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Single shear element in stress tensor: Finding Von Mises

  1. Apr 25, 2016 #1
    When finding the Von Mises of given a stress tensor who's only element is a single shear component (τ):

    0 & τ & 0\\
    τ & 0 & 0\\
    0 & 0 & 0
    the result is simply √3×τ. Is the Von Mises criterion not valid when considering a single component as in this example? I can't seem to reconcile that a calculated shear stress (say by a simple shaft twisting where τ=Tc/J) should be multiplied by √3. I understand that when using the Von Mises yield criterion, it is to be compared to the uniaxial yield allowable, not the shear allowable. However, the yield allowable is not √3 greater than the shear allowable.

    In the example I am actually considering there is also normal component, but I want to see the effect of the two components in isolation as well as combined
  2. jcsd
  3. Apr 25, 2016 #2


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    I think this is consistent with the calculations of the Von Mises criterion. In the case of a pure shear stress, b9639ecbd343db3d22c0be8c5155911c.png according to the load scenario table in the page below.

    See below information from Wikipedia:

    Last edited by a moderator: May 7, 2017
  4. Apr 25, 2016 #3
    Thanks for the reply. How am I to understand this from the point of view of a simple strength of materials problem, where I am asked to find the shear stress in a shaft with an applied torque, and to compare it to the shear allowable for that material.

    For example, a torque of 150 in*lb and a shaft radius of 0.15 in results in an applied shear stress of

    τ = T*c/J = 150*0.15/(Π*.152) = 28,294 psi.

    The shear allowable of the material I am working with is 31000 psi, and the uniaxial yield allowable is 36000 psi.

    Using the simple shear stress calculated, I get a margin of safety of 31,000/28,294 -1 = 0.10

    Using Von Mises, I get √3*28,294 psi = 49,007 psi, with a resulting margin of 36,000/49,007 -1 = -0.27

    Is this simply a matter of choosing how conservative you want to be in design? I wouldn't be comfortable using only the first method knowing the second method results in negative margin.
  5. May 9, 2016 #4


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    Generally speaking I would lean towards the Von Mises criterion for your margin, especially since you have such a small shaft for the torque.
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