Module 18 · Lesson 18.2
Strength versus stability
Two completely different ways for a column to fail.
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.
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
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