Plane stress: principal stresses and oblique plane

Use the plane-stress transformation equations (Rajput, Timoshenko).

Given

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

(a) Principal stresses

$$\begin{aligned} \sigma_{avg} &= \frac{\sigma_x+\sigma_y}{2} = \frac{80-40}{2} = 20.0\ \text{MPa} \\ R &= \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2} = \sqrt{60^2+30^2} = 67.08\ \text{MPa} \\ \sigma_{1,2} &= \sigma_{avg} \pm R = 87.08,\ -47.08\ \text{MPa} \end{aligned}$$

Orientation:

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

Substituting $\theta = 13.28^\circ$ in the $\sigma_n$ equation gives $87.08$ MPa, so this plane carries $\sigma_1$. The other principal plane is at $13.28 + 90 = 103.28^\circ$. Shear stress is zero on both.

(b) Maximum shear stress

$$\tau_{max} = R = 67.08\ \text{MPa}$$

It acts on planes at $\theta_s = \theta_p + 45^\circ = 58.28^\circ$, with a normal stress $\sigma_{avg} = 20$ MPa on those planes. (The out-of-plane principal stress is zero; $(\sigma_1 - 0)/2 = 43.5$ MPa is smaller, so the in-plane value governs.)

(c) Plane at $\theta = 30^\circ$

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 20 + 60\cos 60^\circ + 30 \sin 60^\circ = 20 + 30.00 + 25.98 = 75.98\ \text{MPa} \\[4pt] \tau_{n} &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -60\sin 60^\circ + 30\cos 60^\circ = -51.96 + 15.00 = -36.96\ \text{MPa} \end{aligned}$$

Check: the normal stress on the perpendicular plane ($120^\circ$) is $-35.98$ MPa, and $75.98 + (-35.98) = 40 = \sigma_x + \sigma_y$ (first invariant preserved).

Answer: $\sigma_1 = 87.1$ MPa, $\sigma_2 = -47.1$ MPa at $\theta_p = 13.3^\circ$; $\tau_{max} = 67.1$ MPa; on the $30^\circ$ plane $\sigma_n = 76.0$ MPa and $\tau = -37.0$ MPa.