Two-Plane Bending

Occurs in asymmetrical bending (moment applied on an angle not in one plane) When there are one or two planes of symmetry
$\sigma =\frac{-M_{z}y}{I_z}+\frac{M_{y}z}{I_y}$
Where $I_z$ = $I_y$ and z = y = $\frac{d}{2}$
Maximum occurs when summation gives greatest positive and negative stresses.
For solid circular cross sections, max stress is:
${\sigma }_{\text{max}}=\frac{Mc}{I}=\frac{32}{\pi d^3}{\left(M_y^2+M_z^2\right)}^{1/2}$

Solving

1. Resolve the asymmetrical moment into components along the principle axes and follow steps in Beams and Loading to calculate max stresses in each direction
2. Combine stresses from the stress distributions (sum the two depending on the direction of the moment and how they apply themselves on the x-plane)
   • You take a sum because we’re finding the stress applied on the x-plane in this case which both the y and z moments affect, so its almost like its applying an axial load ${\sigma }_x=\frac{-M_{z}y}{I_z}+\frac{M_{y}z}{I_y}$
3. Determine angle of neutral axis $\tan \varphi =\frac{y}{z}=\frac{I_z}{I_y}$