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Computational Engineering / Second-order, buckling and dynamic methods

Buckling as an eigenvalue problem

Finding the load factor that cancels the stiffness, and reading what comes back without over-claiming.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

  • A computational engineering reviewer must check algorithms, convergence claims, statistical assumptions and numerical examples beyond the cases explicitly recorded.
  • Confirm current adopted code edition and national provisions for stability screening; SCI reference documents are explanatory and not a substitute for the governing standard.
  • Audit buckling/eigenvalue helper implementations and lab mode-classification weighting; translations and rotations must not be combined unscaled.
  • Numerical mesh table and all dynamic modelling choices require specialist verification.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

Read the lesson register and outstanding review notes

Why this matters

An elastic eigenvalue buckling analysis estimates critical multipliers for a specified reference load pattern and returns the associated mode shapes. It can identify local member instability or global sway. Interpretation requires the reference loading, restraint assumptions and mode shapes; the result alone is not a design resistance.

By the end of this lesson you should be able to

  • Set up (K + αKg)u = 0 and say what α and u each mean
  • Explain why a buckling load factor is an amplifier and not a margin
  • Use a single-mode amplification estimate and distinguish movement from the initial imperfect position from total offset
  • Back-calculate an effective length and state what must accompany it
  • Distinguish local from global modes using participation

Eigenvalue buckling

For the proportional elastic load model used here, (K + αKg)u = 0 identifies critical factors α and mode shapes u. Kg is formed from a stated reference force pattern. A generalised eigenproblem may have finite, infinite or negative eigenvalues; inspect the physically relevant modes rather than assuming every degree of freedom gives a usable positive factor.

The eigenvector amplitude is arbitrary. It describes the relative pattern of movement, not the actual deflection under load. A low local mode still needs assessment even when a higher global sway mode is used to screen frame sensitivity.

Amplification estimate

In an idealised elastic single-mode model, lateral-load displacement is amplified by α/(α−1). An initial imperfection δ₀ contributes an additional load-induced movement δ₀/(α−1). The total offset from the perfect reference also includes δ₀ itself. These factors are estimates for general frames, not exact multipliers for every displacement or moment.

State the percentage reference

At α = 10, the second-order result is 1.111 times the first-order result. The increase is 11.1% of the first-order value, while omission represents 10% of the second-order value. A 10% increase relative to first order occurs at α = 11. This arithmetic does not supersede the applicability conditions or thresholds in the governing design standard.

Load-induced movement in an idealised single-mode model
δtotal=αcrαcr−1 δfirst order  +  1αcr−1 δ0\delta_{\text{total}} = \frac{\alpha_{cr}}{\alpha_{cr}-1}\,\delta_{\text{first order}} \;+\; \frac{1}{\alpha_{cr}-1}\,\delta_0

What it calculates: Movement from the initial imperfect position under the applied loading

αcr
Buckling load factor for the combination being checked (—)
δfirst order
Deflection from a linear analysis (m)
δ₀
Initial out-of-straightness or out-of-plumb (m)

This assumes

  • αcr > 1 within the idealised elastic equilibrium branch; the formula cannot assess post-buckling behaviour
  • The response and imperfection are represented by the same dominant mode; general frames may need a full second-order analysis
  • Elastic behaviour throughout; yielding needs the nonlinear methods of Module 11

In plain terms: The first term amplifies the first-order lateral-load displacement. The second gives additional movement caused by the initial imperfection. Add δ₀ when reporting the full offset from the perfect geometry.

Try it

P-delta explorer

A lateral load, an axial load and an initial out-of-straightness. The two amplifiers are different numbers and are reported separately.

kN/m
m
kN
kN
mm

The input nobody measures, and the one that grows fastest as α falls.

αcr
5.00
First-order deflection F/k
100.0 mm
Amplified deflection
125.0 mm

× 1.250

From the imperfection
5.0 mm

20 mm × 0.250

Total second-order
130.0 mm
Under-report if ignored
30.0 %

At αcr = 5.00 the error from ignoring second-order effects is 25.0 %. It must be in the design.

The point of separating the two amplifiers

  • Set the imperfection to zero and the total is just the amplified lateral deflection. Most engineers stop there.
  • Put the imperfection back and watch its contribution grow much faster than the lateral one as α falls. At α = 4 the load amplifier is a modest 1.33 and the imperfection is growing by a third of itself.
  • Set F to zero and the structure still deflects — the imperfection alone is amplified. A perfectly-loaded imperfect column deflects; a perfectly-straight loaded one does not.

What this shows: A buckling load factor is an amplifier, not a margin — and the imperfection amplifier is the one that runs away.

Worked example

Reading αcr = 4.2 correctly

Given

  • An idealised elastic sway model with a dominant critical load factor of 4.2 for the governing combination
  • First-order sway deflection at roof level: 18 mm
  • Erection out-of-plumb allowance: 12 mm over the height

Find

Estimate movement from the initial imperfect position and total offset from the perfect reference

    Effective length

    For a member represented by Euler flexural buckling with an appropriate EI and critical axial force, Le = π√(EI/Pcrit). Record the buckling mode, member force, stiffness assumptions and physical length used in any Le/L ratio. A frame eigenvalue is a multiplier on a load pattern; it is not itself a member critical force. Curved members, non-uniform force and interacting modes require particular care before using this equivalence in a design check.

    Local and global modes

    A mode concentrated in one member suggests local member instability; a mode distributed through the frame suggests global behaviour. Both can matter to safety, and a local mode cannot be dismissed merely because a higher sway mode is available. Altering a member or restraint changes the model and requires reassessment.

    Use mode plots and suitably defined measures of deformation or strain energy. A sum of squared nodal entries is only a diagnostic: rotations and translations have different units, mesh density changes the weighting, and a percentage alone does not determine the governing design mode.

    Try it

    Buckling mode explorer

    A pin-ended strut with EI = 21 000 kN·m². Change how many elements model it and watch the reported buckling load.

    m
    Euler load π²EI/L²
    8290 kN
    Model reports
    10080 kN
    Ratio to Euler
    1.2159

    unsafe — the model is stiffer than the strut

    Free degrees of freedom
    2
    Convergence series. Every value is above Euler, and the sequence decreases monotonically.
    ElementsPcr (kN)RatioError
    1100801.215921.59 %
    283531.00750.75 %
    383041.00160.16 %
    482951.00050.05 %
    682911.00010.01 %
    882911.00000.00 %

    A single element gives exactly 12EI/L². That is 21.6 % above Euler, in the unsafe direction, from a model that would give the exact answer for a static analysis.

    Why it converges from above

    • A finite element model of a continuum is a CONSTRAINED version of it: the displacement field can only take the shapes the elements allow.
    • A constrained structure is stiffer, so the computed buckling load is always too high — and always in the unsafe direction.
    • The same is not true of a static analysis with cubic elements, which is exact for a prismatic member under end loads. Same model, different question, different adequacy.

    What this shows: One element per member is exact for statics and about 22 % unsafe for buckling — because a single element has no node at mid-height to displace.

    Practice

    For the idealised single-mode model, αcr = 6. Estimate the factor on first-order lateral displacement to three decimal places.

    Practice

    A pin-ended strut has E = 210 × 10⁶ kN/m², I = 1 × 10⁻⁴ m⁴ and L = 4 m. Compute its Euler buckling load in kN.

    Check yourself

    Mode 1 at α = 3.1 is concentrated in a roof brace; mode 2 at α = 5.8 is a global sway mode. How should these results be used?

    Check yourself

    Why is the mode shape from a buckling analysis as important as the load factor?

    Check yourself

    A linear buckling analysis gives a load factor of 12 for a real steel frame. Should the frame be designed for that?

    Summary

    • (K + αKg)u = 0 — α is the load multiplier, u is a shape with arbitrary magnitude
    • A buckling analysis reports a factor and a shape, never a displacement
    • Single-mode amplification estimates require a stated displacement reference and appropriate assumptions
    • At α = 10, the increase is 11.1% of first order, or 10% of the second-order result
    • Effective length is recoverable from Euler, and is meaningless without its mode and its L
    • Assess both local member modes and global modes within the complete stability system