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Stress Tensor and Motion

  1. Apr 5, 2010 #1
    I have a couple of questions about the stress tensor. I am not an engineering student, so this is the first time I have dealt with internal forces, stress, shears, and such.

    It is my understanding that the entries in the stress tensor are forces per unit area. I assume this means the total force per unit area. Considering the case of a cylinder with axis along the x axis...

    Say you press the two flat ends towards each other, i.e, exert a force F on the left surface and -F on the right, such that the cylinder does not move. Is the T_xx entry in the stress tensor then zero for any point inside the cylinder, since the surface forces should cancel so that the point remains stationary?

    In other words, if I have an object that experiences only surface forces, no body forces, should the stress tensor have all zeros along the diagonal?
     
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  3. Apr 5, 2010 #2

    tiny-tim

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    There is a (possibly) different stress tensor at each point.

    The area in the definition is an infinitesimally small area in the neighbourhood of that point. We can assume that the force is constant over such a small area. :smile:
    The diagonal is the tension (or compression) forces. Everything else is the shear forces.
     
  4. Apr 5, 2010 #3
    I understand all that. I guess my question is, if I look at a point in the medium, and there is a force from the left pointing toward to point, and a force from the right also pointing towards the point, do the forces cancel such that T_xx is zero...
     
  5. Apr 5, 2010 #4

    tiny-tim

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    That's compression … Txx is not zero.
     
  6. Apr 5, 2010 #5
    I could be mistaken, but my current understanding is that Txx (at each point in space where it's defined) is the limit, as area goes to zero, of the normal force-per-unit-area (in other words, pressure) exerted on a surface of constant x by material from the -x side:

    [tex]\lim_{A \rightarrow 0}\frac{F_x}{A_x}[/tex]
     
    Last edited: Apr 5, 2010
  7. Apr 5, 2010 #6

    Andy Resnick

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    If I understand your question, the answer is that, at equilibrium, the stress tensor is *symmetric*. You seem to be asking about Cauchy's law as applied to equilibrium. T_xx is properly interpreted as a pressure, not as some summation of forces.
     
  8. Apr 5, 2010 #7
    Alright, this sort of helps...

    I am trying to get a better understanding of the stress tensor mainly because we are reviewing the Maxwell stress tensor in my E&M class. The professor talks about the stress tensor as representing "momentum flow", i.e, the amount of momentum crossing a surface per unit time. This has the same units as force per area, so it makes sense to talk about it as such.

    My main hangup, as you might be able to see, is that when something is static, I don't see why there should be any momentum flowing into it. The forces cancel identically such that the object remains at rest. Therefore, when I see a nonzero stress tensor with a static object, I see momentum flow, but it doesn't make sense.
     
  9. Apr 5, 2010 #8

    Andy Resnick

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    Oh. I think part of the problem here is bad nomenclature: specifically, how the concepts used in the Cauchy stress tensor get extended to the stress-energy tensor:

    http://en.wikipedia.org/wiki/File:StressEnergyTensor.svg
    http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor

    In electromagnetism, the Poynting vector is associated with momentum, and the stress-energy tensor now has this extra bit that the stress tensor did not. What matters is the *divergence* of the stress tensor, so in equilibrium, there are no sources or sinks of momentum. The momentum flux across a dividing surface (in equilibrium) is continuous, as well.

    Does that help?
     
  10. Apr 5, 2010 #9
    Ah, I think I see part of my confusion now. It is necessary for the stress tensor field to be divergenceless for equilibrium, not the tensor being zero...

    Perhaps you could answer this to solidify my understanding: Say you have some object that you are accelerating straight up in the z direction. To do this, you apply a contact force with your hand to the bottom. What does the T_zz component of the stress tensor look like at a point on the top surface, in the middle, and on the bottom surface?

    My uncertainty is that at the top surface, the point experiences a force only from the solid beneath it, and not from above it, whereas at the bottom there are forces on either side. It seems arbitrary to pick one force from either side to be placed in the stress tensor, so perhaps it is the total force (F_bottom + F_top)?
     
  11. Apr 6, 2010 #10

    Andy Resnick

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    That's not a simple question to answer- it depends on the mechanical properties of the object being accelerated. Is the object perfectly rigid? An elastic solid? A viscous fluid? Something else?

    The idea is to use Cauchy's law (balance of momentum)

    http://en.wikipedia.org/wiki/Continuum_mechanics
     
  12. Apr 6, 2010 #11
    Alright, that makes sense! Thanks.
     
  13. Apr 6, 2010 #12
    I think you are making things unecessarily complicated.

    The stress tensor is applicable to continuous media. It is not applicable to contact forces. Look up St Venant's Principle.

    Contact forces and stresses are covered by a totally different theory, often regarded as part of tribology.

    What about motion?

    Well you will be familiar with electromagnetic waves in dielectric media, when the waves are the eigenfunctions of the harmonic equation.

    It is similar in continuous physical media and stress waves motion can be represented by elastic waves, which are eigensolutions of the biharmonic equation

    [tex]{\nabla ^2}({\nabla ^2}\phi ) = 0[/tex]

    for some stress function, [tex]\phi [/tex]

    Of course some media are actually in motion and we can relate the analysis to the equations of motion to the stresses in the medium.
     
  14. Apr 6, 2010 #13

    Andy Resnick

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    While it is true that there is no microscopic theory of contact forces, they can be incorporated into continuum mechanics fairly easily by means of jump conditions on a boundary.
     
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