Stress invariants

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?


Arash_Yavari's picture

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?


David C. Kellermann's picture

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.