Plane stress: principal, max shear, normal

Reading of the figure: $\sigma_x = 80$ MPa (left/right faces), $\sigma_y = 60$ MPa (top/bottom faces), $\tau_{xy} = 50$ MPa.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 70.00 \pm \sqrt{(10.00)^2+(50.00)^2} = 70.00 \pm 50.99$$

$\sigma_1 = 120.99\ \text{MPa}$ and $\sigma_2 = 19.01\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(50.00)}{20.00} \Rightarrow 2\theta_p = 78.69^\circ$$

So $\theta_p = 39.35^\circ$ (normal of the plane carrying $\sigma_1$) and $129.35^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 50.99\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 84.35^\circ$ and $174.35^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 70.00\ \text{MPa}$.

Answer: $\sigma_1 = 120.99$, $\sigma_2 = 19.01$ at $\theta_p = 39.35^\circ$ and $129.35^\circ$; $\tau_{max} = 50.99$ with normal stress $70.00$ (MPa).