Module 11 · Lesson 11.2
Orthogonality and normalisation
The property that makes modal analysis possible, derived from the symmetry of M and K alone — and the scaling choices that change every printed number without changing any physics.
Why this matters
Orthogonality is the single most consequential property in this subject. Without it there is no modal analysis, no participation factor, no effective mass, no response-spectrum method — the entire content of Stage C rests on it.
And it is not an assumption or an approximation. It follows from the symmetry of the mass and stiffness matrices and from nothing else, which is why it holds exactly for every structure whose matrices are correctly assembled.
By the end of this lesson you should be able to
- Derive orthogonality through M and through K
- Explain why symmetry is the only ingredient
- Define modal mass and modal stiffness
- Compare the common normalisations and say what each buys
From first principles
Orthogonality of mode shapes
We want to show: Show that φᵢᵀMφⱼ = 0 and φᵢᵀKφⱼ = 0 for two modes with different frequencies, using only the symmetry of M and K.
Two different modes each satisfy the eigenproblem at their own frequency. Take one mode's equation and multiply it by the other mode's shape; then do the same the other way round. Because both matrices are symmetric, the two results contain the same quantities — so subtracting them leaves a statement with the shapes on one side and the difference of the two frequencies on the other. If the frequencies differ, the only way that statement can hold is if the shapes are orthogonal.
Normalisation: four choices, no physics
Because any multiple of a mode shape is also a mode shape, the scale has to be fixed by convention. Four conventions are in common use.
| Normalisation | Condition | What it buys | ||
|---|---|---|---|---|
| Mass | φᵀMφ = 1 | Modal mass is 1, so φᵀKφ = ω² and the modal equation has unit mass | ||
| Unit maximum | max | φᵢ | = 1 | Every ordinate is between −1 and 1; easy to read |
| Unit roof | φtop = 1 | Roof displacement reads directly; natural for buildings | ||
| Unit length | φᵀφ = 1 | Mathematically tidy; little engineering use |
Mass normalisation is what analysis software uses internally, because it makes the modal equations as simple as possible:
q̈ₙ + 2ζₙωₙq̇ₙ + ωₙ²qₙ = φₙᵀp(t)
with no modal mass to divide by anywhere.
Unit-roof normalisation is what engineers use to read results, because the ordinates then mean something directly.
Changing between them changes: every ordinate, the modal mass, the modal stiffness, the participation factor and the modal coordinate. It does not change: the ratios between ordinates, the natural frequency, the effective modal mass, or any physical response.
If a quantity changes when you renormalise, it is not by itself a physical result.
That single sentence resolves most confusion about participation factors, and Module 12 makes it precise.
Worked example
The same mode, normalised four ways
Given
- The three-storey building of the previous lesson: m = 400 t per floor, k = 600 MN/m per storey
- Mode 1, whose shape ratios are 0.445 : 0.802 : 1.000
Find
The mode written under each normalisation, and which quantities move.
Assumptions
- M = mI with m = 400 000 kg per floor
Predict first
Two analysis packages are given the same building. Package A reports a mode-1 modal mass of 736 tonnes; package B reports 1.00. What has happened?
Practice
A mode shape is {0.30, 0.65, 1.00} on a building with 500 tonnes per floor. What is its modal mass, in tonnes?
Practice
A mass-normalised mode has a natural frequency of 24 rad/s. What is its modal stiffness φᵀKφ?
Practice
Two mode shapes are φ₁ = {0.5, 1.0} and φ₂ = {1.0, −0.4} on a system with m₁ = 300 t and m₂ = 375 t. Compute φ₁ᵀMφ₂ in tonnes to test orthogonality.
Check yourself
Which single property of M and K does the orthogonality derivation actually require?
Check yourself
A mode shape is reported with a maximum ordinate of 1.0 in one analysis and 0.043 in another, for the same building. What has changed?
Worked example
Checking modes are orthogonal before trusting them
Given
- A solver returns three mode shapes for a three-storey building
- The mass matrix is diagonal: 120, 120 and 90 tonnes
Find
The check to run before using them for modal superposition
Summary
- Orthogonality follows from the symmetry of M and K, and from nothing else
- φᵢᵀMφⱼ = 0 and φᵢᵀKφⱼ = 0 for i ≠ j; the i = j case defines modal mass and stiffness
- Kₙ = ωₙ²Mₙ, exactly as k = mω² for one degree of freedom
- The argument is silent when two frequencies are equal — orthogonality is then imposed
- Mass normalisation makes Mₙ = 1 and Kₙ = ωₙ², which is why software uses it
- Renormalising changes ordinates, modal mass, Γ and q — and no physics
- If a quantity moves when you renormalise, it is not a physical result on its own
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint