Mechanical · structures

Column Buckling Calculator

Euler's formula for slender columns, Johnson's for short ones.
Pcr = π²EI / (KL)²
λ 240 · Euler

Material

Cross-section

Column

Critical stress vs. slenderness

Euler (long column)

Watch it load — and buckle

P = 0% of Pcr
speed
Buckling result
Detail
Representative numbers Euler and Johnson formulae, verified in column_model.py. End-fixity factors are the idealised textbook values; real end conditions are rarely that clean, and material constants here are representative.
The assumption behind every number above

No column is straight

what an initial bow of a few millimetres costs

Everything above assumes a member that is dead straight and loaded exactly through its centroid. Under that assumption the column stays arrow-straight at every load below Pcr and then buckles suddenly at it — which is what the animation shows, and it is a theoretical result about an object that does not exist. Rolling tolerance for a structural section is around L/1000. A real column is bent before you load it, so the axial force acts at an eccentricity from the first newton, which bends it further, which increases the eccentricity.

L/5000 — near perfectL/200 — visibly bent

The amplification is exact and it is one line. For a pinned column with an initial bow e0, the midspan deflection under load is δ = e0 / (1 − P/Pe). At a quarter of the Euler load the bow is already a third bigger; at half, doubled; at 90%, ten times. The column never reaches Pe because the bending stress that comes with all that deflection reaches yield first. Setting the extreme-fibre stress P/A + Pδc/I to Sy gives a quadratic in stress — the Perry–Robertson formula, and the ancestor of the buckling curves in every modern steel code.

Notice where the penalty lands. Drag the bow slider and the stocky end of the second chart barely moves — a short column is crushing, and a bow of a few millimetres changes almost nothing. The slender end barely moves either, because Euler buckling there is already so far below yield that the bending amplification has little left to take. The damage is concentrated in the middle, around the transition slenderness, which is exactly where most real columns are designed — and exactly where the two-formula curve above is at its most optimistic. For a solid round steel section at the rolling tolerance of L/1000, the penalty is about 3% at a slenderness of 20 and 5% at 300, but reaches 19% around 120 — and at L/250 that worst case is more than 40%.

Worked example

The default column: 50 mm solid steel, 3 m, pinned both ends. The design curve above gives a critical load of 67.3 kN at a slenderness of 240.

Give it the rolling tolerance, a 3 mm bow, and it reaches first yield at 62.6 kN — 93% of the Euler load. By that point the midspan has moved 43 mm sideways: fourteen times the bow it started with, and most of the diameter of the column.

Bend it to L/250 — a 12 mm bow, visible by eye down the length — and first yield arrives at 52 kN, 77%. The lesson is not that the tolerance is dangerous; it is that the deflection, not the load, is what tells you how close you are. A column that has visibly moved is already most of the way there, whatever the load cell says.

How it works

Two formulas meet in the middle

Euler: slender columns

Pcr = π²EI/(KL)². Depends only on stiffness and geometry, not strength: elastic buckling happens before yield.

Johnson: stocky ones

A short column crushes at yield. The parabolic formula models the in-between, and the two meet exactly at the critical slenderness ratio; the tool picks the one that applies.

End conditions (K)

Pinned–pinned 1; fixed–fixed 0.5; fixed–pinned 0.7; fixed–free (a flagpole) 2, by far the weakest.

Radius of gyration

How efficiently material is spread from the centroid. A tube has a much larger r than a solid rod of equal area, and resists buckling far better.

Worked example. A 50 mm diameter, 3 m steel column pinned at both ends has a slenderness ratio of 240, well into the Euler regime, giving a critical load of about 67 kN. Fix both ends and the same column carries exactly 4× that, since fixed ends halve the effective length and K enters squared. A flagpole end condition cuts it to a quarter.
Field notes

Long, short, or in between

Why buckling can surprise engineers

A column can survive its material’s crushing stress in a lab compression test yet fail at a small fraction of that load once it is slender and long: buckling is a geometry problem hiding inside what looks like a strength problem, a stiffness problem not a strength one.

× 4for half the length

Why K matters so much

It appears squared in the denominator, so halving the effective length (fixing rather than pinning an end) quadruples the critical load: end conditions can matter as much as the material or the cross-section.

Why a hollow tube beats a solid rod

For a given area, pushing the material further from the centreline dramatically increases I and so buckling resistance: a tube carries a far higher critical load than a solid rod of the same weight, which is why so many structural columns are tubular.

2 – 4+safety factor

A safety factor on top of the critical load

Pcr is a theoretical failure point for a perfectly straight, perfectly centred load. Use about 2 for well-known static loads and 4 or more for dynamic loads, uncertainty or human safety (classic machine-design guidance), to cover imperfections, eccentric loading, overloads, defects and corrosion.

Local buckling is a separate check

A thin-walled tube can fail by its wall crumpling locally, long before the whole column buckles as this tool assumes. Thin-wall sections need a separate local-buckling check this tool does not perform.

Questions

What people ask

Why doesn't Euler's formula use yield strength at all?

Because elastic buckling happens before the material ever reaches its yield point — a slender column fails by suddenly deflecting sideways under a load its material could easily survive in pure compression, which is exactly why buckling is a stiffness problem, not a strength problem.

Why does the effective length factor K matter so much?

It appears squared in the denominator, so halving the effective length (say, by fixing rather than pinning an end) quadruples the critical load — end conditions can matter as much as the material or the cross-section itself.

What's radius of gyration, intuitively?

A single number, r = √(I/A), that captures how efficiently a cross-section's material is spread away from its centroid — the same area arranged into a wide, thin shape (like a tube) has a much larger r, and therefore resists buckling far better, than a solid compact shape of equal area.

Why use a hollow tube instead of a solid rod?

For a given amount of material (area), pushing that material further from the centreline dramatically increases I and therefore buckling resistance — a hollow tube can carry a far higher critical load than a solid rod of the same weight, which is exactly why so many structural columns are tubular.

Ideal, centred axial load only. Real columns have imperfections, eccentric loading and residual stress that reduce the true buckling load below this idealised value — always apply a safety factor.
Version history · 2 releases
  1. v1.792026-08-09Column Buckling: a rectangle buckled about the axis you typed rather than the one it would actually fail on — 13× unsafe
  2. v3.992026-08-23Buckling deflection reads 0.0 mm right up to the critical load, then 57 mm

Releases in which this page changed, newest last. Derived from the archived copy of every release, not from notes written afterwards — so it reflects what actually shipped. Site-wide passes are left out; they are in the full changelog.

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