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
In reply to Re: Stress invariants by arash_yavari
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 reply to Julie Dear Sir, Sorry by bruno-page
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?
In reply to fourth order tensor invariants by ebarbero
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
In reply to tr(C*) by M. Jahanshahi
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