# Anisotropic stiffness of isotropic material

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Dear colleagues,

Consider a simple non-linear elastic material with stress given as

σ = D(εdev) εdev + B εiso

where εdev is the norm of εdev, D is a function of εdev and B is constant. The material is isotropic since the principal directions of  σ and ε will coincide.

If we differentiate σ wrt ε to obtain the material stiffness the form of the stiffness tensor will be

C = D(εdev) Idev + B Iiso + A(εdev) ε © ε

where A is a function of εdev, © is the tensor/dyadic product, and Idev and Iiso are the fourth order deviatoric and isotropic projection tensor, respectively.

An increment of strain may be in an arbitrary direction (i.e. independent of the current principal directions), and the first two terms in C will produce a stress increment in the same direction; however, the last term will produce a term with principal directions equal to that of the current strain which means that the increment in stress generally will have principal directions that differ from the increment in strain.

The question is now, how can the total stress--strain relation be isotropic?

PS: the rich text stuff (such as subscripts) seem to not be working..

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### Re: Anisotropic isotropic material aka deformation induced aniso

You can look at "On the effects of deformation induced anisotropy in isotropic materials" by Fuller and Brannon, 2013, IJNME 37(9), 1079-1094, DOI: 10.1002/nag.1139  for some insight into the problem.

-- Biswajit

### Spot on!

Thank you so much Biswajit, the article is spot on!

I also found the thesis by Fuller which I'm looking forward to reading :)

I'm surprised that this phenomenon is mentioned so scarsely in the litterature, but very good that Brannon (being the supervisor, I guess) does something about it. BIG fan of her :)

-Espen

### Isotropic Stress-Strain Relation

Actually, the relation between stress and elastic strain remains isotropic. Consider the following relation:

s = 2G (e - ep)

In this relation s is the deviatoric part of the stress and e and ep are respectively total deviatoric and plastic strain tensors. From this simple relation it is evident that s is coaxial (share the same principal directions) with (e - ep) and not e itself. It is possible that we might have plastic strain only in x direction (e11p ≠ 0) while there is no plastic strain in y direction (e22p = 0). According to previous relation we can write:

s11 = 2G (e11 - e11p)     and

s22 = 2G (e22)

Clearly, the stress-strain relations are different for x and y directions showing that the isotropy is violated due to the prescence of plastic strain in x direction only.

Mohsen

### Is the isotropy violated?

Thank you for your response, Mohsen.

I'm not quite sure I follow what you say here, do you mean that the relation isotropic or not?

In my post I didn't mention plasticity, but I agree that there are aspects in common in this case as one gets more or less the same stiffness tensor using J2 plasticity. Also I wrote the total stress--strain relation which could perhaps be interpreted as something to do with total strain; what I meant was the opposite of incremental/differential relation.

never the less, I don't follow the argument for why isotropy is lost when there is plastic strain only in the x direction. The principal directions will still coincide, right?

-Espen

### Isotropy or Coaxiality

Dear Espen,

There are a couple of points about your question that should be clarified. They can be itemized as follows:

1. Isotropy means that constitutive behavior remains identical for all material directions. Therefore stress should be independent of prinipal directions of strain (eigenvectors of strain tensor). To this end stress should be related to strain only through invariants of strain tensor. If this is the case isotropy is preserved for all material directions.

2. I would be so grateful if you could provide explicit definitions for εdev ( = ε - 1/3 tr(ε) I ?) and εiso. Then we can be more specific about elasticity tensor.

3. In previous simple example, it has been shown that for two perpendicular directions, the stress-strain relation (constitutive relation) is different and therefore we have deformation induced anisotropy.

Mohsen

### That is the question

I think we can agree that an isotropic operator (in this case the stiffness tensor) produce a result that is coaxial with the argument, and I have always thought that the converse was also true, i.e. that if the result is coaxial with the argument (for an arbitrary argument) then the operator must be isotropic.

Apparently this is not the case, since in my example we can tell that the stress is coaxial with the (total and elastoc) strain.

This is the 'issue' I addressed in the OP, but according to the article cited by Biswajit there is nothing funny going on. I guess what seems to be a source to non-coaxiality will infact turn out to produce coaxial tensors when integrated, though this is just me guessing now.

Your interpretation of εdev is correct, and εiso  is the isotropic part that you subtracted from the total.

I agree that in your plasticity example there is deformation induced anisotropy when cosidering the total strain, but the stress is still coaxial with the elstic strain, right?

-Espen

### Stress and Elastic Strain Coaxiality

Yes, in this case stress is coaxial with elastic strain.

Mohsen

### small strain v.s. large strain

In small strain case (or linear elasticity), D is a constant instead of a function, which results in A = 0. Hence, the last term of C disappers. However, for nonlinear elasticity, things are different.