n+1

= 2μn+1 _I

C

−1

n+1

− 13

C

−1

n+1

⊗ C

−1

n+1

_ − 23

__trial

n+1

⊗ C

−1

n+1

+ C

−1

n+1

⊗ _trial

n+1_

(6.250)

f0 = 2μn+1_γn+1

_trial

n+1

 

(6.251)

f1 = 1

δ0

− f0 (6.252)

δ0 = 1 + h

eq

2

3μ

+ h

 

1

3μn+1

+ f

 

3μn+1

. (6.253)

300 FINITE STRAIN ELASTOPLASTICITY

δ1 = f12μn+1 −

_ 1

δ0

_

1 + h

eq

2

3μ

_

− 1

_ 43

_γn+1_trial

n+1

 (6.254)

δ2 = 2_trial

n+1

f1. (6.255)

This concludes a long and tedious calculation. Notice that the tangent modulus is usually

not isotropic, although the material is isotropic in the elastic range. Plasticity induces

anisotropy.

The expression for Bn+1 is not symmetric because of the Nn+1 ⊗ DEVn+1(N2

n+1)

term. In practice, this term is often symmetrized:

_Nn+1 ⊗ DEVn+1(N2

n+1)_S

:= 12

_Nn+1 ⊗ DEVn+1(N2

n+1) + DEVn+1(N2

n+1) ⊗ Nn+1_ . (6.256)

This does not lead to wrong solutions, but may decrease the rate of convergence of the

scheme. However, the effect is deemed to be small.

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