CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > Main CFD Forum

second invariant of rate-of-strain tensor

Register Blogs Members List Search Today's Posts Mark Forums Read

Reply
 
LinkBack Thread Tools Display Modes
Old   December 9, 2003, 03:02
Default 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.
  Reply With Quote

Old   December 9, 2003, 11:04
Default 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> = (<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> = 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>*</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>
  Reply With Quote

Old   December 10, 2003, 03:24
Default 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> = (<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>


</div>

</body>

</html>
  Reply With Quote

Old   December 10, 2003, 12:34
Default 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 = (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.
  Reply With Quote

Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
strain rate in UDF Paulina FLUENT 11 November 20, 2014 04:06
strain rate magnitude ahmadbakri STAR-CCM+ 1 June 3, 2010 14:03
About "Shear strain rate" leo1985 CFX 0 April 26, 2008 05:48
Strain rate. MM? Main CFD Forum 1 February 15, 2006 09:55
Rate of Strain Tensor from ANSYS FLOTRAN H Mayer Main CFD Forum 0 June 27, 2005 12:56


All times are GMT -4. The time now is 09:27.