Principal stresses, max shear: 80, 60, 20 MPa

Reading of the figure: $\sigma_x = 80$ MPa (horizontal, on left/right faces), $\sigma_y = 60$ MPa (vertical, on top/bottom faces), shear $\tau_{xy} = 20$ MPa; all tensile.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 70.00 \pm \sqrt{(10.00)^2+(20.00)^2} = 70.00 \pm 22.36$$

$\sigma_1 = 92.36\ \text{MPa}$ and $\sigma_2 = 47.64\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(20.00)}{20.00} \Rightarrow 2\theta_p = 63.43^\circ$$

So $\theta_p = 31.72^\circ$ (normal of the plane carrying $\sigma_1$) and $121.72^\circ$ (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

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 22.36\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 76.72^\circ$ and $166.72^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 70.00\ \text{MPa}$.

Answer: $\sigma_1 = 92.36$, $\sigma_2 = 47.64$ at $\theta_p = 31.72^\circ$ and $121.72^\circ$; $\tau_{max} = 22.36$ with normal stress $70.00$ (MPa).