Plane stress: principal planes and max shear

Reading of the figure: shear stresses on perpendicular faces must be equal, so the labels are read as $\sigma_x = 60$ MPa (left/right faces), $\sigma_y = 15$ MPa (top/bottom faces) and $\tau_{xy} = 30$ MPa.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 37.50 \pm \sqrt{(22.50)^2+(30.00)^2} = 37.50 \pm 37.50$$

$\sigma_1 = 75.00\ \text{MPa}$ and $\sigma_2 = 0.00\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{45.00} \Rightarrow 2\theta_p = 53.13^\circ$$

So $\theta_p = 26.57^\circ$ (normal of the plane carrying $\sigma_1$) and $116.57^\circ$ (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

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 37.50\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 71.57^\circ$ and $161.57^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 37.50\ \text{MPa}$.

Answer: $\sigma_1 = 75.00$, $\sigma_2 = 0.00$ at $\theta_p = 26.57^\circ$ and $116.57^\circ$; $\tau_{max} = 37.50$ with normal stress $37.50$ (MPa).