Principal stresses and planes (90, 120, 50 MPa)

Reading of the figure: $\sigma_x = 120$ MPa (left/right faces), $\sigma_y = 90$ MPa (top/bottom faces), $\tau_{xy} = 50$ MPa, all tensile.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 105.00 \pm \sqrt{(15.00)^2+(50.00)^2} = 105.00 \pm 52.20$$

$\sigma_1 = 157.20\ \text{MPa}$ and $\sigma_2 = 52.80\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(50.00)}{30.00} \Rightarrow 2\theta_p = 73.30^\circ$$

So $\theta_p = 36.65^\circ$ (normal of the plane carrying $\sigma_1$) and $126.65^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Answer: $\sigma_1 = 157.20$, $\sigma_2 = 52.80$ at $\theta_p = 36.65^\circ$ and $126.65^\circ$.