Unsymmetrical bending

When the plane of bending does not coincide or parallel to the plane containing the principal centroidal axis of cross section is called as unsymmetrical bending. It is also known as complex or bi axial bending.

When a section of a beam is not symmetrical about the plane of bending, or the applied load is not coincide with the plane of bending an unsymmetrical bending takes place, i.e., in addition to bending, due to applied loads twisting is observed in the beam.

In unsymmetrical bending, the direction of the neutral axis will not be perpendicular to the plane of bending.

To prevent the torsion, the line of application of the load must pass through the shear center. If it does not, the beam undergoes the combined bending and torsion loading.

Analysis of unsymmetrical bending

If the plane of the bending moment is perpendicular to the neutral surface, we may use simple bending equation. This condition will occurs only if the bending axes X and Y axes are principal axes. Hence, we can tackle bending of beams of non–symmetric cross section by: 

1. Finding the principal axes of the section and principal MI 

2. Resolving moment M into components in the principal axis directions. 

3. Calculating stresses and deflections in each direction 

4.Superimpose stresses and deflections to get the final result

Neutral axis method :

The above method is most useful when the principal axes are known or can be found easily by calculation or inspection. The method is also useful for finding deflections.

Let Y’ and Z’ be the principal axes and let M be the bending moment vector making an angle θ with Z' axis.

Resolving into components with respect to the principal axes we get:

 Mz' = M cosθ 

My' = M sinθ 

If Ix' and Iy' are principal Moment of Inertia, the Stress at any point 

σz = Mz' [y/Iz'] - My [z/Iy'] 

On the neutral axis, the stress σz = 0 

by definition, hence as the point (y’,z’) lies on the neutral axis in this case, we have the neutral axis at angle β with respect to the principal axis OX

Tanβ = [Ix'/Iy'] tanθ  

Note that β ≠ 0 in general. 

Deflections: 

Using the method described above, deflections can be found easily by resolving the applied lateral forces into components parallel to the principal axes and separately calculating the deflection components parallel to these axes. 

The total deflection at any point along the beam is then found by combining the components at that point into a resultant deflection vector. 

Note:

The resulting deflection will be perpendicular to the neutral axis of the section at that point.