Euler Buckling
Critical buckling load, slenderness ratio, and critical stress for a column.
The load at which the column becomes unstable and bows sideways. Apply a safety factor on top — this is failure, not a working load.
Effective length 3.00 m ÷ radius of gyration 12.50 mm. Euler applies above roughly 89 for this material; below that, use a short-column or Johnson formula.
Pcr ÷ area. Compare against the 250 MPa yield strength to see which failure mode governs.
Buckling always happens about the axis with the smallest second moment of area, which is why the weak axis is used here. The result also assumes a perfectly straight column loaded exactly on its centroid — real imperfections mean actual capacity is lower, which is what design codes account for.
A slender column fails by bowing sideways long before it is crushed. Euler's critical load is Pcr = π²EI ÷ (KL)², so a 3 m pinned steel bar 50 mm across buckles at about 67 kN — far below the load its cross-section could otherwise carry.
Failure by geometry, not by strength
Compress a short block and it eventually crushes when the stress reaches the material's yield strength. Compress a long thin one and something quite different happens: at a load well below yield it suddenly bows sideways and collapses. That is buckling, and it is a stability failure rather than a strength failure.
The distinction matters because buckling depends mostly on geometry and stiffness, not on how strong the material is. Swapping mild steel for a high-strength alloy barely helps a slender column, because both have almost the same Young's modulus. Making the section fatter, or the column shorter, helps enormously.
End conditions change everything
How the column is held at its ends changes the effective length over which it can bow. A column fixed at both ends can only buckle in a shorter wave, so its effective length is half the real one — and since the critical load depends on the square of that, fixing both ends makes it four times stronger. A column fixed at one end and free at the other has an effective length twice its real one, making it four times weaker than pinned.
I is the second moment of area about the weakest axis, K the effective-length factor, and r the radius of gyration. Euler applies only while σcr stays below the yield strength.
Worked example: 50 mm steel bar, 3 m, pinned ends
Section properties first, then the critical load:
- 1 Find the second moment of area. For a solid round, I = πD⁴/64 = π × 0.05⁴ / 64 = 3.068 × 10⁻⁷ m⁴.
- 2 Take the effective length. Pinned at both ends means K = 1.0, so KL = 3 m.
- 3 Apply Euler’s formula. π² × 200 GPa × 3.068e−7 ÷ 3² ≈ 67.3 kN.
- 4 Check the slenderness. r = √(I/A) = 12.5 mm, so KL/r = 3000 ÷ 12.5 = 240.
- 5 Confirm Euler applies. Critical stress is 34.3 MPa, well under the 250 MPa yield, so the column buckles elastically as the formula assumes.
- 6 Apply a safety factor. Pcr is the collapse load, not a working load — design codes divide it down substantially.
Effective-length factors
K multiplies the real length. Because the critical load depends on (KL)², halving K quadruples the capacity.
| End conditions | K (theory) | Effect on Pcr |
|---|---|---|
| Fixed – fixed | 0.5 | 4× the pinned case |
| Fixed – pinned | 0.699 | ≈ 2× the pinned case |
| Pinned – pinned | 1.0 | The reference case |
| Fixed – free (cantilever) | 2.0 | ¼ of the pinned case |
Where Euler stops being valid
The formula assumes the column stays elastic all the way to buckling. Below a certain slenderness that stops being true: the predicted critical stress exceeds the yield strength, meaning the column would crush before it could buckle. For steel with a 250 MPa yield and 200 GPa modulus that transition sits near a slenderness of 89.
Below that threshold, Euler over-predicts capacity badly and a short-column approach such as the Johnson parabola or a code-specified curve is required. The calculator flags this rather than returning a confidently wrong number.
Two further practical points. Buckling always occurs about the axis with the smallest second moment of area, which is why a rectangular section bends about its weak axis and why this tool uses that value. And real columns are never perfectly straight nor perfectly centrally loaded; those imperfections mean actual capacity falls below the Euler prediction, which is exactly what the safety factors in design codes exist to cover.