Derivation of plane stress transformation equations

Sign convention

Tensile normal stress positive. Shear stress $\tau_{xy}$ is positive when it acts in the $+y$ direction on the face whose outward normal is $+x$. Angle $\theta$ is measured anticlockwise from the $x$ axis to the normal of the inclined plane.

        sy (top face)
      ^   ^
   +-----------+
   |           |  txy on the +x face points up (+y)
 <-|           |->  sx
   |           |
   +-----------+
      sy (bottom face)

Equilibrium of a wedge

Take a triangular wedge with the inclined face of area $dA$, normal at $\theta$. The areas of the vertical face and horizontal face are $dA\cos\theta$ and $dA\sin\theta$.

Forces along the normal $n$ ($\sum F_n = 0$):

$$\sigma_\theta\,dA = \sigma_x\cos^2\theta\,dA + \sigma_y\sin^2\theta\,dA + 2\tau_{xy}\sin\theta\cos\theta\,dA$$

Forces along the plane ($\sum F_t = 0$):

$$\tau_\theta\,dA = -(\sigma_x-\sigma_y)\sin\theta\cos\theta\,dA + \tau_{xy}(\cos^2\theta-\sin^2\theta)\,dA$$

Using double-angle identities:

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

Principal stresses

Principal planes have $\tau_\theta = 0$:

$$\tan2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y}$$

This gives two values of $2\theta_p$ that differ by $180^\circ$, so the two principal planes are $90^\circ$ apart. Substituting back:

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}$$

(Equivalently $d\sigma_\theta/d\theta = 0$.)

Maximum shear stress

Setting $d\tau_\theta/d\theta = 0$:

$$\tan2\theta_s = -\frac{\sigma_x-\sigma_y}{2\tau_{xy}}$$

so $\theta_s = \theta_p \pm 45^\circ$ (planes of maximum shear are at $45^\circ$ to the principal planes).

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

On these planes the normal stress is $\sigma_{avg} = (\sigma_x+\sigma_y)/2$.