Useful Results In Continuum Mechanics

Some Useful Results Biswajit Banerjee March 22, 2007

1

A spectral decomposition problem • In one of Simo’s 1992 papers on plasticity [1] (p. 76) we find the statement that “An easy calculation then gives the spectral decomposition” 3 X ∂ ∂ b [φ(β1 , β2 , β3 , q)] nA ⊗ nA . [φ(τ , q)] = ∂τ ∂βA

(1)

A=1

Here, τ is the Kirchhoff stress, q is a scalar internal variable, and φ is yield function. The spectral decomposition of the Kirchhoff stress is given by τ =

3 X

βA nA ⊗ nA .

(2)

A=1

Also, isotropy implies that the yield function φ can be expressed in terms of the principal values βi of τ such that b 1 , β2 , β3 , q) . φ(τ , q) = φ(β (3)

Let’s try to work out the easy calculation. Using the chain rule, we have ∂φ ∂ φb ∂β1 ∂ φb ∂β2 ∂ φb ∂β3 = + + ∂τ ∂β1 ∂τ ∂β2 ∂τ ∂β3 ∂τ

or

3 X ∂ φb ∂βA ∂φ = . ∂τ ∂βA ∂τ

(4)

A=1

Also, since τ = β1 n1 ⊗ n1 + β2 n2 ⊗ n2 + β3 n3 ⊗ n3 using the identity (since the eigenvectors nA are orthonormal) ( 0 (ni ⊗ ni ) · (nj ⊗ nj ) = ni ⊗ ni

if i 6= j if i = j

(5)

(6)

we get τ · (n1 ⊗ n1 ) = β1 n1 ⊗ n1 ; τ · (n2 ⊗ n2 ) = β2 n2 ⊗ n2 ; τ · (n3 ⊗ n3 ) = β3 n3 ⊗ n3 . 1

(7)

Using the identity (ni ⊗ ni ) : (ni ⊗ ni ) = 1

(8)

we then have [τ · (n1 ⊗ n1 )] : (n1 ⊗ n1 ) = β1 ; [τ · (n2 ⊗ n2 )] : (n2 ⊗ n2 ) = β2 ; [τ · (n3 ⊗ n3 )] : (n3 ⊗ n3 ) = β3 . (9) We can simplify these further by using the identities (A · B) : C = A : (C · B T ) ; n ⊗ n = (n ⊗ n)T ; (n ⊗ n) · (n ⊗ n) = n ⊗ n

(10)

to get τ : [(nA ⊗ nA ) · (nA ⊗ nA )T ] = τ : [(nA ⊗ nA ) · (nA ⊗ nA )] = τ : (nA ⊗ nA ) = βA ; A = 1, 2, 3

(11)

Taking the derivatives of both sides with respect to τ , we get ∂βA ∂ [τ : (nA ⊗ nA )] = ; A = 1, 2, 3 ∂τ ∂τ

(12)

Note here that as τ varies, the eigenvectors of τ (i.e., the nA s) also vary. So the derivatives will have the form ∂ ∂βA ∂τ : (nA ⊗ nA ) + τ : (nA ⊗ nA ) = ; A = 1, 2, 3 ∂τ ∂τ ∂τ

(13)

or, ∂βA ∂ (nA ⊗ nA ) = ; A = 1, 2, 3 ∂τ ∂τ where Is is the symmetric fourth-order identity tensor. Therefore, Is : (nA ⊗ nA ) + τ :

(14)

∂βA ∂ = nA ⊗ nA + τ : (nA ⊗ nA ) . ∂τ ∂τ

(15)

3 X ∂φ ∂ φb ∂ ∂ φb nA ⊗ nA + τ : = (nA ⊗ nA ) . ∂τ ∂βA ∂βA ∂τ

(16)

Plugging (15) into (4), we get

A=1

Compare equations (16) and (1), shown below for your convenience. 3 X ∂ φb ∂φ = nA ⊗ nA . ∂τ ∂βA

(17)

A=1

You will see that the derivativ)es of the eigenvectors with respect to τ do not appear in Simo’s equation. This is an approximation that Simo does not mention in his paper, i.e., that the values of nA are kept fixed when evaluating the derivatives of φ. Or, am I missing something? Indeed, there is an error in my analysis. Can you find out what it is? Let us now try an alternative proof (hat tip Prof. Andrew Norris). Since φ(τ ) is isotropic, instead of working directly with the eigenvalues βA , we can work with the invariants of τ . 2

Recall that the basic invariants of τ are Iτ = tr(τ )  1 2 IIτ = Iτ − tr(τ 2 ) 2  1 3 IIIτ = tr(τ 3 ) − IA + 3 IA IIA 3

(18)

Therefore, the yield function can also be represented as φ(τ , q) = φ(α1 , α2 , α3 , q) where α1 = tr(τ ) ; α2 =

(19)

1 1 tr(τ 2 ) ; α3 = tr(τ 3 ) . 2 3

(20)

Differentiating with respect to τ , we get

Now,

∂φ ∂φ ∂α1 ∂φ ∂α2 ∂φ ∂α3 = + + . ∂τ ∂α1 ∂τ ∂α2 ∂τ ∂α3 ∂τ

(21)

∂[tr(τ )] ∂α2 1 ∂[tr(τ 2 )] ∂α3 1 ∂[tr(τ 3 )] ∂α1 = =1 ; = =τ ; = = τ2 . ∂τ ∂τ ∂τ 2 ∂τ ∂τ 3 ∂τ

(22)

∂φ ∂φ ∂φ ∂φ 2 = 1+ τ+ τ . ∂τ ∂α1 ∂α2 ∂α3

(23)

Therefore,

Let us now express τ in terms of its spectral representation to get 1 =

τ = τ2 =

3 X A=1 3 X A=1 3 X

3 X

nA ⊗ nA =

0 βA nA ⊗ nA

A=1 1 βA nA ⊗ nA

(24)

2 βA nA ⊗ nA

A=1

Let us also express the αi , i = 1, 2, 3 in terms of the spectral representation of τ , i.e., α1 = tr(τ ) =

3 X

βA tr(nA ⊗ nA ) =

A=1

3 X

βA

A=1

3 3 1X 2 1X 2 1 βA tr(nA ⊗ nA ) = βA α2 = tr(τ 2 ) = 2 2 2

1 1 α3 = tr(τ 3 ) = 3 3

A=1 3 X A=1

3 βA tr(nA ⊗ nA ) =

1 3

A=1 3 X

(25)

3 βA

A=1

Therefore, ∂α2 ∂α3 ∂α1 1 0 2 = 1 = βA ; = βA = βA ; = βA . ∂βA ∂βA ∂βA 3

(26)

Plugging these into (24) gives 1 =

τ = τ2 =

3 X ∂α1 nA ⊗ nA ∂βA

A=1 3 X A=1 3 X A=1

∂α2 nA ⊗ nA ∂βA

(27)

∂α3 nA ⊗ nA ∂βA

Substituting into (23), we get 3 3 3 ∂φ X ∂α2 ∂φ X ∂α3 ∂φ X ∂α1 ∂φ = nA ⊗ nA + nA ⊗ nA + nA ⊗ nA ∂τ ∂α1 ∂βA ∂α2 ∂βA ∂α3 ∂βA

(28)

 3  X ∂φ ∂α2 ∂φ ∂α3 ∂φ ∂φ ∂α1 + + nA ⊗ nA = ∂τ ∂α1 ∂βA ∂α2 ∂βA ∂α3 ∂βA

(29)

3 X ∂ φb ∂φ nA ⊗ nA = ∂τ ∂βA

(30)

A=1

A=1

A=1

or,

A=1

or,

A=1

We finally have the correct expression! • In the operator split algorithm in [1] we have the the following expression for the algorithmic flow rule   ∂φ be = exp −2γ∆t ˙ · (be )trial (31) ∂τ If the spectral decompositions of be and τ are be =

3 X

λ2A nA ⊗ nA ; τ =

A=1

and

3 X

βA nA ⊗ nA

(32)

A=1 3 X ∂φ ∂ φb = nA ⊗ nA ∂τ ∂βA

(33)

A=1

show that (be )trial =

3 X A=1

" λ2A

∂ φb exp 2γ∆t ˙ ∂βA

!# nA ⊗ nA .

(34)

From (31), we have e trial

(b )



∂φ = exp 2γ∆t ˙ ∂τ





∂φ · b = exp κ ∂τ e



· be

(35)

where κ := 2γ∆t ˙ . 4

(36)

Expanding in an infinite series, we have       ∂φ ∂φ κ2 ∂φ 2 κ3 ∂φ 3 exp κ =1 +κ + + + ... ∂τ ∂τ 2 ∂τ 3! ∂τ

(37)

Define αA :=

∂ φb . ∂βA

(38)

Then,  2 X  3 X 3 3 3 X ∂φ ∂φ ∂φ 2 3 = = = αA nA ⊗ nA ; αA nA ⊗ nA ; αA nA ⊗ nA ; . . . ∂τ ∂τ ∂τ A=1

A=1

(39)

A=1

Therefore, 

∂φ exp κ ∂τ

 =1+

3 X

"

A=1

κ2 2 κ3 3 κ αA + αA + α + ... 2 3! A

# nA ⊗ nA .

(40)

Since 1 =

X

nA ⊗ nA

(41)

A

we then have " #   X 3 3 X (κ αA )2 (κ αA )3 ∂φ exp κ = 1 + κ αA + + + . . . nA ⊗ nA = eκ αA nA ⊗ nA . ∂τ 2 3! A=1

(42)

A=1

Using the spectrla decomposition of be , we get " 3 # " 3 #   X X ∂φ 2 e ·b = exp(κ αA ) nA ⊗ nA · λB nB ⊗ nB . exp κ ∂τ A=1

(43)

B=1

From the identity ( 0 (ni ⊗ ni ) · (nj ⊗ nj ) = ni ⊗ ni

if i 6= j if i = j

(44)

we then get 

∂φ exp κ ∂τ



e

κ α1

·b =e

λ21

κ α2

n1 ⊗ n1 + e

λ22

κ α3

n2 ⊗ n2 + e

λ23

n3 ⊗ n3 =

3 X

λ2A eκ αA nA ⊗ nA . (45)

A=1

Therefore, from (35) e trial

(b )

=

3 X A=1

" λ2A

∂ φb exp 2γ∆t ˙ ∂βA

!# nA ⊗ nA .

References [1] J. C. Simo. Algorithms for static and dynamic multiplicative plasticity that preserve the classical return mapping algorithms of the infinitesimal theory. Comp. Meth. Appl. Mech. Engrg., 99:61–112, 1992. 5

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