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second invariant of rate-of-strain tensor

The shear, or strain, rate is often calculated based on the square root of the second invariant of rate-of-strain tensor. The tensor itself is made up of all the possible deformation of a fluid element, which includes volumetric and shear deformation. I would like to invite comments from everyone concerning:

1) if this second invariant, in its general definition, includes both the volumetric-rate of deformation and shear-rate of deformation? Or simply shear-rate of deformation alone?

2) the physical meaning of the 'second invariant' of the strain rate tensor. I take it as a way to 'average' all the strain components in the tensor, thus an 'effective' strain rate.

I look forward for your comments.

Cheers.

Re: second invariant of rate-of-strain tensor

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Chun Min Chew,<o:p></o:p>

<o:p>*</o:p>

The origin of the invariants is from tensor analysis: For any tensor Aik, you may wish to find its eigenvalues, corresponding to the principal components (i.e., the rotation to a coordinate system where Aik becomes a diagonal matrix).

<o:p>*</o:p>

Let us concentrate on symmetric tensor only, such as the stress and strain (or-strain-rate). The eigenvalues (principle componenets), l, are found from the characteristic polynomial

|A-lI| = – l3 + I1 l2 – I2 l + I3 = 0

where the invariants are

I1 = Akk (i.e. the trace of Aik)

I2 = ½ (AiiAkk – AikAik)

I1 = |Aik|(i.e. the determinant of Aik)

<o:p>*</o:p>

If you decompose A to its deviatoric and volumetric parts, say S and v, respectively

v = 1/3 tr(A)

S = A -vI

and substitute in the 2nd invariant, I2, you will notice it becomes

I2 = –½SikSik

i.e., the volumetric part is absent, and only deviatoric components are influencing it.

<o:p>*</o:p>

Returning to your question, it is true that the 2nd invariant of the strain-rate is independent of the volumetric strain. However, if you consider a uniaxial case, where the only non-zero entry is, say A11, there is no shear (but the deviator S is non-zero, having diagonal entries) and* the result in this case is I2 = 2/3 a2.

<o:p>*</o:p>

I hope it answers your question,

Rami

<o:p>*</o:p>

<o:p>*</o:p>

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Chun Min Chew,

Sorry, there was a mistake in my former posting, leading to a wrong conclusion. Here is the corrected version.<o:p></o:p>

The origin of the invariants is from tensor analysis: For any tensor Aik, you may wish to find its eigenvalues, corresponding to the principal components (i.e., the rotation to a coordinate system where Aik becomes a diagonal matrix).

Let us concentrate on symmetric tensor only, such as the stress and strain (or strain-rate). The eigenvalues (principle componenets), l, are found from the characteristic polynomial

|A-lI| = – l3 + I1 l2 – I2 l + I3 = 0

where the invariants are

I1 = Akk (i.e. the trace of Aik)

I2 = ½ (AiiAkk – AikAik)

I3 = |Aik| (i.e. the determinant of Aik)

If you decompose A to its deviatoric and volumetric parts, say S and v, respectively

v = 1/3 tr(A)

S = A -vI

and substitute in the 2nd invariant, I2, you will notice it becomes

I2 = 3v2 – ½SikSik

Therefore, for a pure volumetric tensor (A = vI, and therefore S = 0) we get I2 = 3v2. On the other hand, for pure deviatoric tensor (v=0, A = S), the 2nd invariant of course has no contribution from the volumetric part.

Rami

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Re: Correction...

So, if the tensor A is decomposed into its deviatoric and volumetric tensors, the 2nd invariant of the deviatoric tensor would not contain the volumetric part; while the 2nd invariant for the volumetric tensor would of course contain the voluemtric part.

Then, from the equation for the 2nd invariant: I2 = ½ (AiiAkk – AikAik), does it imply that for a complete tensor, A, the 2nd invariant will include the contributions from both volumetric and deviatoric (non-volume) part?

Thanks again.