Uniaxial, biaxial and pure shear stress systems

All three are special cases of the plane stress equations of transformation:

$$\sigma_\theta = \frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos2\theta+\tau_{xy}\sin2\theta, \quad \tau_\theta = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta+\tau_{xy}\cos2\theta$$

Uniaxial stress ($\sigma_y = \tau_{xy} = 0$)

A bar under a single axial stress $\sigma$.

$$\sigma_\theta = \frac{\sigma}{2}(1+\cos2\theta) = \sigma\cos^2\theta, \qquad \tau_\theta = -\frac{\sigma}{2}\sin2\theta$$

$\tau_{max} = \sigma/2$ at $\theta = 45^\circ$. Principal stresses: $\sigma$ and $0$.

Biaxial stress ($\tau_{xy} = 0$)

Normal stresses $\sigma_x$ and $\sigma_y$ in two perpendicular directions, e.g. a thin plate or shell.

$$\sigma_\theta = \frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos2\theta, \qquad \tau_\theta = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta$$

$\tau_{max} = \dfrac{\sigma_x-\sigma_y}{2}$ at $45^\circ$. The x and y planes are principal planes.

Pure shear ($\sigma_x = \sigma_y = 0$, $\tau_{xy} = \tau$)

Occurs in a twisted circular shaft.

$$\sigma_\theta = \tau\sin2\theta, \qquad \tau_\theta = \tau\cos2\theta$$

Principal planes: $\tan2\theta_p = \infty$, so $\theta_p = 45^\circ$ and $135^\circ$.

$$\sigma_1 = \tau\ (\text{at }45^\circ), \qquad \sigma_2 = -\tau\ (\text{at }135^\circ)$$

So a state of pure shear $\tau$ is the same as tension $\tau$ and compression $\tau$ on planes inclined at $45^\circ$, with no shear on those planes. Maximum shear stress $= \tau$. This is why brittle materials fail along a $45^\circ$ helix in torsion (tension failure), while ductile materials fail by shear on the cross-section.