second invariant of rate-of-strain tensor

jotac · December 9, 2003, 03:02

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.

December 9, 2003, 11:04

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Chun Min Chew,<o

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>

<o

>*</o

>

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

>*</o

>

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

>*</o

>

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

>*</o

>

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

>*</o

>

I hope it answers your question,

Rami

<o

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<o

>*</o

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holzkiste · December 10, 2003, 03:24

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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></o>

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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December 10, 2003, 12:34

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.

December 9, 2003, 03:02	second invariant of rate-of-strain tensor	#1
Chun Min Chew Guest Posts: n/a	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. jotac likes this.

December 9, 2003, 11:04	Re: second invariant of rate-of-strain tensor	#2
Rami Guest Posts: n/a	<html xmlns="urn:schemas-microsoft-comfficeffice" xmlns:w="urn:schemas-microsoft-comffice:word" xmlns="http://www.w3.org/TR/REC-html40"> <head> <meta http-equiv=Content-Type content="text/html; charset=windows-1255"> <meta name=ProgId content=Word.Document> <meta name=Generator content="Microsoft Word 10"> <meta name=Originator content="Microsoft Word 10"> <link rel=File-List href="invariants_files/filelist.xml"> <title>The origin of the invariants is from tensor analysis: For any tensor Aik, you may wish to find its eigenvalues, corresponding </title> <!--[if gte mso 9]><xml> <w:WordDocument> <w:Zoom>BestFit</w:Zoom> <w:SpellingState>Clean</w:SpellingState> <w:GrammarState>Clean</w:GrammarState> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--> <style> </style> <!--[if gte mso 10]> <style> /* Style Definitions / table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman";} </style> <![endif]--> </head> <body lang=EN-US style='tab-interval:.5in'> <div class=Section1> <p class=MsoNormal style='text-align:justify'>Chun Min Chew,<o></o> <p class=MsoNormal style='text-align:justify'><o></o> <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></o> <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> = ½ (<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></o> <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> = –½S<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></o> <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></o> <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></o> <p class=MsoNormal style='text-align:justify'><o>*</o> </div> </body> </html>

December 10, 2003, 03:24	Correction...	#3
Rami Guest Posts: n/a	<html xmlns="urn:schemas-microsoft-comfficeffice" xmlns:w="urn:schemas-microsoft-comffice:word" xmlns="http://www.w3.org/TR/REC-html40"> <head> <meta http-equiv=Content-Type content="text/html; charset=windows-1255"> <meta name=ProgId content=Word.Document> <meta name=Generator content="Microsoft Word 10"> <meta name=Originator content="Microsoft Word 10"> <link rel=File-List href="invariants-2_files/filelist.xml"> <title>The origin of the invariants is from tensor analysis: For any tensor Aik, you may wish to find its eigenvalues, corresponding </title> <!--[if gte mso 9]><xml> <w:WordDocument> <w:Zoom>BestFit</w:Zoom> <w:SpellingState>Clean</w:SpellingState> <w:GrammarState>Clean</w:GrammarState> <w:Compatibility> <w:BreakWrappedTables/> <w:SnapToGridInCell/> <w:WrapTextWithPunct/> <w:UseAsianBreakRules/> </w:Compatibility> <w:BrowserLevel>MicrosoftInternetExplorer4</w:BrowserLevel> </w:WordDocument> </xml><![endif]--> <style> </style> <!--[if gte mso 10]> <style> /* Style Definitions / table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:10.0pt; font-family:"Times New Roman";} </style> <![endif]--> </head> <body lang=EN-US style='tab-interval:.5in'> <div class=Section1> <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></o></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> = ½ (<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> – ½S<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> </div> </body> </html> holzkiste likes this.

December 10, 2003, 12:34	Re: Correction...	#4
Chun Min Chew Guest Posts: n/a	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.

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