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Cauchy green strain tensor

  1. May 14, 2012 #1
    dear all,

    as a newbie in solid mechanics modelling, i always come across these few terms,
    Cauchy-Green strain tensor
    Green Lagrange strain tensor
    isochoric Cauchy green strain tensor.

    Consider a cubic, when we move the top face, while fixing the bottom face, we will able to see the strain occur, then it generate stress. May i know is this fall in which type of tensor?
    is it difference in terms of vector direction or plane motion?

    i am struggling to explain it in to myself.
     
  2. jcsd
  3. May 14, 2012 #2

    AlephZero

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    Be careful about the terminology. Usually the Cauchy-Green tensor means a deformation tensor not a strain tensor. The Green Lagrange strain tensor is the "strain part" of the Cauchy-Green defiormation tensor. The "strain" is what is left when you take away the rigid body translation and rotation from the "deformation".

    Sometimes it is useful to split the strain into two parts, the change in volume (volumetric strain) and the deformation at constant volume (isochoric strain).
     
  4. May 14, 2012 #3
    Thanks for reply.
    Yesterday debate with some newbies, following the reply. I conclude that deformation include rotation and translation,but strain we only have translation, so strain tensor is part of deformation tensor. Here comes new Q, so strain you only can measure in 1D, then times other tensor, to form a deformation tensor in 3D?


    feel free to correct me.
     
  5. May 14, 2012 #4

    AlephZero

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    There are two types of strain. When you say "you can only measure strain in 1D" you are probably thinking about direct strain or elongation. That measures the amount of stretching or compressing of the object in one direction. For a three dimensional object, there are three independent direct strains in directions perpendicular to each other (e.g the amount of stretching in the X Y and Z directions).

    There is also "shear strain" which measures the deformation of a rectangle into a parallelogram, without changing its area. Again there are three independent shear strains, for example in the XY plane, the YZ plane and the ZX plane.

    The 6 strain components (3 direct and 3 shear) make up the strain tensor. There are 6 components not 9, because the strain tensor is always symmetric.
     
  6. May 15, 2012 #5
    What AlephZero said is all correct.

    To be clear, the Lagrangian Strain tensor and the Left C-G deformation tensor both contain information about normal and shear strains.

    However:
    1) The Lagrangian Strain tensor doesn't change under rigid body rotation, nor does the Second Piola-Kirchhoff Stress Tensor.

    2) The Left C-G deformation tensor does change under rigid body rotation, as does the Cauchy Stress.

    Either stress strain pair can be used, generally speaking.
     
  7. May 15, 2012 #6
    Thanks for AlephZero and afreiden input, it give me a deep thought on the physical meaning. because i only know how to calculate,but at the end it give me a question what i am working for.

    i always thought the basic term of stress is Cauchy stress.and according to my understanding, the 2nd PK stress tensor can be derived from the Cauchy stress. What afreiden say is true under rigid body rotation.erm, but i thought rigid body should not deform as it is not elastic material.(pls correct me if i am wrong)

    Previously AlephZero say that Cauchy Green tensor is a deformation tensor, Lagrange strain is part of it. Then only a Cauchy Green strain tensor is enough to describe deformation phenomenan, why suddenly we need Lagrange strain tensor come out to describe the strain part?

    i believe that there should be some reason why these two tensor coexist.anyone to share with me?
     
  8. May 15, 2012 #7

    It's a good question.


    In finite (large strain, nonlinear) "hyper"elasticity, which is where you are presumably headed, you will find that some authors develop a stress strain relationship using pair #1 and others use pair #2.

    Regardless, you will find that you need to switch back and forth between the two pairs at least once when performing the derivation.



    Let me briefly elaborate:

    Consider that the whole point of this branch of continuum mechanics is to develop the equations used in finite element simulations.

    These equations primarily consist of:
    a) some relationship between stress and strain that depends only on the material (F=kx, essentially -- but we're dealing with complicated materials like rubber)

    b) some equilibrium relationship involving stress (F=ma, essentially)


    Since our reference frame might be rotating (rigid body rotation is occurring) in our finite element simulation, the derivation of "a" begins, initially, within a coordinate system that rotates as well. This is because we need the stress to be a function of material strain and not a function of rigid body rotation. So, we always begin, initially, with stress strain relationship #1, since this relationship is in fact invariant under (rotating with) rigid body rotation.

    However: Since we are talking about finite elements, we need adjacent elements to interact with each other properly. In other words, we need a COMMON frame of reference for the purpose of looking at stress equilibrium (Newton's second and third laws, essentially). If the need for a common frame of reference isn't clear, it may be helpful to consider two separate objects that are impacting each other, which is something that is, in fact, commonly considered in computer simulations. Hopefully you can then see why we use the Cauchy Stress for our treatment of equilibrium (i.e. stress strain relationship #2 will be unavoidable when equilibrium between elements is considered).

    If you have other questions, shoot
     
  9. May 15, 2012 #8

    AlephZero

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    There are two different "big ideas" about how to model the motion and deformation of a body.

    One is to refer everything to back to the initial position of the body. The other is to work in a coordinate system (or reference frame) that is attached to a particle of the body and moves with it. Both these are useful for different types of problem, and (without going into any details of the math) it chould be clear they are related to each other even though the details of the equations look different. The two approaches are often described as "Eulerian" and "Lagrangian", after the people who first used them.

    As an example of the difference, suppose there is a thin rod lying along the X axis, and you apply equal and opposite forces to the ends to stretch it. Assuming Poisson's ratio is zero to keep things simple, the only non-zero terms in the stress and strain tensors are ##\sigma_{xx}## and ##\epsilon_{xx}##.

    Now, while keeping it stretched, rotate the rod through 90 degrees so it lies along the Y axis. If you work in a system fixed in space, the stress and strain in the body will change so ##\sigma_{xx}## and ##\epsilon_{xx}## are zero and ##\sigma_{yy}## and ##\epsilon_{yy}## are non zero. If you work in a system attached to the body, the stress tensor and strain tensors are not changed by the rotation through 90 degrees, but the direction of "X" that the tensors refer to changes instead.

    Now suppose this "rod" was actually a strand of carbon fiber in a composite materal. If you want to understand the motion of the object as a whole, the first approach may be simpler. But if you want to know if that strand of carbon fiber will break or not, the second approach makes more sense, because it automatically keeps track of which direction every fiber in the body is pointing in, as the body deforms.

    Just as an example of how complcated it can get keeping track of where everything is, suppose the original rod was flexible like a piece of string, and you tied a knot in it before you pulled on the two ends. You probably won't be solving problems like that in a first course (!!!) but the general formulation of continuum mechanics has to be able to deal with situations like that!
     
    Last edited: May 15, 2012
  10. May 15, 2012 #9
    Thanks for both of you try to keep thing simple and interesting to me.and not asking me stop think too much.
    I assume both of you have experienced in solid continuum mechanics , may i know why we use finite strain theory to describe large strain while infinitesimal to describe small strain? My opinion is when you squeeze a particle, you have infinite thing you can squeeze, hence you can call theory " infinitesimal strain theory" to describe it.

    I am reading book A First course in Continuum mechanics written by Y.C Fung. Throughout the book, it use Cauchy stress -strain tensor to desribe the equilibrium. It actually shows what exactly afreiden said. Then based on the book and afreiden, every material should possess both stress-strain pair, just in terms of Eulerian(absolute) and lagrangian(relative) differences. However, that is not the case i have been taught during my education. in books we use Cachy stress and strain to describe the linear elastic material, while normally people favourite to use 2nd PK to describe nonlineary strain relationship as in hyperelastic material. Actually we can also use Cauchy stress tensor and CG strain tensor to describe hyperelastic material, am i correct?

    But today i come accross a paper, there is not the only way, people present their results in the research paper in stretch ratio. What a suprise, strain means change of length/ original length, strech ratio means after strech the displacement / original length, may i know the reason behind why researchers prefer stretch ratio rather than strain rate?

    Dear AlepthZero, hope it wont make you feel i am picky and troublesome. I assume you are talking about "how to model the model and deformation a free body". If you work in a system fixed in space, while keeping it streched in X axis rotate the rod through 90 degresss and it lies along the Y axis , the stress and strain in the body will change, but in my opinion, σxx and ϵxx should not be zero, rememeber we strech it in x axis, before rotate it, right?

    If we attached ourselves in the body, consider we are the particles, we do not feel change of stress tensor and strain tensor,so here the stress-strain you refer is 2nd PK stress and Lagrangian tensor .
    at your last sentence, what i understand your meaning is if we use 2nd PK stress and lagrangian tensor is complicated, but if we use Cachy stress to deal, it typically can give us convenience in description a movement of rod to others, right?
     
  11. May 16, 2012 #10

    Finite strain involves large strains - anything from 1% to 1000% (e.x. polymers) is fair game.

    "Finite strain elasticity" and "hyperelasticity" are used interchangeably.

    Linear infinitesimal elasticity involves small strains - higher order components of the Lagrangian strain tensor are neglected.

    I hope that answers that part of your question, but I'm not sure I entirely understood the question.



    You are correct that one can use the Cauchy stress and Left C-G tensor in hyperelasticity - a matter of preference. As for linear infinitesimal elasticity, let me make a comment, because I detect some confusion about the stress that is used here.

    Firstly, the 2nd PK stress and the Cauchy stress are NOT the same, even in linear infinitesimal elasticity. Perhaps the magnitude of their components are the same for linear infinitesimal elasticity, but these components are certainly arranged differently when rigid body rotations are present.

    Authors do refer to the linear infinitesimal stress as the Cauchy stress. It may make more sense to think of the linear infinitesimal material stress as the 2nd PK stress if you are figuring out how to deal with rigid body rotations. This "confusion" is mentioned briefly by Arash Yavari here: http://imechanica.org/node/5277



    You mean in finite [hyper] elasticity: the principal cauchy stresses can be defined in terms of the principal stretches (eigenvalues of U or V). Is this what you are talking about? It's just another way of formulating the stress-strain relationship. They are all equally accurate. It is a desirable form for relating stress to strain for certain "material models" such as the "Ogden model."
     
  12. May 16, 2012 #11
    Hi, afreiden,

    Yup, you answer all my curiosity. I agree with what you share here.
    I once be arrogant when i stand in front of solid mechanics. but after seeing your explanation, i realize i am the so small compared to all of you.

    I will dig into solid mechanics deeply. thanks for your input. I will make new post if i still dont understand about it.

    so we wait for AlephZero, i am worrying i may misunderstand what he mention about the example.as we always say," dont understand is fine, if you do not understand, but you think you understand, that is dangerous." Hence i always come to ask people.
     
  13. May 16, 2012 #12

    AlephZero

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    Don't try to read too much into my example. It was meant to get you thinking physically about what this is all about, and realize there are two different ways of looking at the situation. You seem to be doing that pretty well! The diagrams in textbooks are often fairly shapeless "blobs" that don't seem to relate to anything in the "real world".

    Keep an open mind on which description you think is "better". Eventually you might decide they they both have good and bad features.
     
  14. May 17, 2012 #13
    Hi, AlephZero,

    Thanks for your experience sharing. I fully agree with what you say. Continuum seems no that boring for me now after communicating with you all. start to see star shining in the dark. :)

    Again, thanks for you and afreiden input.
     
  15. Sep 21, 2012 #14
    this was useful discussion
     
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