# Stress invariants

Submitted by bruno-page on Sun, 2009-08-23 14:38.

This is a very fundmantal question.

What is the physical significance of stress invariants?

I understand that the stress invariant J2 of the deviatoric stress tensor is used to ecpress the yield criteria-but this is same as Von Mises yield criteria-I want to know what is so special about-stess invariants?

I understand that these invariants remain unaltered by rotation/transformation of the axis-is this the only reason for being so special or there i any other reason as well?

## Re: Stress invariants

Dear Bruno:

Assuming that by stress you mean Cauchy stress (it's symmetric), the following result is known. A scalar function f of stress is invariant under orthogonal transformations if and only if it is a function of the three invariants of stress, i.e. f=f(I_1, I_2, I_3). This means that the number of arguments in f is reduced from 6 to 3. Of course, you can replace Cauchy stress by any symmetric 2-tensor.

In plasticity, J_1 is zero by definition and J_3 is usually ignored. This is why you only see J_2 in defining yield surfaces.

In nonlinear elasticity, the above theorem tells you that internal energy density depends on the invariants of C (right Cauchy-Green tensor) and hence there are only three elastic constants (all explicitly depending on invariants of C) that relate C and S (the second Piola-Kirchhoff stress).

Regards,

Arash

## Julie Dear Sir, Sorry

Julie

Dear Sir,

Sorry for a basic question.

Can you kindly explain,what you mean when you say:

'A scalar function f of stress is invariant under orthogonal transformations if and only if it is a function of the three invariants of stress, i.e. f=f(I_1, I_2, I_3).'What you mean by orthogonal transformation here?

## In this case, "orthogonal

In this case, "orthogonal transformation" refers to rotation, both in the sense of rigid body rotation and of rotation of observational frame.