# Elasticity, free energy of isotropic body mismatch

1. Aug 15, 2012

### malasti

I'm looking at the free energy of a body (theory of elasticity) but I can't really square the general expression with the one usually used for isotropic bodies.

According to wikipedia (http://en.wikipedia.org/wiki/Elastic_energy), Landau & Lifgarbagez etc the general expression for the free (elasticity) energy "F" of a body is

½*Cijkl*uij*ukl (summation implied)

where Cijkl is the elastic modulus tensor and the uij's are the strain tensor, uij=½*((∂ui/∂xj)+(∂uj/∂xi))=uji (small deformations, i.e. no non-linear term).

Considering 2D, the non-zero components of Cijkl for an isotropic medium (as can be seen e.g. on the above wikipedia page) are

Cxxxx=Cyyyy=λ+2μ
Cxyxy=Cyxyx=Cxyyx=Cyxxy
Cxxyy=Cyyxx

where λ and μ are the Lame coefficients (material constants). Now, if I use these above values, I get

(½λ+2μ)*[uxx2+uyy2]+2μ*uxy2+λ*uxx*uyy .

According to Landau & Lifgarbagez, a million papers etc, the free energy of an isotropic body is

F=½λ*uii2+μ*uik2=(½λ+μ)*[uxx2+uyy2]+2μ*uxy2 .

I would expect the general expression to reduce to the isotropic one when using the isotropic elastic modulus tensor. However, I get an extra term "+λ*uxx*uyy" which does not appear in the usual isotropic expression. Moreover, this extra term seems quite unphysical too me, since it does not involve a square, and thus could probably be negative for a non-zero deformation.

It seems to me that this term should go, however, I do not see how.

Last edited: Aug 15, 2012
2. Aug 15, 2012

### malasti

Could it possibly have something to do with the definitions of the axes or something? I really suck at tensors, but if I viewed Cijkluijukl as a four-component tensor and then presented it as a 4x4 matrix (I'm not entirely sure how legit this is), i.e. with the rows (columns), starting from the top (left) and designating the second (first) two indices: "xx", "xy, "yx", "yy", I find (using symmetry properties) that it is symmetric and real, and thus should be diagonalizable. I'm not entirely sure how to interpret such a diagonalization though.

3. Aug 15, 2012

### malasti

Okay, here's a better idea. The initial expression can be written as a matrix of constants (the moduli) times a column vector with the components uxx, uxy, uyx, uyy, times a row vector with the same components (I checked that it sums up to the right thin). Since it's a scalar, it's invariant under coordinate transformations.

I used Mathematica to find the eigenvalues of the matrix (I included the factor ½): 0, μ, μ, λ+μ, then changed everything to the eigenbasis, making the matrix diagonal. Performing the multiplications now yields the scalar (expressed in a different basis, I use "1" and "2" instead of "x" and "y"; also, I assumed I still have the same symmetries, uij=uji) as

(λ+μ)*u222+2μ*u122 .

This seems better to me, even though it still doesn't quite look like the regular expression for the free energy of an isotropic body. I think I'm gonna try messing around with that expression in a similar manner.

Last edited: Aug 15, 2012
4. Aug 15, 2012

### malasti

Now I wrote the "canonical" isotropic expression in the same "matrix-vector-vector"-form, and for it the matrix looks different, but not diagonal.

My matrix:
½λ+μ 0 0 ½λ
0 ½μ ½μ 0
0 ½μ ½μ 0
½λ 0 0 ½λ+μ

The "standard" matrix:
½λ+μ 0 0 0
0 μ μ 0
0 μ μ 0
0 0 0 ½λ+μ

The vectors are as before.

5. Aug 27, 2012

### malasti

Ok, actually, this was way simpler than I though. I had the right expression, I just misinterpreted others' notation.

uii2=uiiujj=(uxx+uyy)2 ,

including the cross term.