**Data:**

$P_x = 80 \ N/mm^2$ (tensile)

$P_y = -20 \ N/mm^2$ (Compressive)

$\zeta = 25 \ N/mm^2$

**A] Magnitude of principal stresses:**

Let $P_1$ and $P_2$ be the principal stresses.

$P_1, \ P_2 = \frac{Px + Py}{2} \pm \sqrt{ (\frac{Px – Py}{2})^2 + \zeta^2}$

$= \frac{80 – 20}{2} \pm \sqrt{ (\frac{80 – (-20)}{2})^2 + 25^2}$

$P_1 , P_2 = 30 \pm \sqrt{50^2 + 25^2}$

$= 30 \pm 25\sqrt{5}$

$P_1 = 85.90 \ Nmm^2$ **(tensile)** $\rightarrow \ 30 + 25 \sqrt{5}$

$P_2 = -25.90 \ N/mm^2$ **(Compressive)** $\rightarrow \ 30 – 25 \sqrt{5}$

Now, the inclination of the principal planes is

$tan \ 2 \ \theta_1 = \frac{2 \zeta}{Px – Py}$

$tan \ 2 \ \theta_1 = \frac{2 \times 25}{80 – (-20)} = 0.5$

$\therefore \ \theta_1 = 13°16’$

$\theta_2 = \theta_1 + 90 = 103°16’$

**B] The maximum shear stress is,**

$\zeta_{max} = \frac{P_1 – P_2}{2} = \frac{85.90 – (- 25.90)}{2}$

$\zeta_{max} = 111.8 \ N/mm^2$

And its inclination is $\theta = \theta_1 + 45° = 13°16’ + 45°$

$\theta = 58°16°$