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 held
K
Both ends fixed
0.5
One fixed, one pinned
0.7
Both ends pinned
1
One fixed, one free
2
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:
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