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Mechanical Engineering
Single shear element in stress tensor: Finding Von Mises
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[QUOTE="jasc15, post: 5454442, member: 49304"] When finding the Von Mises of given a stress tensor who's only element is a single shear component (τ): \begin{bmatrix} 0 & τ & 0\\ τ & 0 & 0\\ 0 & 0 & 0 \end{bmatrix} 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 [/QUOTE]
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Single shear element in stress tensor: Finding Von Mises
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