Principal stresses: 50 T, 20 C, 30 shear

Given: $\sigma_x = 50$ MPa (tension), $\sigma_y = -20$ MPa (compression), $\tau_{xy} = 30$ MPa.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 15.00 \pm \sqrt{(35.00)^2+(30.00)^2} = 15.00 \pm 46.10$$

$\sigma_1 = 61.10\ \text{MPa}$ and $\sigma_2 = -31.10\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{70.00} \Rightarrow 2\theta_p = 40.60^\circ$$

So $\theta_p = 20.30^\circ$ (normal of the plane carrying $\sigma_1$) and $110.30^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Graphical check (Mohr's circle)

• Plot $X(50, 30)$ and $Y(-20, -30)$ and join them; the centre C is at $\sigma = 15$ MPa.
• Radius $CX = 46.10$ MPa; the circle meets the $\sigma$-axis at $P_1 = 61.10$ and $P_2 = -31.10$.
• $\angle XCP_1 = 2\theta_p = 40.60^\circ$, so the principal plane for $\sigma_1$ is at $20.30^\circ$ from the plane on which $\sigma_x$ acts.

Answer: $\sigma_1 = 61.10$ MPa and $\sigma_2 = -31.10$ MPa, acting on planes whose normals are at $20.30^\circ$ and $110.30^\circ$ from the $x$-axis.