useful results in continuum mechanics

8/14/2019 Useful Results in continuum mechanics

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Some Useful Results

Biswajit Banerjee

March 22, 2007

1 A spectral decomposition problem

In one of Simos 1992 papers on plasticity [1] (p. 76) we nd the statement that An easy calculation thengives the spectral decomposition

[( , q)] =3

A =1

A

[ ( 1 , 2 , 3 , q)] n A n A . (1)Here, is the Kirchhoff stress, q is a scalar internal variable, and is yield function. The spectraldecomposition of the Kirchhoff stress is given by

=3

A =1 A n A n A . (2)

Also, isotropy implies that the yield function can be expressed in terms of the principal values i of suchthat

(

, q) = ( 1 , 2 , 3 , q) . (3)

Lets try to work out the easy calculation.

Using the chain rule, we have

= 1 1 + 2 2 + 3 3 or =

3

A =1

A A . (4)Also, since

= 1 n 1 n 1 + 2 n 2 n 2 + 3 n 3 n 3 (5)using the identity (since the eigenvectors n A are orthonormal)

(n i n i ) (n j n j ) =0 if i = jn i n i if i = j

(6)

we get

(n 1 n 1 ) = 1 n 1 n 1 ; (n 2 n 2 ) = 2 n 2 n 2 ; (n 3 n 3 ) = 3 n 3 n 3 . (7)1


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Using the identity(n i n i ) : (n i n i ) = 1 (8)

we then have

[ (n 1 n 1 )] : (n 1 n 1 ) = 1 ; [ (n 2 n 2 )] : (n 2 n 2 ) = 2 ; [ (n 3 n 3 )] : (n 3 n 3 ) = 3 . (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

: [(n A n A ) (n A n A )T ] = : [(n A n A ) (n A n A )] = : (n A n A ) = A ; A = 1 , 2, 3 (11)

Taking the derivatives of both sides with respect to , we get

[ : (n A n A )] = A

; A = 1 , 2, 3 (12)

Note here that as varies, the eigenvectors of (i.e., the n A s) also vary. So the derivatives will have the form

: (n A n A ) + :

(n A n A ) = A

; A = 1 , 2, 3 (13)

or,I s : (n A n A ) + :

(n A n A ) = A

; A = 1 , 2, 3 (14)

where I s is the symmetric fourth-order identity tensor. Therefore,

A

= n A n A + :

(n A n A ) . (15)

Plugging (15) into (4), we get

=3

A =1

A n A n A + A : (n A n A ) . (16)Compare equations (16) and (1), shown below for your convenience.

=3

A =1

A n A n A . (17)You will see that the derivativ)es of the eigenvectors with respect to do not appear in Simos equation. This is anapproximation that Simo does not mention in his paper, i.e., that the values of n A are kept xed when evaluatingthe derivatives of . Or, am I missing something?

Indeed, there is an error in my analysis. Can you nd out what it is?

Let us now try an alternative proof (hat tip Prof. Andrew Norris).

Since ( ) is isotropic, instead of working directly with the eigenvalues A , we can work with the invariants of .

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Recall that the basic invariants of are

I = tr( )

II =12

I 2 tr( 2 )

III =13 tr(

3

) I 3

A + 3 I A II A

(18)

Therefore, the yield function can also be represented as

( , q) = ( 1 , 2 , 3 , q) (19)

where 1 = tr( ) ; 2 =

12

tr( 2 ) ; 3 =13

tr( 3 ) . (20)

Differentiating with respect to , we get

=

1

1

+

2

2

+

3

3

. (21)

Now, 1

= [tr( )]

= 1 ;

2

=12

[tr( 2 )]

= ; 3

=13

[tr( 3 )]

= 2 . (22)

Therefore,

=

11 +

2

+

3 2 . (23)

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

1 =3

A=1

n A n A =3

A=1

0A n A n A

=3

A =1 1A n A n A

2 =3

A =1 2A n A n A

(24)

Let us also express the i , i = 1 , 2, 3 in terms of the spectral representation of , i.e.,

1 = tr( ) =3

A =1 A tr(n A n A ) =

3

A =1 A

2 =12

tr( 2 ) =12

3

A =1 2A tr(n A n A ) =

12

3

A =1 2A

3 =13

tr( 3 ) =13

3

A =1 3A tr(n A n A ) =

13

3

A =1 3A

(25)

Therefore, 1 A

= 1 = 0A ; 2 A

= A = 1A ;

3 A

= 2A . (26)

3


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Plugging these into (24) gives

1 =3

A =1

1 A

n A n A

=3

A =1

2

An A n A

2 =3

A =1

3 A

n A n A

(27)

Substituting into (23), we get

= 1

3

A =1

1 A

n A n A +

2

3

A =1

2 A

n A n A + 3

3

A =1

3 A

n A n A (28)

or,

=

3

A =1

1

1

A+

2

2

A+

3

3

An A n A (29)

or,

=3

A =1

A n A n A (30)We nally have the correct expression!

In the operator split algorithm in [1] we have the the following expression for the algorithmic ow rule

b e = exp 2 t

(b e )trial (31)

If the spectral decompositions of b e and are

b e =3

A =1 2A n A n A ; =

3

A =1 A n A n A (32)

and

=3

A =1

A n A n A (33)show that

(b e

)trial

=

3

A =1 2

A exp 2 t

An

A n

A . (34)

From (31), we have

(b e )trial = exp 2 t

b e = exp

b e (35)

where := 2 t . (36)

4


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Expanding in an innite series, we have

exp

= 1 +

+ 2

2

2

+ 3

3!

3

+ . . . (37)

Dene

A := A . (38)Then,

=3

A =1 A n A n A ;

2

=3

A =1 2A n A n A ;

3

=3

A =1 3A n A n A ; . . . (39)

Therefore,

exp

= 1 +3

A =1 A +

2

2 2A +

3

3! 3A + . . . n A n A . (40)

Since 1 =A

n A n A (41)

we then have

exp

=3

A =11 + A +

( A )2

2+

( A )3

3!+ . . . n A n A =

3

A =1e A n A n A . (42)

Using the spectrla decomposition of b e , we get

exp

b e =3

A =1

exp( A ) n A n A 3

B =1

2B n B n B . (43)

From the identity

(n i n i ) (n j n j ) =0 if i = jn i n i if i = j

(44)

we then get

exp

b e = e 1 21 n 1 n 1 + e 2 22 n 2 n 2 + e 3

23 n 3 n 3 =

3

A =1 2A e

A n A n A . (45)

Therefore, from (35)

(b e) trial =3

A =1 2A exp 2 t

A n A n A . (46)References

[1] J. C. Simo. Algorithms for static and dynamic multiplicative plasticity that preserve the classical returnmapping algorithms of the innitesimal theory. Comp. Meth. Appl. Mech. Engrg. , 99:61112, 1992.

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