---
title: "Cases of symmetry and anti-symmetry"
subject: "Theory of Structures II"
chapter: 3
key: "semester_5/theory_of_structures_ii:symmetry-antisymmetry-cases"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/symmetry-antisymmetry-cases
site: BCT Engineering
---

# Cases of symmetry and anti-symmetry

Theory of Structures II · Chapter 3: Displacement method

- Asked in: 2067 Asar, 5 marks
- Video: [Symmetrical Structure Symmetric and Anti Symmetric loads - Advanced Structural Analysis](https://www.youtube.com/watch?v=8ob3Pi9Om5c) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/symmetry-antisymmetry-cases
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3

## Question

Explain about cases of symmetry and anti symmetry.

## Short answer

Split loading into symmetric and antisymmetric parts. Symmetric loading: axis has no rotation and zero shear, half-structure is a guided support, crossing-member stiffness is 2EI/L. Antisymmetric loading: axis has zero moment and no vertical deflection, half-structure is a roller, crossing-member stiffness is 6EI/L.

## Answer

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
| Case | Axis conditions | Half-structure support | Beam stiffness across the axis |
|---|---|---|---|
| Symmetric loading | rotation = 0, shear = 0 | guided (fixed against rotation, free vertically) | $2EI/L$ |
| Antisymmetric loading | moment = 0, vertical deflection = 0 | roller | $6EI/L$ |

#### 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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For symmetric structure under symmetric loading, what is true on the axis?
   A. Rotation is zero and shear is zero; the half-structure has a guided support
   B. Moment is zero
   C. Vertical displacement is zero
   D. Sway occurs
   Answer: A. Joints on the axis do not rotate and shear is zero, so a guided support models the cut. (Key point: Symmetric loading)

2. Stiffness of a beam across the axis for antisymmetric loading?
   A. 6EI/L, with carry-over +1
   B. 2EI/L, carry-over -1
   C. 4EI/L, carry-over 0.5
   D. 3EI/L
   Answer: A. theta_B = theta_C gives M = 2EI/L (3 theta) = 6EI theta/L. (Key point: Antisymmetric stiffness)

3. What support replaces the axis under antisymmetric loading?
   A. A roller (moment is zero, vertical deflection zero)
   B. A fixed support
   C. A guided support
   D. A hinge with no vertical restraint
   Answer: A. Moment is zero and vertical displacement is zero at the axis, which a roller provides. (Key point: Antisymmetric half-structure)