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# 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?

Forums:

## 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.

## fourth order tensor invariants

Where can I find formulas for the invariants of the fourth order stiffness tensor? hopefully in contracted notation.

I already verified that the I1=trace(C)=C11+C22+C33+2C44+2C55+2C66 is invariant, where the "2"are needed because of contracted notation.

I'd like to write C*=C-tr(C), then find the second invariant J2=1/2(tr(C*)^2-tr(C*^2)).

The first term is tr(C*)=0,... but the second?

## tr(C*)

You should note that tr(C*) is not zero. If by C* you mean, C*=C-tr(C)I, then:

tr(C*)=tr(C)-tr(C)tr(I)=tr(C)- 3×tr(C)=-2tr(C).

Perhaps you mean C*=dev(C)=C-1/3tr(C)I. Then its first invariant is zero as expected. For the second invariant of C* we can write:

II_C* = 1/2[(tr C*)^2-tr(C*^2)]=-1/2 C*:C*

Knowing that C*:C* = tr(C*^2) and tr(C) = I:C. We can write:

C*:C*=[C-1/3tr(C)I]:[C-1/3tr(C)I]=C:C-1/3tr(C)C:I-1/3tr(C)I:C+1/9(tr C)^2I:I=

tr(C^2)-1/3(tr C)^2-1/3(tr C)^2+1/3(tr C)^2=tr(C^2)-1/3(tr C)^2=tr(C^2)-(tr C)^2+2/3(tr C)^2=-2II_C+2/3 I_C

Therefore we have:

II_C* = II_C - 1/3 I_C

where I_C and II_C are the first and second invariants of C.

Mohsen

## Minor Note

In my previous post please note that I_C is actually I_C^2. In other words:

II_C* = II_C - 1/3 I_C^2

Mohsen

## Knowing that the components

Knowing that the components of a fourth orther tensor from one coordinate system to another transform according to

C'_ijkl = Q_ir Q_js Q_kt Q_lu C_rstu,

it is an easy task to show for example:

C'_iikk = Q_ir Q_is Q_kt Q_ku C_rstu = δ_rs δ_tu C_rstu = C_rrtt

In a similar manner one can verify that C_ijij, C_ijji are also invariant.

Mohsen