Principal stresses, stress invariant, equivalent stress

Given: $\sigma_x = 120$ and $\sigma_y = 60\ \text{MN/m}^2$ (tensile), no shear stress ($\tau_{xy} = 0$), $\nu = 1/4$.

Principal stresses and directions

Since there is no shear on the X and Y planes, these planes are already principal planes.

$$\sigma_{1,2} = \frac{120+60}{2} \pm \sqrt{\left(\frac{120-60}{2}\right)^2 + 0} = 90 \pm 30$$ $$\sigma_1 = 120\ \text{MN/m}^2\ \text{(on the plane normal to the X-axis, }\theta_p = 0^\circ), \qquad \sigma_2 = 60\ \text{MN/m}^2\ \text{(}\theta_p = 90^\circ)$$

$\tan 2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y} = 0 \Rightarrow \theta_p = 0^\circ$ and $90^\circ$.

Stress invariant

The sum of normal stresses on any two perpendicular planes is constant (an invariant of plane stress). Check with a plane at, say, $\theta = 30^\circ$ and its perpendicular ($120^\circ$):

$$\sigma_{30} = 90 + 30\cos60^\circ = 105, \qquad \sigma_{120} = 90 + 30\cos240^\circ = 75$$ $$\sigma_x + \sigma_y = 120 + 60 = 180, \qquad \sigma_1 + \sigma_2 = 120 + 60 = 180, \qquad \sigma_{30} + \sigma_{120} = 105 + 75 = 180$$

So $\sigma_x + \sigma_y = \sigma_1 + \sigma_2 = 180\ \text{MN/m}^2$ is verified.

Equivalent stress for the same maximum strain

Maximum strain (in the direction of $\sigma_1$):

$$\varepsilon_{max} = \frac{1}{E}(\sigma_1 - \nu\sigma_2) = \frac{1}{E}(120 - 0.25\times60) = \frac{105}{E}$$

A single stress $\sigma_e$ acting alone gives the strain $\sigma_e/E$. Equating:

$$\sigma_e = 105\ \text{MN/m}^2$$

Answer: $\sigma_1 = 120$, $\sigma_2 = 60$ MN/m² (at $0^\circ$ and $90^\circ$); the invariant is $180$ MN/m²; equivalent stress $= 105$ MN/m².