Stresses on inclined plane, bi-axial loading

Expression for stresses on an inclined plane (bi-axial loading)

Consider a small element subjected to tensile stresses $\sigma_x$ and $\sigma_y$ on two perpendicular faces (no shear stress). Cut it by a plane AB whose normal makes angle $\theta$ with the $x$-axis (the plane makes $\theta$ with the vertical face). Let the area of AB be $dA$ (unit thickness).

        sy
     B  |  |  |
     |\ v  v  v
     | \
  sx |  \ AB (area dA)
 ----> |___\ A
       (BA vertical face = dA cos(theta))

Areas: vertical face $= dA\cos\theta$; horizontal face $= dA\sin\theta$.

Forces on the wedge: $\sigma_x\,dA\cos\theta$ (horizontal) and $\sigma_y\,dA\sin\theta$ (vertical).

Normal stress $\sigma_n$ (resolve along the normal to AB):

$$\sigma_n\,dA = \sigma_x\,dA\cos\theta\cdot\cos\theta + \sigma_y\,dA\sin\theta\cdot\sin\theta$$ $$\sigma_n = \sigma_x\cos^2\theta + \sigma_y\sin^2\theta = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta$$

Shear (tangential) stress $\tau$ (resolve along AB):

$$\tau\,dA = \sigma_x\,dA\cos\theta\sin\theta - \sigma_y\,dA\sin\theta\cos\theta$$ $$\tau = \frac{\sigma_x-\sigma_y}{2}\sin 2\theta$$

Resultant stress: $\sigma_R = \sqrt{\sigma_n^2 + \tau^2}$, at obliquity $\tan\phi = \tau/\sigma_n$ to the normal.

Maximum $\sigma_n = \sigma_x$ at $\theta = 0^\circ$; maximum shear $= \dfrac{\sigma_x-\sigma_y}{2}$ at $\theta = 45^\circ$.

Mohr's graphical method

• Draw a horizontal axis for normal stress. Mark $OA = \sigma_y$ and $OB = \sigma_x$ to scale.
• Bisect AB at C and draw a circle with diameter AB (centre C, radius $\frac{\sigma_x-\sigma_y}{2}$).
• From C draw a line making angle $2\theta$ with CB, cutting the circle at Q.
• The horizontal distance of Q from O is $\sigma_n$ and the vertical distance of Q from the axis is $\tau$, because $OC + CQ\cos2\theta = \sigma_n$ and $CQ\sin2\theta = \tau$.
• The line OQ is the resultant stress; $\angle QOC$ is its obliquity.