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Difference of 1st, 2nd ,3rd principal stretch

  1. Jul 9, 2012 #1
    Dear all,

    I have problem on defining first principal stretch, second principal stretch and third principal stretch.

    Does it means in x axis we definite it as first, y axis is second, third is z axis?

    what if my load is applied on a cubic from the -y direction in a uniaxial testing tension, so my first principal stretch remains as x axis, or will it change?

    I hope someone can understand my statement, if not, do post it. i try to make myself clear.
     
  2. jcsd
  3. Jul 9, 2012 #2

    berkeman

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    Staff: Mentor

    Are you talking about 1st 2nd and 3rd principal stresses in Engineering? If so, I can move this thread to the Engineering forums. If not, it can stay here.
     
  4. Jul 9, 2012 #3
    Hello Mike, i think FK is referring to principal strains not stresses.

    Edit:


    Fruitkiwi,

    Continuing with this guess, I am further guessing that you are referring to the tensor shear strain

    εij = 0.5(εij + εji) i,j = 1,2,3

    and not the engineering strains γ which are twice this.

    Please confirm which system we are working in.

    Hopefully you realise that the axes of principal strain do not, in general, coincide with the normal x,y,z axes?
     
    Last edited: Jul 9, 2012
  5. Jul 9, 2012 #4
    Hi, Berkeman,

    I think it is related to continuum mechanics so i choose to post it here.
    I have not problem on engineering application, however, in terms of physic definition, erm, I have problem.

    Hi, Studiot,

    Thanks for save my post :)
    The system you refer to is Lagrangian or Eulerian system, am i right?
    I am in Eulerian system,
    "since axes of principal strain do not, in general, coincide with the normal x,y,z axes, "
    May i know how can i determine the axes of principal strain? as looks like it is determined by the stretch you prescribed on the material.
     
  6. Jul 10, 2012 #5
    Ok,

    Firstly you again used the term 'stretch'. Further you seem to be working is 3 dimensions.

    We are not sure what you mean by this and I have guessed you mean the conventional word 'strain'.

    But just to be sure we have deformable body with a system of loads (applied forces).

    The applied forces result in stresses within the body. The stress system is the bodies response to these load system.

    The stresses within the body result in strains within that body. Strains are the formal numerical descripion of the deformations.

    However it is also possible for the body to be strained (deformed) without being loaded or stressed. This might occur for instance if the body is unrestrained and heated up so it expands (deforms or strains).

    Are you clear about the difference between stress and strain?

    Please tell us whether you means stress or strain by your term stretch.

    The point of all this is that the direct analysis of strain is more difficult than the direct analysis of stress so we often prefer to analyse the stress and convert to strain via the elastic moduli.

    So we start with a base system of corordinates (X,Y,Z), for convenience.

    The coordinates (x,y,z) identify any point within the body and a set of axes parallel XYZ centred on that point.

    In the unstrained condition consider a small sphere around (x,y,z). All radii are of equal length.

    This sphere is deformed into an ellipsoid, where there are three principal radii. these conform to the directions of the principal strains.

    If and only if the material of the body is isotropic then the principal axes of stress coincide with those of strain.

    Calculating these directions involves solving a cubic equation for either stress or strain. This is not so bad because it can be split into three simultaneous equations.

    So over to you, where do you want to go next?

    BTW what is your first language?
     
  7. Jul 11, 2012 #6
    Dear Studiot,

    Thanks for your detail evaluations.
    I think I can differentiate stress and strain, they have clear distinction, :)
    The one I make mistake is strain and stretch.

    Currently read about Holzapfel "A continuum Approach for engineering" , that's why i use the term stretch. stretch =1+strain, I overlook it.

    My first language is chinese...now I see the reason why we need the coordinate system to describe, ad indicated in the book which is full of equation.

    I will derive it to strengthen the understanding.thanks again.
     
  8. Jul 12, 2012 #7
    Sorry I know nothing of chinese texts, I was hoping that you were going to say you were from a French influenced part of the East, I would then have recommended

    Mecanique de Materiaux Solides by Lemaitre and Chaboche
    Which has an english translation from Cambrideg University Press.

    I don't know Holzapfel and can't find direct reference to 'stretch', however from you comments I think it corresponds to the deformation of a line segment.

    In continuum mechanics the analysis is pointwise - that is strains refer to the limiting displacement and deformation shrunk to a point.
    Note strain and dispalcement are different tensors.

    If, however we consider a short line segment ds between P (x,y,z) and Q (x+dx,y+dy,z+dz) in the undeformed body

    Then P moves to P*(x*,y*,z*) and Q moves to Q*(x*+dx*,y*+dy*,z*+dz*) and ds becomes ds*, so the components of the change are u,v,w

    x* = x+u, y*=y+v, z*=z+w

    and define the quantity εE as


    [tex]{\varepsilon _E} = \frac{{d{s^*} - ds}}{{ds}}[/tex]


    Then some manipulation of differential coefficients leads to


    [tex]\begin{array}{l}
    \left( {1 + {\varepsilon _E}} \right){l^*} = \left( {1 + \frac{{\partial u}}{{\partial x}}} \right)l + \frac{{\partial u}}{{\partial y}}m + \frac{{\partial u}}{{\partial z}}n \\
    \left( {1 + {\varepsilon _E}} \right){m^*} = \frac{{\partial v}}{{\partial x}}l + \left( {1 + \frac{{\partial v}}{{\partial y}}} \right)m + \frac{{\partial v}}{{\partial z}}n \\
    \left( {1 + {\varepsilon _E}} \right){n^*} = \frac{{\partial w}}{{\partial x}}l + \frac{{\partial w}}{{\partial y}}m + \left( {1 + \frac{{\partial w}}{{\partial z}}} \right)n \\
    \end{array}[/tex]

    Where l, m and n are direction cosines of the principal strain axes.

    I wonder if your stretch quantity is (1+εE)

    εE is the engineering strain.

    Edit
    I have found some definitions in my continuum mechanics books defining 'stretch' as εE so there is some difference between authorities about this one.
     
    Last edited: Jul 12, 2012
  9. Jul 12, 2012 #8
    Dear Studiot,

    Thank for agree with me on the stretch term i used.
    I think the strech is 1+εE.

    Finally I can start to read the books.
    "In continuum mechanics the analysis is pointwise - that is strains refer to the limiting displacement and deformation shrunk to a point.
    Note strain and dispalcement are different tensors."

    Thank you.
     
    Last edited: Jul 12, 2012
  10. Jul 12, 2012 #9
    Now we are getting somewhere would you like to refine you question?

    I note that your wiki reference is to stress.

    Does that mean you wish to approach principal axes via stress analysis?

    That would make things much easier.
     
  11. Jul 12, 2012 #10
    Thanks for points out my mistake.
    I edited the post. My mistake, use the wrong image.

    Ya, my intention is to understand the stretch, not stress.
    You answered my question.
     
  12. Jul 13, 2012 #11


    Hi fruitkiwi --

    I've read through this thread, and some good points were made, as well as some confusion. Let me first address general confusion -- I will address your main question shortly.

    If you were to do a tension test on your material, for example, and no rigid body rotation is present, then the "principal stretch" in the longitudinal direction would be "1" plus the quantity of engineering strain that is measured, as you know.

    Strictly speaking, recall (from "polar decomposition") that V (the left stretch tensor) and U (the right stretch tensor) have the same eigenvalues and note that these scalar values (the eigenvalues, λ) are the principal stretches. This is standard terminology in continuum mechanics, as you know.

    The eigenvalues of V and the eigenvalues of U would also be in the same order if no rigid body rotation is present.

    Your question relates to the direction of these principal stretches.
    The eigenvectors of V and U differ when rigid body rotation is present. The direction of principal stretch depends on whether you are working in "spatial" (V) or "material" (U) coordinates.

    In other words, the values of λ are in a certain order and this order will differ, if rigid body rotation is present, depending on how you obtain your values of λ.


    It will be in the y direction.


    This isn't really the important question though, in my opinion.




    My question to you: what direction do the principal stretches act if you perform that tension test while simultaneously performing a rigid body rotation on the specimen?

    The principal stretch magnitude along the length of the specimen is always "1" plus the measured engineering strain. This is one of your eigenvalues.

    However, the direction of principal stretch (the order of your eigenvalues) will depend on how you go about calculating them (due to the rigid body rotation).

    I recommend you solve such a problem (make one up). For example, start with x1=-1/2X1, x2=1/2X2, x3=2X2. Draw what this looks like -- should look like a tension test with a 90degree rigid body motion. Find F, C, B, their eigenvalues and eigenvectors -- etc. This is the great thing about strains (or "stretches") - you can see them!






    I think that the comments regarding stress and isotropy, etc., are important too, but I won't hijack. I have a feeling you already understand which strain measure and stress measures form appropriate pairs anyway.
     
  13. Jul 14, 2012 #12
    Thank you anyway afreiden,

    Quality contributions are always welcome.
     
  14. Jul 14, 2012 #13
    Was a mistake in my little exercise - plus I wanted to write it more clearly:

    Fruitkiwi -- start with:

    [itex]x_1=-\frac{1}{2}X_1[/itex]
    [itex]x_2=\frac{1}{2}X_3[/itex]
    [itex]x_3=2X_2[/itex]

    Also - Studiot:

    I guess what I would mention just to address everything in the thread is that in large strain isotropy some people pair up the Cauchy stress tensor, σ, with the strain tensor B since they behave the same under rigid body motion.

    Similarly, in spatial principal stress space (the main topic of this thread), these folks (or FEM codes) may then choose to pair up the principal Cauchy stresses, [itex]\sigma_i[/itex], with the principal stretches, [itex]\lambda_i[/itex]. In this case, those [itex]\lambda_i[/itex] values are the eigenvalues of V (recall V2=B), which would put the eigenvalues, [itex]\lambda_1, \lambda_2, \lambda_3[/itex], in the order that corresponds to the principal Cauchy stress values, [itex]\sigma_1, \sigma_2, \sigma_3[/itex].

    In linear infinitesimal isotropy, the stress used by the FEM codes is analogous to the Second-Piola-Kirchhoff stress tensor, [itex]\hat{\sigma}[/itex] and the corresponding strain is analogous to C (or U -- since U2=C). I think I talked more about this in another thread awhile back.
     
  15. Jul 14, 2012 #14
    Hello afreiden,

    Thank you again for your comments.

    However as far as I can see fruitkiwi wants to attack strain directly, without necessarily relating it to stress.

    This is more difficult (although in real life we actually measure strain or load not stress) and why I have been feeling my way.

    Further, in the uniaxial tension situation outlined by both afreiden and fruitkiwi there is more to it than meets the eye. Some authors use a magnification factor for the uniaxial stress depending upon the restraint conditions on the other two axes.
     
  16. Jul 14, 2012 #15
    Too much reality for the classical physics forum :-) :-p
     
  17. Jul 14, 2012 #16
    Sometimes it is good to understand the difference between theory and reality.

    Tensor stress analysis describes stress at point. You can't shrink (or expand) a point.

    Consider this.

    Let a bar, of modulus E and poissons ratio 1/m, be stretched by stress s1 in the first principal direction in such a manner as to prevent all lateral strain (e2 and e3).

    Then


    [tex]\begin{array}{l}
    {e_1} = \frac{{{s_1}}}{E} - \frac{{{s_2}}}{{mE}} - \frac{{{s_3}}}{{mE}} \\
    {e_2} = \frac{{{s_2}}}{E} - \frac{{{s_1}}}{{mE}} - \frac{{{s_3}}}{{mE}} = 0 \\
    {e_3} = \frac{{{s_3}}}{E} - \frac{{{s_2}}}{{mE}} - \frac{{{s_1}}}{{mE}} = 0 \\
    \end{array}[/tex]

    A little algebra leads to


    [tex]{s_1} = {e_1}E\frac{{m\left( {m - 1} \right)}}{{\left( {m - 2} \right)\left( {m + 1} \right)}}[/tex]

    Now if the bar is steel with a typical poissons ration of 1/3, then m=3

    This means that the effective modulus in this situation along the first principal axis is modified by a factor of


    [tex]\frac{{3\left( {3 - 1} \right)}}{{\left( {3 - 2} \right)\left( {3 + 1} \right)}} = \frac{6}{4} = 1.5[/tex]

    So to calculate the strain from the stress we need to use an effective modulus of

    Eeffective = 1.5E

    Quite a difference.
     
  18. Jul 16, 2012 #17
    Hi, Afreiden,

    Thanks for physic of classical mechanics. now progressing in this thought
    "what direction do the principal stretches act if you perform that tension test while simultaneously performing a rigid body rotation on the specimen?"

    Hi, Studiot,
    Thanks for the application of continuum mechanics. i thought it always a black hole in continuum mechanics :P

    update your guys after i finish the thought.

    Thanks a lot.
     
  19. Jul 16, 2012 #18
    You need the axis of rotation to answer that question.

    Let us say that the test specimen is a cylindrical bar with the test tension applied along the cylinder axis.

    The principal axes will be

    e1 will be along the cylinder axis

    e2 & e3 will be any pair of orthogonal lines in a disk through the cylinder at right angles to the axis.
    If the spin axis is e1 then the centrifugal effects will be along these axes (e2 & e3) and will therefore subtract from the stresses and strains on these axes since they are of opposite sign.
    That is the poisson strains represent a contraction, the centrifugal strain represents an expansion.

    http://www.roymech.co.uk/Useful_Tables/Mechanics/Rotating_cylinders.html

    edit : Of course since the rotation imposes real external loads and stresses there will be a poisson effect in the e1 direction ie against the applied tension.
     
    Last edited: Jul 16, 2012
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