Why is shear stress equal to half of yield stress

Why is the value for acceptable shear stress equals to half of yield stress? $$\tau = \frac{\sigma_{yield}}{2}$$ P.s Along the math behind it would be possible to explain this with visual consepts?
P.p.s. edits are welcome

• It is just an acceptable safety factor. Yield stress is where something fails. You don't want to design too close to that because of unexpected loads or protection from mistakes or bumps or what have you. – mmesser314 Oct 29 '17 at 3:17
• "Why is the value for acceptable shear stress equals to half of yield stress?" - The answer should be covered in most introductory textbooks on materials science. The answer comes almost directly from straightforward definitions of the yield stress and shear stress. – Samuel Weir Oct 29 '17 at 4:07
• @mmesser314 Hussein gives the correct answer below. – Chemomechanics Oct 29 '17 at 14:35
• @Chemomechanics - So he does. My bad. Thanks for letting me know. – mmesser314 Oct 29 '17 at 14:39

Plastic deformation occurs when the applied stress causes atomic planes to slide with respect to one another. You can see then why shear is a more "stressful" state than traction: it is more likely to cause sliding. For the same reason, a hydrostatic pressure (usually) does not contribute to the yield stress and only the deviatoric does.

With that in mind, say you have a (2D) traction stress state $$\begin{bmatrix} \sigma & \\ & 0 \end{bmatrix}$$ which is equivalent, as far as yield stress is concerned, to $$\begin{bmatrix} \sigma/2 & \\ & -\sigma/2 \end{bmatrix}$$ Now turn your head by $\pi/4$ and this is the same as $$\begin{bmatrix} 0 & \sigma/2\\ \sigma/2 & 0 \end{bmatrix}$$ All in all, a traction stress of $\sigma$ is equivalent to a shear stress of $\sigma/2$.

• This may be a silly question, but can you clarify why you split stress_xx into half, and made stress_yy equal the other half? – Jek Denys Oct 29 '17 at 19:32
• In calculating the yield stress, you can subtract any hydrostatic stress state to your convenience (again, usually. This is empirical as far as I know.) In the example, I subtracted $(\sigma/2) I$ where $I$ is identity. – Hussein Oct 29 '17 at 19:55