Module 12 · Lesson 12.2
Participation, effective mass and truncation
Two quantities that are constantly confused, one criterion that is widely misapplied, and an honest account of what stopping at n modes actually costs.
Why this matters
Nobody keeps every mode. A fifty-storey model has 150 modes and an analysis retains perhaps twenty. Deciding how many is the most consequential routine judgement in modal analysis, and the criterion everyone uses — 90% effective mass — is both useful and widely misunderstood.
It does not catch a wrong mass source. It does not guarantee acceleration has converged. And it is stated in terms of a quantity that is routinely confused with a different one that means something else entirely.
By the end of this lesson you should be able to
- Derive the participation factor and the effective modal mass
- Show that effective mass is scale-independent and participation is not
- Prove that effective masses sum to the total mass
- Apply the 90% criterion and state precisely what it does not catch
Where both quantities come from
Under uniform base excitation the load vector is p = −Mι üg, where ι is the influence vector — the displacement of each degree of freedom under a unit static movement of the base. For a shear building shaken along one axis it is a vector of ones.
The generalised force for mode n is then
φₙᵀp = −(φₙᵀMι) üg = −Lₙ üg
and the modal equation becomes
q̈ₙ + 2ζₙωₙq̇ₙ + ωₙ²qₙ = −(Lₙ/Mₙ) üg = −Γₙ üg
That ratio is the participation factor:
Γₙ = Lₙ/Mₙ = φₙᵀMι / φₙᵀMφₙ
Why it is not what people think it is
Scale the mode shape by α. Then Lₙ scales by α and Mₙ scales by α², so
Γₙ → Γₙ/α
The participation factor depends on the arbitrary normalisation. It is not a percentage, it is not bounded by 1, and it can be negative. A mode-1 participation factor of 1.22 does not mean 122% of anything.
The quantity that IS meaningful
The effective modal mass is
M*ₙ = Lₙ²/Mₙ
Scale by α and the numerator gains α² while the denominator gains α² — they cancel exactly. Effective modal mass is a property of the structure and the direction of excitation, and of nothing else.
It has units of mass, it is never negative, and — the key property —
Σ M*ₙ = ιᵀMι = total mass
Every kilogram belongs to some mode. That is why it can be expressed as a percentage and accumulated, and it is why every design code is written in terms of effective mass and none is written in terms of participation factors.
What the 90% criterion catches
It catches too few modes extracted. If the retained modes account for less than 90% of the mass, the base shear will be too small because some of the building's inertia is simply not represented.
What it does NOT catch — and this matters
A wrong mass source. Effective mass is a percentage OF THE MASS IN THE MODEL. A model containing a quarter of the building's real mass will happily report 95% cumulative effective mass while producing a base shear four times too small. The percentage is reassuring and meaningless. This is Case 2 of the Module 20 audit, and no modal check finds it — only comparing the total mass against a hand estimate does.
Unconverged accelerations. Effective mass measures the contribution to global INERTIA, which is the base shear. Floor accelerations are dominated by higher modes that carry very little effective mass, so 90% can be reached while acceleration is far from converged.
Local demands. A mode with negligible effective mass can still govern a slender element, a parapet or a plant item. It contributes nothing globally and everything locally.
From first principles
Effective modal mass, and why the masses sum to the total
We want to show: Derive Γₙ and M*ₙ, show that effective mass is independent of mode-shape scaling, and prove that the effective masses sum to the total mass.
Base excitation shakes every mode, but not equally. How hard it shakes a particular mode depends on how well that mode's shape resembles the rigid-body movement the ground is imposing. A mode in which the whole building moves together resembles it closely and is driven hard; a mode in which the floors move in opposite directions largely cancels against it and is driven weakly. Effective mass turns 'how well it resembles' into a number of kilograms.
Try it
Participation factor is not effective mass
Rescale every mode shape by the same factor and see which quantity moves.
Physically meaningless. Watch which column reacts.
- Effective mass fraction
- Cumulative
- 90% target
| Mode | T (s) | Γ (scale-dependent) | M_eff (tonne) | % of total | Cumulative |
|---|---|---|---|---|---|
| 1 | 0.753 | 1419.871 | 2016.0 | 79.75% | 79.75% |
| 2 | 0.292 | -557.551 | 310.9 | 12.30% | 92.05% |
| 3 | 0.182 | 327.340 | 107.2 | 4.24% | 96.28% |
| 4 | 0.135 | 219.452 | 48.2 | 1.91% | 98.19% |
| 5 | 0.110 | -154.438 | 23.9 | 0.94% | 99.13% |
| 6 | 0.096 | 109.667 | 12.0 | 0.48% | 99.61% |
| 7 | 0.087 | -78.953 | 6.2 | 0.25% | 99.85% |
| 8 | 0.081 | 60.660 | 3.7 | 0.15% | 100.00% |
- Total mass
- 2530 tonne
- Sum of effective masses
- 2530 tonne
- Modes to reach 90%
- 2
- Mode 1 Γ
- 1419.871
- Mode 1 effective mass
- 2016.0 tonne
Must equal the total. Every kilogram belongs to some mode.
Changed by the rescaling — it is now 1.0× smaller than at scale 1.
Unchanged by the rescaling, whatever you set the slider to.
Set the rescaling slider anywhere you like. The Γ column moves and the effective-mass column does not. Γ is only meaningful alongside the mode shape it was computed with; effective mass is meaningful on its own — and that is the entire reason codes are written in terms of it.
What a low effective mass means
- A mode with almost no effective mass contributes almost no base shear, however dramatic its shape looks in the software.
- It can still matter for local response — a slender element, a plant item, a parapet — because effective mass measures the contribution to GLOBAL inertia, not to local demand.
- Introduce a soft storey and watch the mass concentrate into the first mode. That is not a good sign; it means one storey is doing all the deforming.
What this shows: The participation factor depends on how the mode shapes happen to be scaled. The effective modal mass does not — which is why design codes accumulate effective mass and never participation factors.
Worked example
Participation and effective mass for a three-storey building
Given
- The uniform three-storey building: m = 400 t per floor, k = 600 MN/m
- Mode shapes, normalised to unit roof: φ₁ = {0.445, 0.802, 1.000}, φ₂ = {−1.247, −0.555, 1.000}, φ₃ = {1.802, −2.247, 1.000}
- Influence vector ι = {1, 1, 1} for horizontal shaking
Find
Γ and effective mass for each mode, and how many modes reach 90%.
Assumptions
- M = mI; uniform base excitation along the frame axis
Try it
How many modes do you actually need?
Add modes one at a time and watch where the missing response was hiding.
Ground motion
- All modes
- First 1 mode
Show the numbers behind this plot
- All modes
- First 1
| Mode | T (s) | Effective mass | Cumulative | Peak |q_n| |
|---|---|---|---|---|
| 1 | 0.590 | 83.1% | 83.1% | 33.3 |
| 2 | 0.221 | 11.2% | 94.4% | 3.19 |
| 3 | 0.139 | 3.5% | 97.9% | 0.623 |
| 4 | 0.106 | 1.4% | 99.3% | 0.261 |
| 5 | 0.090 | 0.5% | 99.8% | 0.0844 |
| 6 | 0.081 | 0.2% | 100.0% | 0.0486 |
- Peak roof displacement, all modes
- 35.1 mm
- Peak roof displacement, 1 mode
- 34 mm
- Error from truncation
- 3.09%
- Cumulative effective mass retained
- 83.1%
- Modes needed for 90%
- 2
- SRSS of the retained modal peaks
- 34 mm
- CQC of the same peaks
- 34 mm
Close to SRSS here, because the modes of a regular shear building are well separated.
Only 83.1% of the mass is represented. That is below the 90% a design code would normally require, and the missing mass would show as a base shear that is too small.
What this shows: Truncation error is not spread evenly. The first mode gets the roof displacement nearly right and the storey forces near the top badly wrong, because higher modes contribute little displacement and a great deal of acceleration.
Predict first
A model reports 96% cumulative effective mass with four modes. A hand check shows the model's total mass is 2 100 tonnes against a floor-area estimate of 8 400 tonnes. What follows?
Rayleigh damping
The modal equations need a damping ratio for each mode. Where does it come from?
Sometimes it is simply assigned — 5% in every mode — which is what a response-spectrum analysis usually does. But direct integration needs an actual matrix C, and the standard construction is Rayleigh damping:
C = a₀M + a₁K
This is guaranteed classical, because both M and K are orthogonalised by the modes, so any combination of them is too. Pre-multiplying by φₙᵀ and dividing by 2Mₙωₙ gives the damping ratio it delivers:
ζₙ = a₀/(2ωₙ) + a₁ωₙ/2
Two coefficients, so two conditions can be satisfied. Choose two frequencies and the damping wanted at each, and solve.
What it actually delivers
Exactly right at the two anchor frequencies, and wrong everywhere else:
- Below the lower anchor, the mass-proportional term a₀/(2ω) blows up. A very long-period mode receives absurd damping. This is a real trap for base-isolated structures, whose isolated mode has a long period.
- Between the anchors, the delivered damping dips below the target.
- Above the upper anchor, the stiffness-proportional term a₁ω/2 grows linearly, so high modes are progressively suppressed.
That high-frequency suppression is sometimes exactly what is wanted — it quietens spurious numerical modes in a large model. But it must be a decision, not an accident.
Choose the anchors to bracket the modes that carry meaningful effective mass. Then read the ζ the model actually delivers in each mode, rather than assuming it got the target.
Where every mode's damping must be right, a modal damping matrix can be built instead:
C = MΦ[2ζₙωₙ]ΦᵀM
This delivers exactly the specified ratio in every mode. It produces a full matrix with no physical dashpot behind it — which is a reasonable price, given that Rayleigh damping has no physical dashpot behind it either and is wrong in most of the modes retained.
Try it
What Rayleigh damping actually delivers
Choose two modes to anchor the damping at, then look at what every other mode gets.
- ζ delivered
- Mass-proportional part a₀/(2ω)
- Stiffness-proportional part a₁ω/2
- Target
| Mode | ω (rad/s) | ζ delivered | vs target | Effective mass |
|---|---|---|---|---|
| 1 ⚓ | 8.35 | 5.00% | 1.00× | 79.7% |
| 2 | 21.54 | 4.07% | 0.81× | 12.3% |
| 3 ⚓ | 34.56 | 5.00% | 1.00× | 4.2% |
| 4 | 46.51 | 6.14% | 1.23× | 1.9% |
| 5 | 56.91 | 7.22% | 1.44× | 0.9% |
| 6 | 65.40 | 8.14% | 1.63× | 0.5% |
| 7 | 71.97 | 8.85% | 1.77× | 0.2% |
| 8 | 77.49 | 9.46% | 1.89× | 0.1% |
- a₀ (mass-proportional)
- 0.672 1/s
- a₁ (stiffness-proportional)
- 0.00233 s
- Anchor frequencies
- 8.35 and 34.6 rad/s
- Lowest damping delivered
- 4.07%
- Highest damping delivered
- 9.46%
- Worst-damped mode
- mode 8 at 9.46%
The anchors have been placed so that every mode carrying meaningful mass is damped close to the target. That is the right way to choose them: bracket the modes that matter, not the first and the last.
How to choose the anchors
- Bracket the modes that carry the mass. Anchoring on mode 1 and the highest mode with meaningful effective mass is the usual advice.
- Never anchor on a very long-period mode. The mass-proportional term goes as 1/ω, so a near-rigid-body mode receives absurd damping — and a base-isolated structure is exactly where this bites.
- If the damping the model delivers in an important mode is wrong by a factor of two, so is that mode's contribution to the answer. Check the delivered values; do not assume the target was achieved.
- Where every mode's damping must be right, use a modal damping matrix instead. It is not physically realisable as dashpots, but neither is Rayleigh damping.
What this shows: Rayleigh damping is exactly right at two frequencies and wrong at every other. Where the anchors are placed decides which modes are over-damped and which are under-damped — and that is an analyst's choice, not a property of the structure.
Practice
A mode has L = φᵀMι = 1 800 tonnes and modal mass M = 1 500 tonnes. What is its participation factor?
Practice
For that same mode, what is the effective modal mass, in tonnes?
Practice
Rayleigh damping is anchored at 5% on ω = 4 rad/s and ω = 40 rad/s. What damping does it deliver at ω = 12 rad/s? Use a₀ = 2ζω₁ω₂/(ω₁+ω₂) and a₁ = 2ζ/(ω₁+ω₂).
Check yourself
Cumulative effective mass reaches 92% with six modes. What has this established?
Check yourself
What exactly does classical damping buy you?
Worked example
How many modes is enough?
Given
- A twelve-storey building analysed by response spectrum
- Effective mass in the first three modes: 68 %, 14 % and 6 %
- The code requires 90 % participation
Find
Whether three modes suffice, and what to do if not
Worked example
A participation factor that should have been zero
Given
- A symmetric building analysed for ground motion in the X direction
- Mode 4 is a pure torsional mode
- The solver reports 7 % participating mass for mode 4 in X
Find
What that tells you
Summary
- Γₙ = Lₙ/Mₙ says how hard the ground drives a mode, and it depends on scaling
- M*ₙ = Lₙ²/Mₙ says how much mass the mode mobilises, and it does not
- Σ M*ₙ = total mass, which is why effective mass can be accumulated as a percentage
- Higher modes have low effective mass because their ordinates cancel in Lₙ
- 90% catches too few modes; it does NOT catch a wrong mass source or unconverged acceleration
- Rayleigh: ζₙ = a₀/(2ωₙ) + a₁ωₙ/2, exact at two frequencies and wrong elsewhere
- Anchor the damping across the modes that carry the mass, then check what was delivered
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