Stresses on inclined plane and principal stresses

Use $\sigma_x = 80$, $\sigma_y = -40$, $\tau_{xy} = 30$ MPa.

$$\frac{\sigma_x+\sigma_y}{2} = 20\ \text{MPa}, \qquad \frac{\sigma_x-\sigma_y}{2} = 60\ \text{MPa}$$

(a) Stresses on the 30 degree plane

$2\theta = 60^\circ$, $\cos60^\circ = 0.5$, $\sin60^\circ = 0.866$.

$$\sigma_\theta = 20 + 60(0.5) + 30(0.866) = 20 + 30 + 25.98 = 76.0\ \text{MPa}$$ $$\tau_\theta = -60(0.866) + 30(0.5) = -51.96 + 15 = -37.0\ \text{MPa}$$

The negative sign means the shear acts opposite to the positive direction (clockwise sense on this face).

Check on the perpendicular plane ($\theta = 120^\circ$): $\sigma = 20 - 30 - 25.98 = -36.0$ MPa. The sum $76.0 - 36.0 = 40 = \sigma_x+\sigma_y$ confirms the calculation.

(b) Principal stresses

$$R = \sqrt{60^2 + 30^2} = \sqrt{4500} = 67.08\ \text{MPa}$$ $$\sigma_1 = 20 + 67.08 = 87.1\ \text{MPa (tension)}, \qquad \sigma_2 = 20 - 67.08 = -47.1\ \text{MPa (compression)}$$

Principal planes:

$$\tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{60}{120} = 0.5 \;\Rightarrow\; 2\theta_p = 26.57^\circ,\ \theta_p = 13.28^\circ$$

The plane at $13.28^\circ$ anticlockwise from the $x$ axis carries $\sigma_1 = 87.1$ MPa (check: $20+60\cos26.57^\circ+30\sin26.57^\circ = 87.1$). The other principal plane is at $13.28^\circ+90^\circ = 103.28^\circ$ and carries $\sigma_2$.

(c) Maximum shear stress

$$\tau_{max} = R = 67.1\ \text{MPa}$$

acting on planes at $\theta_p \pm 45^\circ$, i.e. $58.3^\circ$ and $148.3^\circ$, with normal stress $20$ MPa.

Answer: (a) $\sigma_\theta = 76.0$ MPa, $\tau_\theta = -37.0$ MPa; (b) $\sigma_1 = 87.1$ MPa, $\sigma_2 = -47.1$ MPa at $13.28^\circ$; (c) $\tau_{max} = 67.1$ MPa.