Principal stresses and planes (12, 8, 6 kN/cm2)

Reading of the figure: $\sigma_x = 12$ kN/cm² (tensile, left/right faces), $\sigma_y = -8$ kN/cm² (compressive, top/bottom faces), $\tau_{xy} = 6$ kN/cm².

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 2.00 \pm \sqrt{(10.00)^2+(6.00)^2} = 2.00 \pm 11.66$$

$\sigma_1 = 13.66\ \text{kN/cm}^2$ and $\sigma_2 = -9.66\ \text{kN/cm}^2$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(6.00)}{20.00} \Rightarrow 2\theta_p = 30.96^\circ$$

So $\theta_p = 15.48^\circ$ (normal of the plane carrying $\sigma_1$) and $105.48^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Answer: $\sigma_1 = 13.66$ kN/cm², $\sigma_2 = -9.66$ kN/cm²; principal planes at $15.48^\circ$ and $105.48^\circ$ from the $x$-axis.