Stresses on 40 degree plane; principal stresses

Reading of the figure: both normal stresses are tensile: $\sigma_x = 50$ MPa (left/right faces), $\sigma_y = 30$ MPa (top/bottom faces), and the shear stress is $\tau_{xy} = 30$ MPa. The plane is at $40^\circ$ to the vertical, so its normal is at $40^\circ$ to the $x$-axis.

Stresses on the inclined plane ($\theta = 40^\circ$)

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 40.00 + (10.00)\cos 80^\circ + (30.00)\sin 80^\circ = 71.28\ \text{MPa} \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(10.00)\sin 80^\circ + (30.00)\cos 80^\circ = -4.64\ \text{MPa} \end{aligned}$$

Resultant stress on the plane $= \sqrt{71.28^2+(-4.64)^2} = 71.43\ \text{MPa}$, inclined at $3.72^\circ$ to the normal of the plane.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 40.00 \pm \sqrt{(10.00)^2+(30.00)^2} = 40.00 \pm 31.62$$

$\sigma_1 = 71.62\ \text{MPa}$ and $\sigma_2 = 8.38\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{20.00} \Rightarrow 2\theta_p = 71.57^\circ$$

So $\theta_p = 35.78^\circ$ (normal of the plane carrying $\sigma_1$) and $125.78^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress (extra)

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 31.62\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 80.78^\circ$ and $170.78^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 40.00\ \text{MPa}$.

Answer: $\sigma_n = 71.28$, $\tau = -4.64$ on the inclined plane; $\sigma_1 = 71.62$, $\sigma_2 = 8.38$ at $\theta_p = 35.78^\circ$ and $125.78^\circ$; $\tau_{max} = 31.62$ with normal stress $40.00$ (MPa).