Cases of symmetry and anti-symmetry

When a structure is geometrically symmetric (same geometry, supports and member properties on both sides of an axis), the analysis can be reduced to one half. Any loading can be split into a symmetric part and an antisymmetric part, the two halves are analysed separately and the results are added.

Symmetric structure, symmetric loading

• The deformed shape is symmetric: joints on the axis of symmetry do not rotate and (in a frame) there is no sway; the vertical displacement may occur.
• Internal forces: the bending moment and the axial force are symmetric; the shear on the axis is zero (for a member crossing the axis).
• Half-structure: replace the cut at the axis by a fixed (guided) support: rotation and horizontal displacement restrained, vertical displacement free.
• Modification of stiffness: for a member crossing the axis (for example the beam of a symmetric portal frame) the end rotations are equal in magnitude and opposite in sense, $\theta_C=-\theta_B$, so $M_{BC}=\dfrac{2EI}{L}(2\theta_B-\theta_B)=\dfrac{2EI\theta_B}{L}$. The member stiffness becomes $\dfrac{2EI}{L}$ (instead of $\dfrac{4EI}{L}$) and the carry-over factor is $-1$.

Symmetric structure, antisymmetric loading

• The deformed shape is antisymmetric: points on the axis have no vertical displacement; rotation and horizontal displacement can occur (a frame sways).
• The bending moment and axial force are antisymmetric; the moment at the axis is zero, and the shear is not zero.
• Half-structure: the axis section is replaced by a roller (vertical displacement restrained, horizontal and rotation free).
• A member that crosses the axis (a beam between two joints B and C placed symmetrically) has equal end rotations in the same sense, $\theta_B=\theta_C$, so $M_{BC}=\dfrac{2EI}{L}(2\theta+\theta)=\dfrac{6EI\theta}{L}$. The modified stiffness is $\dfrac{6EI}{L}$ and the carry-over factor is $+1$.

Summary

Example

A portal frame (columns equal, beam of span $L$, symmetric) under two equal downward point loads placed symmetrically on the beam is a symmetric case: no sway, so a single unknown rotation $\theta_B=-\theta_C$ is enough. Under a horizontal load at the top of one column the loading is the sum of a symmetric part and an antisymmetric part; the antisymmetric part produces sway with zero moment at the centre of the beam (point of contraflexure at mid-span), which reduces the frame to a half-frame with a roller at the axis.

Advantages: fewer unknowns, smaller equations, and a check on the final results (symmetry of the BMD or zero moment at the axis).