Bending stress in unsymmetrical bending

Meaning

Bending is unsymmetrical when (a) the load plane does not coincide with a principal axis of the cross-section (e.g. an inclined load on a rectangular beam), or (b) the section has no axis of symmetry (angle, Z-section). The beam then deflects in a plane different from the plane of loading.

Derivation

Let $x, y$ be centroidal axes, $z$ the beam axis, $M_x$ the moment about $x$ (positive when it causes tension at $+y$) and $M_y$ the moment about $y$ (positive moment vector along $+y$, causing compression at $+x$). Plane sections remain plane, so the longitudinal strain varies linearly and

$$\sigma_z = a + bx + cy$$

Conditions of equilibrium:

$$\begin{aligned} \int \sigma_z\,dA &= 0 \ \Rightarrow\ a = 0 \quad (x, y \text{ centroidal}) \\ \int \sigma_z\,y\,dA &= M_x \ \Rightarrow\ b\,I_{xy} + c\,I_x = M_x \\ -\int \sigma_z\,x\,dA &= M_y \ \Rightarrow\ b\,I_y + c\,I_{xy} = -M_y \end{aligned}$$

Solving for $b$ and $c$ with $D = I_xI_y - I_{xy}^2$:

$$\boxed{\sigma_z = \frac{(M_xI_y + M_yI_{xy})\,y - (M_yI_x + M_xI_{xy})\,x}{I_xI_y - I_{xy}^2}}$$

Special cases:

• Principal axes ($I_{xy} = 0$): $\sigma_z = \dfrac{M_x y}{I_x} - \dfrac{M_y x}{I_y}$ (the two bending stresses simply superpose).
• Only one moment, symmetric section: the ordinary $\sigma = My/I$.

Neutral axis

Put $\sigma_z = 0$:

$$\frac{y}{x} = \tan\alpha = \frac{M_yI_x + M_xI_{xy}}{M_xI_y + M_yI_{xy}}$$

This is a straight line through the centroid. In the principal-axes form ($I_{xy}=0$): $\tan\alpha = \dfrac{M_y I_x}{M_x I_y}$. If the axis of the applied moment makes angle $\phi$ with the x-axis (so $M_y/M_x = \tan\phi$, and the load plane makes angle $\phi$ with the y-axis), then

$$\tan\alpha = \tan\phi\,\frac{I_x}{I_y}$$

Why the neutral axis is not perpendicular to the load plane

The neutral axis would be perpendicular to the load plane only if it coincided with the axis of the applied moment, that is $\alpha = \phi$. This needs $I_x = I_y$ (a circle or square), or a load along a principal axis. Otherwise the ratio $I_x/I_y$ turns the neutral axis towards the axis of least moment of inertia. The deflection is perpendicular to the neutral axis, so the beam deflects in a plane different from the loading plane.

     y
     |   load plane
     |  /
     | / phi
  ---+/------- x
    /|
   / | neutral axis (alpha > phi if Ix > Iy)

The maximum stress occurs at the point of the section farthest from the neutral axis.