Why is shear stress equal to half of yield stress?

Question

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?

Answer 1

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 $\sigma/2$.

Answer 2

A yield value, like all material parameters, doesn't mean anything on it's own: They are just parameters of a specific material model. The von Mises yield criterion is so common, it's often taken for granted. In school book examples, the Tresca yield criterion is also often use (similar, but quite impractical).

This yield criteria states that yielding starts when $\sigma_{\text{effective}} \geq \sigma_y$. You can compute a (scalar) von-Mises or Tresca effective stress for any stress tensor. This effective stress was carefully defined so that the case of 1D stress ($\sigma_{11} = \sigma$), then we simply obtain $\sigma_{\text{eff}} = \sigma$ (because that's how you do your experiments to measure $\sigma_y$)

For pure 1D shear stress ($\tau_{12} = \tau_{21} = \tau$), and using Trescas effective stress measure, it works out to $\sigma_{\text{eff}} = 2\tau$.

So for these 2 simple pure cases you get $$ \text{pure tension} \implies \sigma_{\text{eff}} = \sigma < \sigma_y\\ \text{pure shearing} \implies \sigma_{\text{eff}} = 2\tau < \sigma_y $$ Then someone presumably felt clever and thought "lets divide by 2 and define a new $\tau_y$"

$$ \text{pure shearing} \implies \frac12\sigma_{\text{eff}} = \tau < \frac12\sigma_y =:\tau_y $$

Note Using von Mises effective stress, you would actually get $\sigma_{text{eff}}=\sqrt3\tau$ instead! The use of Tresca effective stress (which is more conservative) is the reason you get the factor $\frac12$.

Hossein's answer is wrong when he removes the hydrostatic part. There is no such thing as 2D stress, so removing the hydrostatic part you must divide by 3; so this is not where you get a factor of $\frac12$. Wikipedia has complete derivations of all the special cases for von Mises yield criterion.

For the general case you will almost always have multiple stress components and you can't use either of these simplifications; you will simply have to compute the effective stress for your particular stress tensor (and it is a very simple forward computation) and in those cases, you can't really ever make use of $\tau_y$.

Answer 3

In case of tensile only loading, the failure occurs when the normal stress is equal to Yield strength of the material i.e. σ1 = Sy. As we recall, the yielding occurs only due shear (at 45°) when the specimen reaches the yield strength of the material.

Using Mohr's circle, for a 2D stress (where σ1 and σ2 are +ve, and σ1 > σ2 and σ3), then τmax = (σ1 − σ3)∕2. In case of tensile test or tensile only load, σ2 and σ3 are zero. So substituing in the τmax equation, the radius of the circle (which gives the shear stress) is (σ1)∕2. So τmax = (Sy/2) ==> (σ1)∕2 (for 1D stress state)