Buckling

Why long, thin members bow sideways before they break, and how to work out when they will

Last updated 2026-09-25

Push the two ends of a plastic ruler or a piece of dry spaghetti towards each other. At first nothing happens. Push a little harder and it suddenly bows out sideways, long before the material itself breaks. That's buckling.

Same push: the short column holds, the long one bows sideways

Two ways to fail when pushed

A member that's being pushed together (in compression) can fail in two different ways:

  • It gets crushed. The material itself runs out of strength. This is what limits short, thick members. Truzme's strength check looks at this one.
  • It buckles. The member bows out sideways and gives way, even though the material is nowhere near its limit. This is what limits long, thin members.

Buckling only happens in compression. A member in tension is being pulled straight, so there's nothing to make it bow.

The Euler formula

The biggest push a straight member can take before it buckles is its critical load, written Ncr. Leonhard Euler worked it out in 1744:

Ncr = π² · E · I / (K · L)²

  • E: how stiff the material is (its modulus of elasticity). Steel is much stiffer than wood, so a steel member buckles much later. It's stiffness that counts here, not strength. The Materials & sections guide explains both.
  • I: how hard the cross-section is to bend (its moment of inertia). It depends on the shape, not just on how much material there is: a tube is much harder to bend than a solid bar made from the same amount of material.
  • L: the member's length. It's squared, so length matters a lot: make a member twice as long and it can only take a quarter of the push.
  • K: how the ends are held. A member pinned at both ends has K = 1. Holding the ends more firmly makes K smaller and the member stronger against buckling. K · L is the length of the part that actually bows out.
How the ends are heldK
Both ends fixed0.5
One fixed, one pinned0.7
Both ends pinned1
One fixed, one free2

It always buckles the easy way

A flat bar has two very different I values; it buckles in the direction of the smaller one

A ruler bends easily across its thin side, and hardly at all across its wide side. A column does the same: it buckles in whichever direction is easiest. So in the formula, you use the smallest I of the cross-section, often written Imin.

A round bar or a round tube is the same in every direction, so it has no weak side.

A worked example

A solid steel bar, 20 mm across and 1 m long, pinned at both ends (K = 1). Steel's E is about 200 000 N/mm².

Keep every length in mm and every force in N, so the units work out. First the I of a solid round bar with diameter d:

I = π · d⁴ / 64 = π · 20⁴ / 64 ≈ 7 854 mm⁴

Then the critical load:

Ncr = π² · 200 000 · 7 854 / (1 · 1 000)² ≈ 15 500 N ≈ 15.5 kN

Now make the same bar 2 m long. The length is squared, so the critical load drops to a quarter:

Ncr ≈ 15.5 kN / 4 ≈ 3.9 kN

Try it

Compare two columns side by side. Column A is your starting point. Change anything on column B, or press Copy B from A and change just one thing, and the bars show how their critical loads compare. Show the math opens the full calculation for both columns.

This interactive demo needs a bigger screen — open this page on a computer to try it

A

Starting column

B

Compare with

K·L

16.3 kN

K·L

4.07 kN

Critical load (Ncr)

A

B

Diameter

Diameter

Length

Length

Both ends pinned

Both ends pinned

Truzme always calculates with K = 1.0 (pinned at both ends)

Things to try:

  • Make B twice as long as A. Its bar drops to a quarter.
  • Keep the length, but pick 0.5 (both ends fixed) for B's end supports. B can now carry four times as much.
  • Change only B's material. The critical load changes in the same proportion as E.
  • Give B a pipe with the same outer diameter as A's round bar. The pipe carries less, but not by nearly as much as the material it saves.
  • Make both columns a flat rectangle, then swap width and height on B. Nothing changes: a rectangle always buckles across its thinner side.

Will it buckle?

Compare the push the member actually carries (its normal force, N) with its critical load:

u = |N| / Ncr

  • u below 1: the member carries less than its critical load.
  • u = 1 or more: the push has reached the critical load, and the member could buckle.

Real designs stay well below 1, because real members are never perfectly straight and loads are never known exactly.

Common mistakes

  • Using the larger I. The member buckles the easy way, so always use the smallest one.
  • Forgetting that the length is squared. Twice as long isn't half as strong; it's a quarter.
  • Mixing units. A length in m with an I in mm⁴ gives a result that's off by a million. Keep everything in mm and N.
  • Checking a member in tension. Only members being pushed can buckle.

Check it in Truzme

Truzme can run this check on every member of your truss that's in compression. The Buckling guide shows how to turn it on and read the results.

Example

Buckling gallery

20 columns, each loaded to right about 100% buckling utilisation

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