Module 10 · Lesson 10.2

Buckling as an eigenvalue problem

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

Why this matters

The Euler formula handles one strut with known end conditions. A real frame has hundreds of members, several load cases and end conditions nobody can name. The eigenvalue formulation solves that: it finds the multiplier on the whole applied load pattern at which the structure loses stiffness, and it tells you where.

It also produces the number most often misread in structural computing. This lesson is as much about interpretation as about method.

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
Compute both amplification factors and say which applies to what
Back-calculate an effective length and state what must accompany it
Distinguish local from global modes using participation

The formulation

Buckling is the load at which the effective stiffness reaches zero. In matrix terms, there is a multiplier α on the applied loads for which the elastic and geometric stiffnesses cancel:

(K + αKg) u = 0

A non-trivial solution exists only for particular values of α. Those are the eigenvalues, and the corresponding u are the eigenvectors — the buckling mode shapes.

There are as many as there are degrees of freedom. The lowest positive one is usually what matters, but not always: a mode confined to a single member may be dealt with by member design, while a higher mode that sways the whole frame may be the one that governs. Look at the shapes, not only the numbers.

What the mode shape is not

An eigenvector is a shape with arbitrary magnitude. The solver normalised it however it chose. Only the ratios between its entries carry meaning, plus the location of the movement.

So a buckling analysis reports:

a load factor — how much the applied load pattern must be multiplied by;
a shape — where the structure loses stiffness, and in what pattern;

and it does not report displacements. Any number that looks like a displacement in a buckling output is the eigenvector's arbitrary normalisation, and reading a millimetre value off it is reading a number the analysis never computed.

The load factor is an amplifier, not a margin

This is the misreading the module exists to prevent. An α_cr of 4.2 does not mean the structure is four times stronger than it needs to be.

Every structure has second-order effects. α_cr tells you how much they are magnifying what is already there:

Lateral deflections, and the moments that come with them, are amplified by α/(α−1).
Initial imperfections grow by 1/(α−1) of themselves.

At α = 4.2 the first is 1.31 — the moments are already 31 % larger than the linear analysis reports. That is not a margin. It is an error in the results you are holding.

Where the 10 % line actually is

The threshold quoted in practice is α_cr = 10. Worth being exact about why, because the algebra and the convention do not agree.

The error from ignoring second-order effects is 1/(α−1). Setting that to 0.10 gives α = 11, not 10. At α = 10 the error is 11.1 %.

The convention is a rounded, memorable number on the conservative side of nothing in particular. This course states the algebra and names the convention as a convention, because a learner who believes 10 was derived will be surprised the day they check.

Second-order amplification

δ_{total} = \frac{α _{cr}}{α _{cr} - 1} δ_{first order} + \frac{1}{α _{cr} - 1} δ_{0}

What it calculates: Total lateral deflection including second-order effects

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

This assumes

α_cr > 1, or the structure has already buckled and the expression is meaningless
The amplification applies only where the mode is active — it is local to the part of the structure that is starting to buckle
Elastic behaviour throughout; yielding needs the nonlinear methods of Module 11

In plain terms: Two amplifiers, not one. The load amplifier α/(α−1) approaches 1 for a stiff structure and is easy to dismiss; the imperfection amplifier 1/(α−1) approaches zero but is unbounded as α approaches 1. At α = 4 the first is a modest 1.33 while the second is already a third of the imperfection — and imperfections are the thing nobody measured.

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.

Lateral stiffness kkN/m

Bar length Lm

Axial compression PkN

Lateral load FkN

Initial out-of-straightnessmm

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

α_cr: 5.00
First-order deflection F/k: 100.0 mm
Amplified deflection: 125.0 mm
From the imperfection: 5.0 mm
Total second-order: 130.0 mm
Under-report if ignored: 30.0 %

At α_cr = 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 α_cr = 4.2 correctly

Given

A braced steel frame, lowest buckling load factor 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

The total sway, and whether a second-order analysis is required

Effective length, and what it needs beside it

Design codes for compression members are written in terms of effective length. A buckling analysis does not give one — it gives a load. But Euler's formula rearranges:

L_e = π√(EI/P_crit)

So a member's effective length can be recovered from its critical load. This is genuinely useful for members whose end conditions defy the standard cases — arch ribs, grid shells, columns in moment frames where the restraint depends on the stiffness of everything attached.

Two cautions that must travel with the number:

It belongs to a mode. An effective length recovered from mode 1 says nothing about mode 3. Quote which.

It needs a length. L_e/L is only meaningful once you have said what L is, and for a curved member or a continuous column through several storeys that is a judgement, not a measurement. The points of contraflexure in the mode's bending diagram are the usual guide.

Local and global modes

A mode with 90 % of its movement in one member is local: that member is the problem, and increasing its section or restraining it will fix it.

A mode spread across the frame is global: no single member is at fault, and stiffening one will move the mode rather than remove it. Global modes need a stability system, not a bigger section.

The way to tell them apart is to sum the squared mode entries over each member or group and see how concentrated the result is. That is what classifyMode in the library does, and it is worth doing explicitly rather than by eye — a mode drawn at an arbitrary scale can look dramatic and be almost entirely confined to one brace.

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.

Elements per member

Length Lm

Euler load π²EI/L²: 8290 kN
Model reports: 10080 kN
Ratio to Euler: 1.2159
Free degrees of freedom: 2

Convergence series. Every value is above Euler, and the sequence decreases monotonically.
Elements	P_cr (kN)	Ratio	Error
1	10080	1.2159	21.59 %
2	8353	1.0075	0.75 %
3	8304	1.0016	0.16 %
4	8295	1.0005	0.05 %
6	8291	1.0001	0.01 %
8	8291	1.0000	0.00 %

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

A frame has α_cr = 6. By what factor are its first-order lateral deflections amplified? Give the factor 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

A buckling analysis reports mode 1 at α = 3.1 with 94 % of its movement in a single roof brace, and mode 2 at α = 5.8 sweeping the whole frame sideways. Which governs the stability of the building?

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
α_cr is an amplifier: deflections by α/(α−1), imperfections by 1/(α−1)
The 10 % error line is at α = 11; 10 is a rounded convention
Effective length is recoverable from Euler, and is meaningless without its mode and its L
A concentrated mode is a member problem; a spread mode is a frame problem

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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