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Airy stress function

  1. Apr 1, 2016 #1
    Hello to everyone, I have a problem with the solution of plane elasticity problems with the method of Airy stress functions.

    For instance I can solve a problem of uniaxial or biaxial uniform tension with a 2nd order polynomial, but if I add shear on only two opposite sides the problem seems to have no solution. Is it possible that I have to formulate the no shear boundary conditions (on the free shear sides) in a weak form? If so I cannot understand the mathematical reason for this. Someone can help me? thank you! (In the attached file a little sketch to clarify my question)

    Attached Files:

  2. jcsd
  3. Apr 1, 2016 #2


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    Is that no solution at all, or no "nice" solutions in the form of polynomials?

    Often times, elasticity problems don't have "nice" solutions in terms of polynomials, but can be solved using numerical techniques. That's one reason why finite element and boundary element techniques are used.
  4. Apr 1, 2016 #3
    I mean nice solution in polynomial form. What I cannot really realize is why simple uniaxial tension admits such a solution, while the application of uniform shear on two opposite sides does not!
  5. Apr 1, 2016 #4
    In the third case, the body is not in equilibrium (unless some kinematic boundary conditions are applied), because it can rotate anticlockwise. In other words, this problem falls outside of the scope of the mechanics of deformable solid bodies.

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