Module 2 · Lesson 2.1
The model is not the structure
What a degree of freedom is, why every dynamic model has far fewer than the real thing, and what is lost each time one is removed.
Why this matters
A twelve-storey frame has tens of thousands of degrees of freedom if every node is free to move in every way. Nobody analyses that dynamically, and nobody needs to: the answer to 'how does this building sway in an earthquake?' is carried by about twelve numbers.
Getting from one to the other is modelling, and it is a sequence of explicit decisions. The danger is never that a decision was made. It is that nobody wrote down what it removed — so when the model is later asked a question it cannot answer, it answers anyway.
By the end of this lesson you should be able to
- Define a degree of freedom precisely
- Count the degrees of freedom of a frame before and after each simplification
- Explain the transformation u = Tq and what TᵀMT means
- Say what a shear-building model can and cannot be asked
What you should already know
- Natural frequency from mass and stiffness (Module 1)
- The stiffness of a column in sway, 12EI/h³
What a degree of freedom is
A degree of freedom is one independent coordinate needed to specify the position of every part of the structure at a given instant.
Two words in that sentence carry the weight. Independent: if one coordinate can be worked out from the others, it is not a separate degree of freedom. And every part: a coordinate set that leaves some of the structure's position undetermined is not complete.
A plane frame joint has three: two translations and a rotation. A space frame joint has six. So a small ten-storey, three-bay, three-bay building has 10 × 16 = 160 joints and 960 degrees of freedom before anything is done to it.
What the simplifications do
Lump the mass at the floors. The mass of a building is overwhelmingly in its floors and the things on them; the columns contribute a few per cent. Concentrating it at floor level makes the mass matrix diagonal, which is a large computational simplification — and it is a good approximation for a building precisely because the physical mass distribution really is like that. It is a poor one for a chimney, where the mass is distributed along the member itself.
Ignore axial deformation of columns. A column shortens by a fraction of a millimetre under a load that sways it tens of millimetres. Setting that shortening to zero removes a degree of freedom per column per floor. It also removes every vertical mode of the building — which matters if you were going to ask about the vertical component of an earthquake, or about a column losing support.
Ignore joint rotations. Condensing them out — either by neglecting rotational inertia or by static condensation — removes the local bending modes of individual beams. For a lateral sway analysis nothing is lost. For a floor-vibration analysis everything is: the mode you were looking for was a local beam mode.
Assume rigid floor diaphragms. This is the largest reduction of all. Every node on a floor is tied to three master coordinates: two translations and a rotation about the vertical. A floor with forty nodes goes from 120 in-plane degrees of freedom to three.
What that discards is in-plane flexibility of the floor plate. Usually negligible, for a compact plan with a solid slab. Not negligible for a long narrow plan, for a plate perforated by large openings, or for a floor of precast units with modest in-plane connection.
The shear building
Apply all four and a regular building becomes a shear building: one lateral coordinate per floor, a diagonal mass matrix and a tridiagonal stiffness matrix. A ten-storey building has ten degrees of freedom instead of 960.
It is an excellent model for the lateral sway of a regular building, and it is wrong for almost everything else. It cannot represent vertical response, floor vibration, torsion, diaphragm flexibility or local member modes — not approximately, but at all, because the coordinates that would describe them no longer exist.
What it calculates: The mass and stiffness of a model expressed in fewer coordinates
- u
- The full set of displacements (m)
- q
- The retained (generalised) coordinates (m)
- T
- Transformation matrix — how the full set follows from the retained set (—)
- M̃, K̃
- Reduced mass and stiffness matrices (kg, N/m)
This assumes
- The assumed relationship u = Tq is exactly true — that is what the constraint asserts
- The reduction is a CONSTRAINT, so the reduced model is stiffer than the real one
In plain terms: Reduction does not throw the structure away; it constrains it to deform only in ways the retained coordinates allow. The TᵀMT and TᵀKT forms come from insisting that kinetic and strain energy are the same whichever coordinates they are written in — which is why the reduced model always overestimates stiffness and therefore frequency.
Try it
From a physical frame to a dynamic model
Choose what the model keeps. Every simplification buys size and costs behaviour, and both sides are reported.
Two orthogonal frame directions rather than one plane.
Only meaningful in a three-dimensional model.
- Joints in the physical model
- 9
- Degrees of freedom, unreduced
- 27
- Retained coordinates
- 3
- Reduction factor
- 9.0×
- Transformation
- u = T q
- Planes analysed
- 1
T is 27 × 3. The reduced mass is TᵀMT and the reduced stiffness is TᵀKT — the model does not lose the structure, it constrains it.
Retained
- Nothing has been simplified away.
No longer representable
- Joint rotations are no longer degrees of freedom, so no rotational inertia is carried and the local bending modes of individual beams cannot appear.
- Column axial deformation is ignored, so vertical modes and the vertical component of an earthquake cannot be represented at all.
- A rigid diaphragm ties every joint on a floor to one master coordinate. In-plane floor flexibility disappears — which matters for a long, narrow or heavily perforated floor plate. 9 coordinates became 3.
This is the classical shear-building model: one lateral coordinate per floor. It is the right first model for a regular building under horizontal excitation, and the wrong model for a floor-vibration problem, a vertical earthquake component or anything with a soft or perforated diaphragm.
What this shows: Model reduction is a series of explicit decisions, each removing a class of behaviour. The reduced coordinates are what you can see; the discarded ones are what you have agreed not to look for.
Predict first
A shear-building model of a ten-storey block reports a first period of 1.1 s. The engineer now wants to check whether the floors vibrate under footfall. What will the model tell them?
Worked example
Counting degrees of freedom through a chain of decisions
Given
- A four-storey, three-bay plane frame
- Four columns per storey, four joints per floor level
- Plane frame joints: two translations and one rotation each
Find
How many degrees of freedom survive each simplification, and what each one removes.
Assumptions
- Base joints are fully fixed and are not counted
Practice
A six-storey building is modelled as a shear building. How many degrees of freedom does the model have?
Practice
A three-dimensional building model uses rigid diaphragms and includes floor torsion. How many degrees of freedom does an eight-storey model have?
Check yourself
What does a rigid diaphragm assumption remove from a model?
Practice
A twelve-storey building is modelled with a rigid diaphragm at every floor. How many dynamic degrees of freedom does the model have?
Practice
The same twelve-storey building is reduced to a plane shear model in one direction. How many degrees of freedom now?
Practice
A three-dimensional frame model has 40 nodes, of which 4 are fully fixed at the base. Taking six degrees of freedom per free node, how many degrees of freedom does the model have?
Check yourself
A reduced model has been built by lumping mass at floor levels. Which question can it NOT answer?
Worked example
A degree-of-freedom budget, before choosing a model
Given
- A six-storey hospital, plan 32 m × 22 m, regular in plan and elevation
- Solid concrete slabs with no large openings
- The questions to be answered: first period, base shear, inter-storey drift
Find
How many degrees of freedom each candidate model has, and which is appropriate.
Assumptions
- Regular building; diaphragms genuinely rigid in plane
Worked example
Deciding whether one lumped mass is enough
Given
- A 34 m steel chimney with a 2.1 tonne platform at the top
- Chimney self-mass 9.4 tonnes, distributed over the height
- The question is the along-wind response at the top
Find
Whether a single lumped mass at the top is a defensible model.
Assumptions
- Cantilever behaviour; first mode dominant for along-wind response
Worked example
How many degrees of freedom does this need?
Given
- A four-storey braced steel frame, regular in plan, with concrete floors
- The question is the base shear under an earthquake spectrum
Find
The smallest model that can answer the question honestly
Summary
- A degree of freedom is an independent coordinate needed to fix the deformed shape
- Modelling is a chain of decisions, each removing a class of behaviour entirely
- u = Tq, with reduced matrices TᵀMT and TᵀKT, from equating energy in either coordinate set
- Reduction is a constraint, so a reduced model is always stiffer and its frequencies higher
- The shear building answers lateral sway well and cannot answer anything else
- Record what a model cannot represent, because it will answer those questions anyway
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