Module 7 · Lesson 7.1

From Euler to a design resistance

Five steps. The first three are mechanics, the last two are calibration, and knowing where the join falls is the point of the lesson.

Why this matters

The Euler load is one of the few results in structural engineering that a student can derive completely and then never use directly. No real column reaches it. What a designer actually uses is a reduction factor read off a calibrated curve — and the relationship between the two is the most misrepresented thing in steel design. Presented carelessly, the buckling curve looks like the Euler load with a safety factor. It is not, and the difference matters.

By the end of this lesson you should be able to

State the Euler load and the squash load, and what each assumes
Compute relative slenderness and interpret it
Explain why real columns fall below both ideal curves
Identify which parts of the chain are derived and which are fitted

What you should already know

Euler buckling (Structural Analysis Fundamentals, Module 10)
Residual stress and where it comes from (Module 1)
Cross-section classification, for Class 4 columns (Module 5)
Cross-section compression resistance (Module 6)

The chain, in five steps

1. The ideal column. Perfectly straight, perfectly elastic, no residual stress, load exactly on the centroid. It buckles at

N_cr = π²EI / L_cr²

This is mechanics — the solution of a differential equation, exact for the ideal member. Notice what is not in it: f_y. An ideal column's buckling load does not know what grade it is made of.

2. The squash load. The other ideal failure: the section simply yields.

N_pl = A f_y

Also exact, and this one does not know how long the member is.

3. Relative slenderness. The ratio of the two, square-rooted:

λ̄ = √(N_pl / N_cr)

Dimensionless, and the single most useful number in member design. It says which of the two ideal failures is closer.

4. Reality. The member is bowed from rolling and handling. It has residual stresses from cooling. It yields progressively, not all at once, so it loses stiffness before it loses strength. It fails below both ideal curves — and worst where they cross.

5. The design curve. A calibrated reduction factor χ, fitted so that χ N_pl sits below the scatter of physical tests.

Steps 1 to 3 are derivable. Steps 4 and 5 are not.

Take care

χ is not the Euler load with a factor on it

The algebra behind χ is an Ayrton–Perry form — the solution for an initially bowed elastic column reaching first yield at its extreme fibre. So the shape has a mechanical ancestry, and that is worth knowing.

But the imperfection factor α is not the bow of any real member. It is a single fitted parameter standing in for three quite different things at once:

initial crookedness from rolling, transport and erection;
residual stress, which makes part of the section yield early;
progressive yielding, which reduces stiffness before it reduces strength.

α was chosen so the resulting curve envelopes test data. You cannot measure it on a member, and you cannot derive it.

The shape has an ancestry. The numbers are calibration. Both halves of that sentence matter, and dropping either one produces a misunderstanding.

The practical consequence is about margin. A design at 0.99 of a derived limit is defensible. A design at 0.99 of a fitted curve, on a member unlike those it was fitted to, is a different act.

Try it

The five steps, on one plot

The squash line, the Euler curve and the design curve together. The design curve is not the Euler curve with a factor on it — watch where it sits relative to BOTH ideal curves as you move along.

Relative slenderness λ̄0.80

Buckling curve

Relative slenderness λ̄: 0.80
Buckling curve: b
Imperfection factor α: 0.34
Squash line N_pl/N_pl: 1.000
Euler ratio N_cr/N_pl: 1.562
Lower ideal curve: 1.000
Design factor χ: 0.724
Gap below the ideal: 27.6 points
Governing behaviour: interaction

λ̄ = 0.80 sits in the worst region, where yielding and buckling interact most strongly. χ = 0.724 — the member reaches neither its squash load nor its Euler load, and the gap between the design curve and both ideal curves is at its widest here.

Things worth trying

Set λ̄ to 0.2. The design curve sits exactly on the squash line — χ = 1, no reduction at all. Stocky members simply yield, and buckling is irrelevant.
Now move to λ̄ = 1.0. The two ideal curves MEET there, both saying the member carries its full squash load, and the design curve says 0.60. That 40-point gap is the widest anywhere, and it is where yielding and buckling interact most strongly.
Keep going to λ̄ = 3. The design curve converges on the Euler curve — the gap falls to about 11%. A very slender member fails nearly elastically, so imperfections have less to work with.
Change the curve from a0 to d at a fixed λ̄. Everything about the member is the same; only the assumed severity of its imperfections has changed. That is what the five curves encode.
Note the labels on the right. Two of these curves are exact results for ideal members and one is fitted to test data. The design curve does not sit at a constant fraction of either — which is the evidence it was not derived from them.

Worked example

A 4 m column, both axes

Given

254 × 254 × 73 UC in S355: A = 9172 mm², I_y = 11 240 cm⁴, I_z = 3907 cm⁴
Height 4.0 m, pinned at both ends about both axes
Class 2 in compression, so the gross area applies

Find

The design buckling resistance, and which axis governs.

Assumptions

Section properties computed from plate geometry, ignoring root radii — about 1.5% below published
Buckling curves b and c, from the selection rules; these are calibrated and unverified here

Predict first

At which relative slenderness is the gap between the design curve and the ideal curves widest?

Why there are five curves

The buckling curves — a0, a, b, c, d — look like arbitrary lettering on a selection table. They are a residual-stress table in disguise, and Module 1 supplied the physics.

A member's stiffness starts falling once part of its section yields. Residual compression at the flange tips means those fibres reach f_y before the rest, at perhaps 70% of the squash load. Once yielded they contribute nothing further to stiffness — and they are the material furthest from the minor axis.

So:

A hot-finished hollow section cools fairly evenly and is efficient about both axes. Curve a.
A rolled I section about its major axis — the residual-compression fibres are close to that axis and matter less. Curve a or b.
The same section about its minor axis — those same fibres are now the ones that matter most. One curve worse.
A welded I section has sharp residual tension at the weld line balanced by compression in the flange tips. Worse again.
Thick flanges cool slowly and unevenly, giving more severe residual stress. Worse again — and thick product also has a lower f_y, so the penalty compounds.

Each step down the table is a statement about how badly the member's stiffness degrades before it reaches yield on average.

Practice

A column has I = 3907 cm⁴ and a buckling length of 4.0 m. What is its Euler load, in kN? Take E = 210 000 N/mm².

Practice

That column has A = 9172 mm² in S355. What is its relative slenderness λ̄?

Practice

At λ̄ = 1.0 on buckling curve b, χ = 0.597. By what percentage does the design resistance fall below BOTH ideal curves, which coincide there?

Check yourself

What does the imperfection factor α in the buckling curves represent?

Summary

N_cr = π²EI/L_cr² and N_pl = A f_y are exact; neither describes a real member
λ̄ = √(N_pl/N_cr) says which ideal failure is nearer
Real members fail below BOTH ideal curves, worst where they cross
At λ̄ = 1 the design curve is 40% below both — the widest gap on the plot
χ has an Ayrton–Perry shape but α is fitted, standing for three effects at once
The five curves are a residual-stress table in disguise
The weak axis is penalised twice: worse λ̄ and a worse curve
A stocky column at λ̄ = 0.8 still loses a third of its cross-section resistance

Progress is kept in this browser only.

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