Module 10 · Lesson 10.3
Diaphragms, rotational inertia and eccentricity
A floor has three degrees of freedom, not two — and the third one is the entry most often left at zero, which does not make the torsional mode wrong but removes it.
Why this matters
Every real building is three-dimensional and every real building twists. Whether a model can SEE that twisting depends on two matrix entries, and both are easy to leave out.
One is the rotational inertia of the floor. Omit it and the torsional mode does not become inaccurate — it disappears, and nothing on the output says so.
The other is the coupling that mass eccentricity introduces. Omit that and the model reports pure translation and pure rotation as independent, when the real building does both at once.
By the end of this lesson you should be able to
- Write the mass matrix of a rigid floor diaphragm
- Compute the mass moment of inertia of a rectangular floor
- Explain how eccentricity couples translation and rotation
- Recognise the modelling faults that silently remove behaviour
Three degrees of freedom, not two
A rigid floor plate can do exactly three things in its own plane: translate in x, translate in y, and rotate about the vertical axis. Every point on the floor follows from those three numbers.
So its mass matrix is three by three:
M = diag(m, m, I₀)
The first two entries are the floor's mass, resisting translation. The third is its mass moment of inertia about the vertical axis through the centre of mass, resisting rotation.
The rotational inertia
For a rectangular floor of plan dimensions a × b:
I₀ = m(a² + b²)/12
Its units are kg·m², and it is the rotational analogue of mass: it says how much torque is needed to produce a given angular acceleration.
A useful way to hold it is through the radius of gyration, r = √(I₀/m). For a 30 × 18 m floor, r = √((900 + 324)/12) = 10.1 m — about a third of the long plan dimension.
What happens if it is left at zero
With I₀ = 0 the rotational degree of freedom has stiffness but no inertia. Its natural frequency ω = √(Kθ/I₀) becomes infinite, so its period becomes zero.
The consequences on the output:
- The torsional mode does not appear in the modal list at all, or appears at a period near zero among the numerical noise.
- The torsional effective mass is nil, so a mass-participation check does not flag it.
- The first two modes are reported as pure translations, and they look entirely reasonable.
- No warning is issued of any kind.
A whole class of response has been removed from the model and nothing on the output says so. This is Case 3 of the Module 20 software audit, and it is among the commonest faults in three-dimensional building models.
Eccentricity couples translation and rotation
If the centre of MASS is offset from the origin of the coordinate system by e, the mass matrix acquires off-diagonal terms:
M = [ m 0 −me ] [ 0 m 0 ] [ −me 0 I₀+me²]
Those −me entries are inertial coupling. Physically: push the floor through a point that is not its centre of mass and it does not merely translate — it twists as well.
Combine that with a centre of STIFFNESS that is also offset and you have the general case, in which every mode is a mixture of translation and rotation. That is why seismic codes require an accidental eccentricity even in a nominally symmetric building: real buildings are never quite symmetric, and the consequences of a small eccentricity are not small.
A free check on any three-dimensional model
A model with N rigid-diaphragm levels has exactly 3N degrees of freedom, and therefore exactly 3N modes.
Count them. If a twelve-storey model returns 41 modes instead of 36, five degrees of freedom exist that you did not intend — usually nodes left outside a diaphragm constraint. That check needs no engineering judgement whatsoever, only arithmetic, and it is Case 6 of the Module 20 audit.
Try it
A floor plate as three degrees of freedom
Translation in two directions and rotation about the vertical axis. The third one is the one models forget.
Offset of the centre of mass from the centre of stiffness.
Turn it off to see what a model with a missing I₀ reports.
Diaphragm mass matrix, no eccentricity(tonne / tonne·m²)
| 486 | 0 | 0 |
| 0 | 486 | 0 |
| 0 | 0 | 49572 |
- Floor mass m
- 486 tonne
- I₀ = m(a² + b²)/12
- 49600 tonne·m²
- Radius of gyration r = √(I₀/m)
- 10.1 m
- I about the offset axis
- 49600 tonne·m²
- Torsional stiffness Kθ
- 2.4e5 MN·m/rad
- Mode 1 period
- 0.155 s
- Mode 2 period
- 0.155 s
- Mode 3 period
- 0.0894 s
predominantly translation in x
predominantly translation in y
predominantly torsional
With no eccentricity the mass matrix is diagonal and the three modes are pure: two translations and one rotation, entirely independent. Add eccentricity and watch them mix.
What this shows: A rigid diaphragm has three degrees of freedom, not two. Leaving out its mass moment of inertia does not make the torsional period slightly wrong — it deletes the torsional mode.
Worked example
A floor diaphragm, with and without its rotational inertia
Given
- Floor plate 32 m × 20 m, mass 900 kg/m²
- Lateral stiffness 500 MN/m in each direction, provided by four cores at the plan corners
- Torsional stiffness about the vertical axis: 2.6 × 10¹¹ N·m/rad
Find
The three modal periods, and what happens if I₀ is omitted.
Assumptions
- Rigid diaphragm
- Centre of mass coincides with centre of stiffness — no eccentricity, for now
- Single-storey, so this is a three-degree-of-freedom problem
Predict first
A three-dimensional model of a nominally symmetric building reports its first mode as predominantly torsional. What is the finding?
Practice
A rectangular floor is 24 m by 15 m with a mass of 400 tonnes. What is its mass moment of inertia about the vertical axis through its centre of mass, in kg·m²?
Practice
For that floor, what is the radius of gyration in metres?
Practice
A three-dimensional model has 9 storeys with rigid diaphragms including torsion. How many modes should a complete modal analysis return?
Check yourself
What happens to a model's torsional mode if the diaphragm rotational inertia is left at zero?
Check yourself
A rigid-diaphragm model is being used for a floor with a large stair opening and a long narrow plan. What is the risk?
Summary
- A rigid floor has three degrees of freedom: two translations and a rotation
- M = diag(m, m, I₀), with I₀ = m(a² + b²)/12 for a rectangular plate
- Check I₀ via the radius of gyration — about a third of the larger plan dimension
- I₀ left at zero removes the torsional mode rather than making it wrong
- No warning is issued, and no mass-participation check catches it
- Mass eccentricity introduces −me terms and couples translation with rotation
- N rigid-diaphragm levels give exactly 3N modes — count them before reading them
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