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Module 10 · Lesson 10.3

What these analyses need from the model

Mesh density for buckling, mass for dynamics, and the modelling decisions that a static analysis lets you get away with and these do not.

Why this matters

A model built for a static analysis will run a buckling analysis and a modal analysis without complaint, and give wrong answers to both. Not because the solver is at fault, but because those analyses ask questions the model was never built to answer.

This lesson is the bridge between the methods and the modelling. Structural Dynamics teaches the dynamics; what belongs here is only what the model must contain.

By the end of this lesson you should be able to

  • Explain why one element per member is adequate for statics and not for buckling
  • State where a vibration model needs refinement and why it is the opposite of static
  • Explain why mass is a load-case decision rather than a material property
  • List what a dynamic model needs that a static one does not

One element is enough for statics and not for buckling

A modern beam element uses cubic shape functions, which represent the exact deflected shape of a prismatic member under end loads. So a single element gives the exact static answer for a simple beam — subdividing it changes nothing but the number of points you can plot.

Buckling is different, and the reason is precise. A buckling mode needs a lateral displacement at mid-height. With one element, the only free degrees of freedom are the end rotations; there is no node in the middle to move sideways. The model is therefore constrained relative to the real strut, and a constrained structure is stiffer.

The result is the classic 12EI/L² for a single-element pin-ended strut, against Euler's π²EI/L² ≈ 9.87EI/L². That is 21.6 % too high — an error in the unsafe direction, produced by a model that looks entirely reasonable.

Two elements bring it within about 1 %. More than four buys nothing measurable. The rule that comes out of this is worth stating generally:

A finite element model of a continuum is a constrained version of it, so it is always stiffer, and buckling loads always converge from above.

Vibration meshing is the reverse of static meshing

Static results are driven by stress gradients, which are steepest where the structure is interrupted: supports, columns, openings, load points, re-entrant corners. That is where a static mesh is refined.

A mode shape is a displacement field, and its curvature is greatest mid-bay. Next to a column, a floor barely moves. So a vibration mesh is refined where a static mesh would be coarsened, and the two are genuinely in tension — one model cannot serve both well without being finer than either needs.

Mass is a decision, not a property

In a static analysis, load is what you apply. In a dynamic analysis, mass is what resists — and the mass that is present during the event is not the ultimate design load.

For a footfall analysis you want the mass actually on the floor: the structure, the finishes, the services, the partitions, and a realistic fraction of the imposed load. Too little mass and the floor comes out too stiff, too high in frequency and too responsive. Too much and it comes out sluggish.

For a seismic analysis the same question has a different answer, and it is set by the relevant code. Using the ultimate imposed load is not conservative in any reliable way: more mass gives more force, which looks safe, but it also lengthens the period, which may move the structure to a lower spectral acceleration. Whether the net effect is safe depends on where you land on the spectrum, and assuming is not analysis.

The worst case for a vibration problem is often the lightest one. An empty floor has the least mass to resist excitation. That is the opposite of every static intuition and it is why a sensitivity study on mass is not optional.

What a dynamic model needs that a static one does not

  • Mass, distributed correctly. Lumped at the floors is right for a building's sway and wrong for a floor's own vibration.
  • Connections modelled as they behave at that strain level. A nominally pinned steel connection acts as continuous under vibration strains, because they are too small to overcome the friction in the joint.
  • Non-structural stiffness that matters at small amplitude. Balustrades and cladding do nothing at ultimate and can change a floor's liveliness.
  • Enough modes. Participating mass is the first number a modal result should be judged on, before any response value is read.
  • A time step, if the analysis marches in time. Convergence in time is the same discipline as convergence in space: halve it until the answer stops moving.

The detail of all five belongs to the Structural Dynamics course. What belongs here is the recognition that they are modelling decisions, taken before the analysis, and that a model inherited from the static work will have made all of them by default and none of them deliberately.

Try it

Modal mass explorer

A floor whose frequency follows the mass present. Move the imposed load and watch the response — it does not move monotonically.

kN/m²
kN/m²

What is actually on the floor when it vibrates — not the ultimate design value.

1.0 is the base design.

Mass present
5.70 kN/m²
Deflection under it
9.12 mm
Frequency f ≈ 18/√δ
5.96 Hz
In the plausible band?
yes, 4–10 Hz
Nearest walking harmonic
5.4 Hz
Detuning from it
10.4 %
Response factor
10.3

The floor is within 10 % of the 5.4 Hz harmonic and the response is elevated. Move the imposed load either way and it falls — which means neither end of the range is the worst case.

Why this breaks bracketing

  • Sweep the imposed load from one end of its range to the other and the response rises, peaks, and falls again. It is not monotonic.
  • So the three-run bracket of Module 14 — worst case at a corner — does not apply. Two runs at the ends would report a comfortable answer and miss the peak entirely.
  • The worst case for a vibration problem is often the LIGHTEST one, because there is least mass to resist the excitation. That is the opposite of every static intuition.

What this shows: In a vibration model, mass is a load-case decision, and the worst case is often not at either end of its range.

Worked example

How many elements does this strut need?

Given

  • A pin-ended strut, E = 210 × 10⁶ kN/m², I = 1 × 10⁻⁴ m⁴, L = 5 m
  • Modelled first with one beam element, then with two, four and eight

Find

The buckling load reported at each mesh density, against the Euler value

    Practice

    A single beam element models a pin-ended strut with EI = 21 000 kN·m² and L = 5 m. Using the one-element result P = 12EI/L², what buckling load does the model report, in kN?

    Practice

    A floor's first natural frequency is 5.2 Hz with the design mass. The mass is then found to be 30 % too low. Frequency goes as 1/√m. What is the corrected frequency, in Hz?

    Check yourself

    Why is participating mass the first number to read on a modal result?

    Check yourself

    What does a dynamic analysis need from the model that a static one does not?

    Summary

    • One element per member is exact for statics and 22 % unsafe for buckling
    • Buckling loads converge from above, because a finite element model is a constrained continuum
    • Vibration meshes refine mid-bay, where static meshes coarsen
    • Mass is a load-case decision; the ultimate imposed load is not the seismic mass
    • The worst vibration case is often the lightest one
    • A model inherited from the static work has made every dynamic decision by default
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    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