Single shear element in stress tensor: Finding Von Mises

[jasc15]

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

 

[Mech_Engineer]

I think this is consistent with the calculations of the Von Mises criterion. In the case of a pure shear stress,

according to the load scenario table in the page below.

See below information from Wikipedia:
https://en.wikipedia.org/wiki/Von_Mises_yield_criterion

In the case of pure shear stress, [PLAIN]https://upload.wikimedia.org/math/4/4/d/44dcfff99c0c0c4fc2ab5177bb18be1f.png, while all other [PLAIN]https://upload.wikimedia.org/math/2/5/d/25d584aefa115a4f46e0d761fef717d8.png, von Mises criterion becomes:

[PLAIN]https://upload.wikimedia.org/math/3/c/3/3c3d2ee47e2ecaab57039f34a5cedcb9.png.
This means that, at the onset of yielding, the magnitude of the shear stress in pure shear is

times lower than the tensile stress in the case of simple tension.

 

[jasc15]

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.

 

[Mech_Engineer]

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