Principal stresses, max shear: 80, 60 N/mm2

Reading of the figure: the face parallel to the wall carries a stress of 80 N/mm² acting at $30^\circ$ to the wall (an oblique resultant), and the face at right angles to it carries a horizontal normal stress of 60 N/mm². Resolving the 80 N/mm² stress into components:

$$\sigma_y = 80\sin 30^\circ = 40\ \text{N/mm}^2\ \text{(normal)}, \qquad \tau_{xy} = 80\cos 30^\circ = 69.28\ \text{N/mm}^2\ \text{(shear)}$$

So $\sigma_x = 60$, $\sigma_y = 40$, $\tau_{xy} = 69.28$ (all in N/mm²).

Principal stresses

$$\sigma_{1,2} = \frac{60+40}{2} \pm \sqrt{\left(\frac{60-40}{2}\right)^2 + 69.28^2} = 50 \pm 70.00$$ $$\sigma_1 = 120.00\ \text{N/mm}^2, \qquad \sigma_2 = -20.00\ \text{N/mm}^2$$

Orientation of principal axes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(69.28)}{20} = 6.928 \Rightarrow 2\theta_p = 81.79^\circ$$

$\theta_p = 40.89^\circ$ for $\sigma_1$ and $130.89^\circ$ for $\sigma_2$, measured from the horizontal (the direction of the 60 N/mm² stress).

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = 70.00\ \text{N/mm}^2$$

It acts on planes at $\theta_s = 85.89^\circ$ and $175.89^\circ$ from the horizontal, i.e. $45^\circ$ from the principal planes, with a normal stress of $50$ N/mm² on them.

Answer: $\sigma_1 = 120.0$, $\sigma_2 = -20.0$ N/mm² at $40.9^\circ$ from the horizontal; $\tau_{max} = 70.0$ N/mm² at $85.9^\circ$.