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Chun Min Chew December 9, 2003 03:02

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.

Rami December 9, 2003 11:04

Re: second invariant of rate-of-strain tensor
 
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<p class=MsoNormal style='text-align:justify'>Chun Min Chew,<o:p></o:p>


<p class=MsoNormal style='text-align:justify'><o:p>*</o:p>


<p class=MsoNormal style='text-align:justify'>The origin of the invariants is from tensor analysis: For any tensor <span class=SpellE>A<sub>ik</sub></span>, you may wish to find its <span class=SpellE>eigenvalues</span>, corresponding to the principal components (i.e., the rotation to a coordinate system where <span class=SpellE>A<sub>ik</sub></span> becomes a diagonal matrix).


<p class=MsoNormal style='text-align:justify'><o:p>*</o:p>


<p class=MsoNormal style='text-align:justify'>Let us concentrate on symmetric tensor only, such as the stress and strain (or-strain-rate). The <span class=SpellE>eigenvalues</span> (principle <span class=SpellE>componenets</span>), <span style='font-family:Symbol'>l</span>, are found from the characteristic polynomial


<p class=MsoNormal style='margin-left:.5in'>|A-<span style='font-family:Symbol'>l</span>I| = <span style='font-family:Symbol'>l</span><sup>3</sup> + I<sub>1</sub><span style='font-family:Symbol'> l</span><sup>2</sup> I<sub>2</sub><span style='font-family:Symbol'> l</span> + I<sub>3</sub> = 0


<p class=MsoNormal style='text-align:justify'><span class=GramE>where</span> the invariants are


<p class=MsoNormal style='margin-left:.5in;text-align:justify'>I<sub>1</sub> = <span class=SpellE>A<sub>kk</sub></span> (i.e. the trace of <span class=SpellE>A<sub>ik</sub></span>)


<p class=MsoNormal style='margin-left:.5in;text-align:justify'>I<sub>2</sub> = (<span class=SpellE>A<sub>ii</sub>A<sub>kk</sub></span> <span class=SpellE>A<sub>ik</sub>A<sub>ik</sub></span>)


<p class=MsoNormal style='margin-left:.5in;text-align:justify'>I<sub>1</sub> = |<span class=SpellE>A<sub>ik</sub></span><span class=GramE>|(</span>i.e. the determinant of <span class=SpellE>A<sub>ik</sub></span>)


<p class=MsoNormal style='text-align:justify'><o:p>*</o:p>


<p class=MsoNormal style='text-align:justify'>If you decompose A to its <span class=SpellE>deviatoric</span> and volumetric parts, say S and v, respectively


<p class=MsoNormal style='margin-left:.5in;text-align:justify'>v = 1/3 <span class=SpellE><span class=GramE>tr</span></span><span class=GramE>(</span>A)


<p class=MsoNormal style='margin-left:.5in;text-align:justify'>S = A -<span class=SpellE><span class=GramE>vI</span></span>


<p class=MsoNormal style='text-align:justify'><span class=GramE>and</span> substitute in the 2<sup>nd</sup> invariant, I2, you will notice it becomes


<p class=MsoNormal style='margin-left:.5in;text-align:justify'>I<sub>2</sub> = S<sub>ik</sub>S<sub>ik</sub>


<p class=MsoNormal style='text-align:justify'>i.e., the volumetric part is absent, and only <span class=SpellE>deviatoric</span> components are influencing it.


<p class=MsoNormal style='text-align:justify'><o:p>*</o:p>


<p class=MsoNormal style='text-align:justify'>Returning to your question, it is true that the 2<sup>nd</sup> invariant of the strain-rate is independent of the volumetric strain. However, if you consider a <span class=SpellE>uniaxial</span> case, where the only non-zero entry is, say A<sub>11</sub>, there is no shear (but the deviator S is non-zero, having diagonal entries) and<span style='mso-spacerun:yes'>* </span>the result in this case is I<sub>2</sub> = 2/3 a<sup>2</sup>.


<p class=MsoNormal style='text-align:justify'><o:p>*</o:p>


<p class=MsoNormal align=center style='text-align:center'>I hope it answers your question,


<p class=MsoNormal align=center style='text-align:center'><span class=SpellE>Rami</span>


<p class=MsoNormal style='text-align:justify'><o:p>*</o:p>


<p class=MsoNormal style='text-align:justify'><o:p>*</o:p>


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Rami December 10, 2003 03:24

Correction...
 
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<p class=MsoNormal style='margin-top:12.0pt;text-align:justify'>Chun Min Chew,


<p class=MsoNormal style='margin-top:12.0pt;text-align:justify'><span style='color:red'>Sorry, there was a mistake in my former posting, leading to a wrong conclusion. Here is the corrected version.<o:p></o:p></span>


<p class=MsoNormal style='margin-top:12.0pt;text-align:justify'>The origin of the invariants is from tensor analysis: For any tensor <span class=SpellE>A<sub>ik</sub></span>, you may wish to find its <span class=SpellE>eigenvalues</span>, corresponding to the principal components (i.e., the rotation to a coordinate system where <span class=SpellE>A<sub>ik</sub></span> becomes a diagonal matrix).


<p class=MsoNormal style='margin-top:12.0pt;text-align:justify'>Let us concentrate on symmetric tensor only, such as the stress and strain (or strain-rate). The <span class=SpellE>eigenvalues</span> (principle <span class=SpellE>componenets</span>), <span style='font-family:Symbol'>l</span>, are found from the characteristic polynomial


<p class=MsoNormal style='margin-top:12.0pt;margin-right:0in;margin-bottom: 0in;margin-left:.5in;margin-bottom:.0001pt'>|A-<span style='font-family: Symbol'>l</span>I| = <span style='font-family:Symbol'>l</span><sup>3</sup> + I<sub>1</sub><span style='font-family:Symbol'> l</span><sup>2</sup> I<sub>2</sub><span style='font-family:Symbol'> l</span> + I<sub>3</sub> = 0


<p class=MsoNormal style='margin-top:12.0pt;text-align:justify'><span class=GramE>where</span> the invariants are


<p class=MsoNormal style='margin-top:12.0pt;margin-right:0in;margin-bottom: 0in;margin-left:.5in;margin-bottom:.0001pt;text-align:justify'>I<sub>1</sub> = <span class=SpellE>A<sub>kk</sub></span> (i.e. the trace of <span class=SpellE>A<sub>ik</sub></span>)


<p class=MsoNormal style='margin-top:12.0pt;margin-right:0in;margin-bottom: 0in;margin-left:.5in;margin-bottom:.0001pt;text-align:justify'>I<sub>2</sub> = (<span class=SpellE>A<sub>ii</sub>A<sub>kk</sub></span> <span class=SpellE>A<sub>ik</sub>A<sub>ik</sub></span>)


<p class=MsoNormal style='margin-top:12.0pt;margin-right:0in;margin-bottom: 0in;margin-left:.5in;margin-bottom:.0001pt;text-align:justify'>I<sub>3</sub> = |<span class=SpellE>A<sub>ik</sub></span>| (i.e. the determinant of <span class=SpellE>A<sub>ik</sub></span>)


<p class=MsoNormal style='margin-top:12.0pt;text-align:justify'>If you decompose A to its <span class=SpellE>deviatoric</span> and volumetric parts, say S and v, respectively


<p class=MsoNormal style='margin-top:12.0pt;margin-right:0in;margin-bottom: 0in;margin-left:.5in;margin-bottom:.0001pt;text-align:justify'>v = 1/3 <span class=SpellE><span class=GramE>tr</span></span><span class=GramE>(</span>A)


<p class=MsoNormal style='margin-top:12.0pt;margin-right:0in;margin-bottom: 0in;margin-left:.5in;margin-bottom:.0001pt;text-align:justify'>S = A -<span class=SpellE><span class=GramE>vI</span></span>


<p class=MsoNormal style='margin-top:12.0pt;text-align:justify'><span class=GramE>and</span> substitute in the 2<sup>nd</sup> invariant, I<sub>2</sub>, you will notice it becomes


<p class=MsoNormal style='margin-top:12.0pt;margin-right:0in;margin-bottom: 0in;margin-left:.5in;margin-bottom:.0001pt;text-align:justify'>I<sub>2</sub> = 3v<sup>2</sup> S<sub>ik</sub>S<sub>ik</sub>


<p class=MsoNormal style='margin-top:12.0pt;text-align:justify'>Therefore, for a pure volumetric tensor (A = <span class=SpellE>vI</span>, and therefore S = 0) we get I<sub>2</sub> = 3v<sup>2</sup>. On the other hand, for pure <span class=SpellE>deviatoric</span> tensor (v=0, A = S), the 2<sup>nd</sup> invariant of course has no contribution from the volumetric part.


<p class=MsoNormal align=center style='margin-top:12.0pt;text-align:center'><span class=SpellE>Rami</span>


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Chun Min Chew December 10, 2003 12:34

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 = (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.


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