Stress on oblique section with direct and shear stress

Problem: a rectangular element carries direct stresses $\sigma_x$ and $\sigma_y$ (tensile) on two perpendicular faces, together with a shear stress $\tau_{xy}$ on all four faces. Find the normal stress $\sigma_n$ and shear stress $\tau$ on an oblique section whose normal makes angle $\theta$ with the $x$-axis (the section makes $\theta$ with the face on which $\sigma_x$ acts).

            sy
        B ------- 
          |\
   txy <- | \  AB = oblique
     sx <-|  \  section
          |___\ A
    (wedge ABC, unit thickness)

Let the oblique face AB have area $dA$. Then the vertical face has area $dA\cos\theta$ and the horizontal face has area $dA\sin\theta$.

Forces on the wedge

• On the vertical face: horizontal force $\sigma_x\,dA\cos\theta$ and vertical force $\tau_{xy}\,dA\cos\theta$.
• On the horizontal face: vertical force $\sigma_y\,dA\sin\theta$ and horizontal force $\tau_{xy}\,dA\sin\theta$.

The shear stress $\tau_{xy}$ is taken so that the shear forces on the two faces act in the senses required by complementary shear.

Normal stress

Resolve all forces along the normal to AB (direction $\theta$ from the $x$-axis):

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

Using $\cos^2\theta = \frac{1+\cos2\theta}{2}$, $\sin^2\theta = \frac{1-\cos2\theta}{2}$ and $2\sin\theta\cos\theta = \sin2\theta$:

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

Shear stress

Resolve along the plane AB (perpendicular to the normal):

$$\tau\,dA = -\sigma_x\,dA\cos\theta\sin\theta + \sigma_y\,dA\sin\theta\cos\theta + \tau_{xy}\,dA\left(\cos^2\theta - \sin^2\theta\right)$$ $$\tau = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta + \tau_{xy}\cos2\theta$$

(the sign of $\tau$ depends on the direction chosen as positive along AB; reversing that direction only changes the sign, not the magnitude).

Result

$$\sigma_n = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\theta + \tau_{xy}\sin2\theta, \qquad \tau = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta + \tau_{xy}\cos2\theta$$

Principal stresses: setting $\tau = 0$ gives $\tan2\theta_p = \dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}$ and $\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}$.