Sum of normal stresses equals sum of principal stresses

To prove: $\sigma_x + \sigma_y = \sigma_1 + \sigma_2$.

The principal stresses for a state of plane stress $(\sigma_x, \sigma_y, \tau_{xy})$ are

$$\sigma_1 = \frac{\sigma_x+\sigma_y}{2} + \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}, \qquad \sigma_2 = \frac{\sigma_x+\sigma_y}{2} - \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$$

Adding the two equations, the square-root terms cancel:

$$\sigma_1 + \sigma_2 = 2\times\frac{\sigma_x+\sigma_y}{2} = \sigma_x + \sigma_y$$

Proof from the transformation equations (any plane): the normal stress on a plane at $\theta$ is

$$\sigma_\theta = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\theta + \tau_{xy}\sin2\theta$$

and on the perpendicular plane ($\theta + 90^\circ$), $\cos 2(\theta+90^\circ) = -\cos2\theta$ and $\sin 2(\theta+90^\circ) = -\sin2\theta$:

$$\sigma_{\theta+90^\circ} = \frac{\sigma_x+\sigma_y}{2} - \frac{\sigma_x-\sigma_y}{2}\cos2\theta - \tau_{xy}\sin2\theta$$ $$\sigma_\theta + \sigma_{\theta+90^\circ} = \sigma_x + \sigma_y = \text{constant}$$

The principal planes are a pair of perpendicular planes, so $\sigma_1 + \sigma_2 = \sigma_x + \sigma_y$.

Mohr's circle view: $\sigma_x$ and $\sigma_y$ are the abscissas of the two ends of a diameter, whose mid-point (centre) is at $\frac{\sigma_x+\sigma_y}{2}$; $\sigma_1$ and $\sigma_2$ are the ends of the horizontal diameter, with the same centre. Hence twice the centre abscissa equals both sums.

This sum is called the first stress invariant: it does not change with the orientation of the axes.