Maximum Shear Stress Theory

Maximum-shear-stress theory predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension-test specimen of the same material when that specimen begins to yield. Also known as Tresca stress.

In general, MSS is a conservative predictor of failure; the predicted yield strength in shear is about 15% lower than actuality.

3D Stress

Recall that for 3D stress we have 3 principal stresses ${\sigma }_1,{\sigma }_2,{\sigma }_3$:

$$\begin{array}{rl}& {\sigma }^3-({\sigma }_x+{\sigma }_y+{\sigma }_z){\sigma }^2\\ & +({\sigma }_x{\sigma }_y+{\sigma }_x{\sigma }_z+{\sigma }_y{\sigma }_z-{\tau }_{xy}^2-{\tau }_{yz}^2-{\tau }_{zx}^2)\sigma \\ & -({\sigma }_x{\sigma }_y{\sigma }_z+2{\tau }_{xy}{\tau }_{yz}{\tau }_{zx}-{\sigma }_x{\tau }_{yz}^2-{\sigma }_y{\tau }_{zx}^2-{\sigma }_z{\tau }_{xy}^2)=0\end{array}$$

and 3 principal shears:

$${\tau }_{1/2}=\frac{{\sigma }_1-{\sigma }_2}{2},{\tau }_{2/3}=\frac{{\sigma }_2-{\sigma }_3}{2},{\tau }_{1/3}=\frac{{\sigma }_1-{\sigma }_3}{2}$$

For plane stress, the stress-free surface will have one of the principal stresses equal zero.

General Case

For a general state of stress, MSS predicts yielding when

$${\tau }_{max}=\frac{{\sigma }_1-{\sigma }_3}{2}\ge \frac{S_y}{2}\text{or}{\sigma }_1-{\sigma }_3\ge S_y$$

where $S_y$ is the yield strength, which is the stress at which a material begins to deform plastically. Naturally, this means that the yield strength in shear is given by:

$$S_{xy}=0.5S_y$$

which is about 15% lower than in actuality, as mentioned above.

We can also incorporate a factor of safety $n$:

$${\tau }_{max}=\frac{{\sigma }_1-{\sigma }_3}{2}=\frac{S_y}{2n}\text{or}{\sigma }_1-{\sigma }_3=\frac{S_y}{n}$$

Plane Stress

In plane stress, we have two principal stresses in the plane stress, ${\sigma }_A\ge {\sigma }_B$. The third principal stress is zero. Thus, there are 3 cases to consider:

	Case 1: ${\sigma }_A\ge {\sigma }_B\ge 0$	Case 2: ${\sigma }_A\ge 0\ge {\sigma }_B$	Case 3: $0\ge {\sigma }_A\ge {\sigma }_B$
Yield	${\sigma }_A\ge S_y$	${\sigma }_A-{\sigma }_B\ge S_y$	${\sigma }_B\le -S_y$
Design	${\sigma }_A=\frac{S_y}{n}$	${\sigma }_A-{\sigma }_B=\frac{S_y}{n}$	${\sigma }_B=-\frac{S_y}{n}$