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Structural Analysis Fundamentals / Structural instability

Strength versus stability

Two completely different ways for a column to fail.

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

  • An independent structural reviewer must check remaining derivations, diagrams, sign conventions and the interactive models across their full input ranges.
  • Independent stability review must cover imperfections, local/flexural-torsional modes, restraint, material nonlinearity and code resistance; ideal elastic examples are not design capacities.

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.

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Why this matters

A long ruler pushed end-on does not crush — it springs sideways at a load far below what the material could stand. That is buckling, and it catches people out because nothing about the material strength predicts it.

By the end of this lesson you should be able to

  • Describe buckling as a stability problem
  • Use the Euler formula
  • Say what governs the critical load

A short, stocky column fails when the stress reaches the material's strength — it squashes or yields. That is a strength failure, and P/A tells you when it happens.

A long, slender column does something quite different. Below a certain load it stays straight. At a particular load it suddenly bows sideways and loses all usefulness, even though the stress in the material may be nowhere near the strength. That is a stability failure, and it depends on geometry and stiffness rather than strength.

Euler critical load
Pcr=π2EILe2Le=KLP_{cr} = \frac{\pi^2 EI}{L_e^2} \qquad L_e = K L

What it calculates: The axial load at which an ideal slender column buckles.

Pcr
Critical (buckling) load (N)
E
Young's modulus (N/mm²)
I
Smaller (weak axis) second moment of area (mm⁴)
Le
Effective length (mm)
K
Effective length factor from the end conditions

This assumes

  • The column is perfectly straight and the load is perfectly axial
  • The material is elastic right up to buckling
  • The column is slender enough that it buckles before it yields
  • No initial imperfections or residual stresses

In plain terms: Length is squared on the bottom, so doubling the length quarters the capacity. Notice that strength does not appear at all — a higher grade of steel does nothing whatever for Euler buckling.

Predict first

A pin-ended column's length is doubled, everything else unchanged. What happens to its Euler load?

Practice

A pin-ended column has E = 205 000 N/mm², I = 2.0 × 10⁷ mm⁴ and length 6.0 m. What is its Euler critical load, in kN?

Practice

The same column is now held by fixed ends instead of pins, so its effective length is halved. By what factor does the Euler load increase?

Practice

A column has a squash load (A × fy) of 900 kN and an Euler load of 1124 kN. Which of these two values is the better guide to its failure load? Enter the smaller of the two, in kN.

Summary

  • Short columns squash; slender columns buckle
  • Pcr = π²EI/Lₑ²
  • Strength does not appear — stiffness and geometry govern