Max shear for 150 MPa stress limit; principal stresses

Reading: the maximum (principal) stress must not exceed 150 MPa. Let $\sigma_x = 100$ MPa (tension), $\sigma_y = -60$ MPa (compression) and $\tau$ the unknown shear stress.

Shear stress that can be applied

$$\frac{\sigma_x+\sigma_y}{2} = \frac{100-60}{2} = 20\ \text{MPa}, \qquad \frac{\sigma_x-\sigma_y}{2} = 80\ \text{MPa}$$

The maximum principal stress is

$$\sigma_1 = \frac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau^2} = 150$$ $$\sqrt{80^2 + \tau^2} = 150 - 20 = 130 \Rightarrow \tau^2 = 130^2 - 80^2 = 10500$$ $$\tau = 102.47\ \text{MPa}$$

Maximum shear stress in the material

$$\tau_{max} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau^2} = 130\ \text{MPa}$$

Principal stresses and planes

$$\sigma_1 = 20 + 130 = 150\ \text{MPa}, \qquad \sigma_2 = 20 - 130 = -110\ \text{MPa}$$ $$\tan 2\theta_p = \frac{2\tau}{\sigma_x-\sigma_y} = \frac{2(102.47)}{160} = 1.2809 \Rightarrow 2\theta_p = 52.02^\circ,\ \theta_p = 26.01^\circ$$

$\sigma_1$ acts on the plane whose normal is at $26.01^\circ$ to the direction of the 100 MPa stress, and $\sigma_2$ on the plane at $116.01^\circ$ (anticlockwise, with $\tau$ taken positive as in the standard sign convention).

Answer: shear stress $= 102.47$ MPa; $\tau_{max} = 130$ MPa; $\sigma_1 = 150$ MPa, $\sigma_2 = -110$ MPa at $26.01^\circ$ and $116.01^\circ$.