Stresses on plane at 60 deg to major principal

When the two principal stresses are known, the plane is located by the angle $\theta$ between its normal and the major principal stress ($\sigma_1$) axis. The plane itself is inclined at $60^\circ$ to the axis of $\sigma_1$, so its normal makes $\theta = 90^\circ - 60^\circ = 30^\circ$ with that axis. Here $\sigma_1 = 100$ MPa and $\sigma_2 = -40$ MPa.

Normal stress

$$\begin{aligned} \sigma_n &= \frac{\sigma_1+\sigma_2}{2} + \frac{\sigma_1-\sigma_2}{2}\cos 2\theta \\ &= \frac{100-40}{2} + \frac{100+40}{2}\cos 60^\circ = 30 + 70(0.5) = 65.00\ \text{MPa} \end{aligned}$$

(positive: tensile).

Shear stress

$$\tau = \frac{\sigma_1-\sigma_2}{2}\sin 2\theta = 70\sin 60^\circ = 60.62\ \text{MPa}$$

Resultant stress and its direction

$$\sigma_R = \sqrt{\sigma_n^2+\tau^2} = \sqrt{(65.00)^2+(60.62)^2} = 88.88\ \text{MPa}$$ $$\tan\phi = \frac{\tau}{\sigma_n} = \frac{60.62}{65.00} \Rightarrow \phi = 43.00^\circ$$

The resultant makes $43.00^\circ$ with the normal to the plane, i.e. $47.00^\circ$ with the plane itself.

Answer: $\sigma_n = 65.00$ MPa (tensile), $\tau = 60.62$ MPa, resultant $= 88.88$ MPa at $43.00^\circ$ to the normal of the plane.