Principal stresses in a shaft under combined loads

Stresses at the surface point

$$\sigma_x = \frac{32M}{\pi d^3} = \frac{32\times1.5\times10^6}{\pi\times60^3} = 70.7\ \text{MPa (tensile)}$$ $$\tau_{xy} = \frac{16T}{\pi d^3} = \frac{16\times2.0\times10^6}{\pi\times60^3} = 47.2\ \text{MPa}$$

and $\sigma_y = 0$ (no stress transverse to the axis).

Principal stresses

$$\frac{\sigma_x}{2} = 35.4\ \text{MPa}, \qquad R = \sqrt{35.4^2 + 47.2^2} = 58.9\ \text{MPa}$$ $$\sigma_1 = 35.4 + 58.9 = 94.3\ \text{MPa (tension)}$$ $$\sigma_2 = 35.4 - 58.9 = -23.6\ \text{MPa (compression)}$$

Maximum shear stress

$$\tau_{max} = R = 58.9\ \text{MPa}$$

(The same value follows from $\frac{16}{\pi d^3}\sqrt{M^2+T^2} = \frac{16\times2.5\times10^6}{\pi\times60^3} = 58.9$ MPa, which is the basis of the equivalent torque $T_e = \sqrt{M^2+T^2} = 2.5$ kN m.)

Principal planes

$$\tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x} = \frac{2\times47.2}{70.7} = 1.335 \;\Rightarrow\; 2\theta_p = 53.1^\circ,\ \theta_p = 26.6^\circ$$

The plane carrying $\sigma_1$ has its normal at $26.6^\circ$ to the shaft axis, so the plane itself is inclined at $26.6^\circ$ to the cross-section, i.e. at $63.4^\circ$ to the shaft axis. The plane carrying $\sigma_2$ is $90^\circ$ from it: its normal is at $116.6^\circ$ to the axis, so the plane makes $26.6^\circ$ with the shaft axis.

Answer: $\sigma_1 = 94.3$ MPa, $\sigma_2 = -23.6$ MPa, $\tau_{max} = 58.9$ MPa; $\theta_p = 26.6^\circ$ and $116.6^\circ$ (angles of the plane normals to the shaft axis).