Principal stresses and orientation: 80 and 60 N/mm2

Given: $\sigma_x = 80$ N/mm², $\sigma_y = 80$ N/mm² (tensile) and $\tau_{xy} = 60$ N/mm² on all four faces.

Principal stresses

$$\sigma_{1,2} = \frac{80+80}{2} \pm \sqrt{\left(\frac{80-80}{2}\right)^2 + 60^2} = 80 \pm 60$$ $$\sigma_1 = 140\ \text{N/mm}^2, \qquad \sigma_2 = 20\ \text{N/mm}^2$$

Orientation

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{120}{0} = \infty \Rightarrow 2\theta_p = 90^\circ,\ \theta_p = 45^\circ$$

So $\sigma_1 = 140$ N/mm² acts on the plane whose normal is at $45^\circ$ anticlockwise from the $x$-axis (the shear stresses are positive, so the major principal direction lies between the +x and +y axes). $\sigma_2 = 20$ N/mm² acts on the plane whose normal is at $135^\circ$ (perpendicular to it).

Maximum shear

$$\tau_{max} = \frac{140-20}{2} = 60\ \text{N/mm}^2\ \text{(on planes at 0° and 90°, with normal stress 80 N/mm}^2)$$

Sketch of the principal element

                    y
                    ^
  sigma_2 = 20      |        sigma_1 = 140
  (135 deg)  \      |       /  (45 deg)
              \     /\     /
               \   /  \   /
                  /    \
                  \    /     element rotated 45 deg
                   \  /      anticlockwise;
                    \/       no shear on its faces
  ------------------+-------------------> x

Check: $\sigma_x+\sigma_y = 160 = \sigma_1+\sigma_2$.

Answer: $\sigma_1 = 140$ N/mm² and $\sigma_2 = 20$ N/mm², the principal planes being at $45^\circ$ and $135^\circ$ to the $x$-axis.