Module 5 · Lesson 5.1
Why a plate buckles before the steel yields
A section is not a solid shape. It is an assembly of thin plates, and a thin plate in compression has its own opinion about how much stress it will take.
Why this matters
Every resistance calculation so far has assumed the section can reach its yield stress. For a great many real sections that is simply untrue: a slender web or a wide flange outstand buckles out of plane long before the material is in any distress, and the section never delivers the capacity its area suggests. Cross-section classification is the mechanism by which that possibility enters a design at all — and it is the only one, because no beam element in any analysis will ever mention it.
By the end of this lesson you should be able to
- Write down the elastic critical stress of a compressed plate and read what it depends on
- Explain the effect of edge restraint through kσ
- Compute λ̄p and say which side of 1 a plate is on
- Separate the plate mechanics from the code limits that stand in for it
What you should already know
- Euler buckling and the idea of a critical stress (Structural Analysis Fundamentals, Module 10)
- Steel as a material, and ε = √(235/fy) (Module 1)
- Section geometry and the plates a section is made of
A section is an assembly of plates
A universal beam looks like a single object. Structurally it is three thin plates welded or rolled together: two flanges and a web. Each of those plates is thin compared with its width, and a thin plate in compression can buckle out of its own plane.
The elastic critical stress of a long compressed plate is a standard result of plate theory:
σcr = kσ · π²E / [12(1 − ν²)] · (t/b)²
This is mechanics. It is exact for an ideal plate, and it would be true if no code had ever been written.
Read what is in it, and what is not.
(t/b)² — the geometry, squared. Doubling the thickness quadruples the critical stress. Doubling the width quarters it. Thickness is the powerful lever; width is the dangerous one.
E — the stiffness. Buckling is a stiffness problem, as it always is.
kσ — the edge restraint. More on this below, and it is the largest single factor in the whole subject.
fy does not appear. The stress at which a plate buckles does not depend on what grade of steel it is made of.
That absence is what makes the next lesson's central result possible, and it surprises almost everyone.
Worked example
The same plate, restrained two different ways
Given
- A steel plate 300 mm wide and 10 mm thick, in uniform compression
- E = 210 000 N/mm², ν = 0.3
- S355 steel, so fy = 355 N/mm²
Find
The stress at which it buckles as an internal element and as an outstand.
Assumptions
- A long plate, so the buckling coefficient takes its asymptotic value
- Ideal plate theory — perfectly flat, elastic, no residual stress
Try it
Thickness, width and restraint
The elastic critical stress of a compressed plate, against the stress the steel is being asked to reach. Change the grade and watch σcr refuse to move — it depends on E and geometry, and not at all on fy.
Edge restraint
Steel grade
- Buckling coefficient kσ
- 0.43
- Plate c/t
- 30.0
- Critical stress σcr
- 91 N/mm²
- Yield strength fy
- 355 N/mm²
- Plate slenderness λ̄p
- 1.98
- So the plate
- BUCKLES first
- The other restraint case
- 844 N/mm²
- Ratio between the two
- 9.3
- ε for this grade
- 0.814
λ̄p = 1.98, above 1, so the plate buckles at 91 N/mm² — only 26% of the stress the steel could have taken. The rest of the material is present and not working.
Things worth trying
- Change the grade and watch σcr. It does not move at all — buckling depends on E and geometry, and E is the same for every structural grade. What moves is the yield bar, and therefore λ̄p.
- Switch between internal and outstand. The critical stress changes by a factor of about 9.3, which is exactly 4.0/0.43 — the ratio of the two buckling coefficients. Nothing else in the expression differs.
- Double the thickness. σcr quadruples, because it goes with t². Thickness is by far the most powerful lever available.
- Double the width instead. σcr falls to a quarter. Width is the dangerous direction, and it is the one that grows when a section is made deeper.
- Find the width at which λ̄p crosses 1 for your grade. That is the plate that yields and buckles at the same instant — the worst place to be, and roughly where the class limits sit.
- Set S235 and then S460 at a fixed geometry. The plate becomes relatively MORE slender in the stronger steel, because the stress it must reach has risen while the stress at which it buckles has not.
λ̄p: the number the class limits stand in for
Two stresses now matter for every plate: the stress at which it would yield, fy, and the stress at which it would buckle, σcr. Their ratio decides everything, and it is conventional to take the square root:
λ̄p = √(fy / σcr)
- λ̄p < 1 — the plate yields before it buckles. Local buckling is not the issue.
- λ̄p > 1 — the plate buckles first, at a stress below yield, and only part of it can be relied on.
That single dimensionless number is what cross-section classification is a proxy for. The class limits do not compute λ̄p; they are calibrated boundaries on c/t that stand in for it, chosen so that a plate inside them can be relied on to do a stated job.
Why the code uses c/t instead
Because λ̄p requires σcr, which requires kσ, which requires a judgement about the restraint the neighbouring plates actually provide — and that varies with the stress distribution, the proportions of the section, and what the adjacent plates are doing. Reducing the whole thing to a c/t limit multiplied by ε is an enormous practical simplification, and it is why classification takes thirty seconds rather than a plate-buckling analysis.
The price of that simplification is that the limits are fitted, not derived. A section one unit over a limit is not unsafe. It has simply left the range over which the simplification was calibrated.
Practice
A plate 240 mm wide and 12 mm thick is an internal element in uniform compression (kσ = 4.0). What is its elastic critical stress, in N/mm²? Take π²E/[12(1−ν²)] = 189 800 N/mm².
Practice
The same plate as an outstand (kσ = 0.43). What is its critical stress now, in N/mm²?
Practice
That outstand is in S355. What is its plate slenderness λ̄p?
Check yourself
Two plates have the same width and thickness and are in the same stress state. One is S235 and the other S460. Which buckles at the higher stress?
Summary
- σcr = kσ · π²E/[12(1−ν²)] · (t/b)² — mechanics, exact for an ideal plate
- It goes with (t/b)²: thickness is the strong lever, width the dangerous one
- fy does NOT appear — a plate's buckling stress does not know its grade
- kσ ≈ 4.0 for an internal element, 0.43 for an outstand — a factor of 9.3
- λ̄p = √(fy/σcr) decides which comes first, yielding or buckling
- The class limits are calibrated boundaries on c/t that stand in for λ̄p
- A section one unit over a limit is not unsafe — it has left the calibrated range
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