Euler Buckling & the Slender Column

Pick a material, a cross-section, a length and how the ends are held. The page works out the load at which the column buckles, the load at which it would simply be crushed, and which of the two happens first.

Worked example — Euler Buckling & the Slender Column

Cross-sectional area	706.86 mm²
Second moment of area	3.976 × 10⁻⁸ m⁴
Radius of gyration, √(I ÷ A)	7.500 mm
Slenderness, KL ÷ r	400.0
Load at which it buckles	8.72 kN
Load at which it would be crushed	176.71 kN
Which of the two happens first	8.72 kN
So it fails by	buckling

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Euler Buckling & the Slender Column

A short column loaded along its axis fails by crushing: the stress reaches the material's strength and it gives way, at a load of area times strength. A long one does not get that far. It fails by buckling — bowing sideways and then folding — at the Euler critical load, P = π²EI ÷ (KL)², where E is the stiffness, I the second moment of area, L the length and K a factor for how the ends are held. Which failure happens is simply whichever load is lower. The two are usually compared through the slenderness ratio λ = KL ÷ r, where r = √(I ÷ A) is the radius of gyration — a length that says how far the material sits from the centre.

What each symbol means

E: stiffness, in pascals. How hard the material is to stretch, and the only material property in Euler's formula
I: second moment of area, in m⁴. A property of the SHAPE alone, and the reason a tube beats a bar of the same weight
r: radius of gyration, √(I ÷ A), in metres. How far from the centre the section's material effectively sits
λ: slenderness, KL ÷ r. One number that decides which of the two failures you are going to get
K: the end-condition factor. 1 for pinned ends, 0.5 with both fixed, 2 with one end free

Assumes: A perfectly straight column, perfectly centred load, uniform section, and a material that stays elastic right up to the moment it goes. Every one is generous. Euler's load is the point at which a perfect column stops being stable, not the load at which a real one breaks: a real column is slightly bent and slightly off-centre to begin with, so it starts bending immediately and gives way below this line. Treat the number as a ceiling that cannot be exceeded rather than a load that can be used, which is why real design codes replace the sharp corner between the two curves with a smooth transition fitted to tests.
Breaks when: Near the crossover, where neither idealisation is right: the column is stocky enough that the material starts yielding on the compressed side before the elastic buckling load is reached, and the true failure is below both curves. It breaks on the I-section too, and visibly — a rolled beam is about thirteen times stiffer about one axis than the other, so an unbraced column made from one buckles sideways about the weak axis, and the strong-axis figure this page also prints is the wrong one to design from. And it breaks on timber in a way the table cannot show: the figure quoted is a characteristic compressive strength from a graded population, not a yield point, because wood has no yield point at all — it fails by cell walls collapsing.

Stronger steel does not buckle at a higher load 🖖

Press A slender strut and then The same, in stronger steel. Same bar, same length, same ends, and a material with 2.76 times the strength. The buckling load is 8.72 kN before and 8.72 kN after — not close, identical, to every digit the computer holds. Euler's load is π²EI ÷ (KL)², and the strength of the material is not in it anywhere; only E, the stiffness, and E is the same for both steels because all steels are about equally stiff. What the stronger steel did buy is a crushing load that went from 176.71 kN to 487.73 kN — a load this column will never reach. The row for how much of the strength you use goes the wrong way, from 4.93% down to 1.79%: the better material returns the worse number.

There is a length at which the two failures change places 🖖

Crushing does not care about length and buckling cares about almost nothing else, so the two curves in the first panel have to cross. Where they cross is the crossover slenderness, and it is λ = π√(E ÷ strength) — no length in it, no area, no shape, just the two material properties. Structural steel crosses at 88.86 and the high-strength steel at 53.49, which is the same fact from the other side: the stronger the material, the sooner it becomes the wrong thing to be paying for. Below the crossover you are buying strength and above it you are buying stiffness and geometry, and the material's own strength has stopped being the question.

Thickness buys at the fourth power, strength at the first 🖖

If strength does nothing up there, what does? The second panel sweeps the size of the section and measures the slope rather than assuming it: on the buckling branch a solid round bar's critical load goes as the fourth power of its diameter, measured at 4.000000, and on the crushing branch as the square, measured at 2.000000. Doubling the diameter multiplies the buckling load by sixteen. A hollow tube does better still for its weight, because moving the same metal outwards raises I and barely touches A — which is why a scaffold pole is a tube and a bicycle frame is a tube and the solid bar of the first preset is a poor way to spend steel.

Problem solved in full

Choosing between two steels for a 3 m strut 6 steps
A 3 m strut, pinned at both ends, is to be a 30 mm solid round bar. You are offered structural steel at 250 MPa and high-strength steel at 690 MPa for more money. Work out what each would carry, and say which to buy. Press A slender strut to see the first half on the tool.

Open this problem in the tool
1. Start with the geometry, because it is the same for both and it is all Euler needs. A 30 mm round bar has an area of 706.86 mm² and a second moment of 3.976 × 10⁻⁸ m⁴.
  
  $A = \frac{π d ^{2}}{4} = 706.86 m m^{2}, I = \frac{π d ^{4}}{64} = 3.976 \times 1 0^{- 8} m^{4}$
2. The radius of gyration follows, and for a solid round bar it is simply a quarter of the diameter: 7.500 mm.
  
  $r = \frac{I}{A} = \frac{d}{4} = 7.500 mm$
3. Pinned at both ends means K = 1, so the slenderness is the bare length over that: 3 ÷ 0.0075 = 400. That is a very slender column, and the crossover for structural steel is 88.86.
  
  $λ = \frac{K L}{r} = \frac{1 \cdot 3 . 0}{0 . 0075} = 400.0 λ_{c} = π \frac{E}{σ _{y}} = 88.86$
4. Now Euler, with E = 200 GPa for both steels because that is what steel's stiffness is:
  
  $P_{E} = \frac{π ^{2} E I}{( K L ) ^{2}} = \frac{π ^{2} \cdot 200 \times 1 0 ^{9} \cdot 3 . 976 \times 1 0 ^{- 8}}{3 . 0 ^{2}} = 8.72 kN$
5. And the crushing loads, which is the only place the two steels differ at all:
  
  $σ_{y} A σ_{y} P_{E} 250 MPa 176.71 kN 8.72 kN 690 MPa 487.73 kN 8.72 kN$
6. Compare the pairs. The strut buckles at 8.72 kN either way, and it will never see a load near either crushing figure — so the stronger steel contributes nothing whatsoever to what this member can carry.
  
  $\frac{8 . 72}{176 . 71} = 4 . 93%, \frac{8 . 72}{487 . 73} = 1 . 79%$
Answer

Buy the cheap steel. Both buckle at 8.72 kN, and the 690 MPa steel's only effect is to lower the fraction of its strength you use from 4.93% to 1.79% — you would be paying for strength and receiving nothing. What would help is anything that enters Euler's formula: a shorter strut, a restraint at midspan, fixed rather than pinned ends, or the same steel arranged further from the centre. Replace the solid 30 mm bar with a 78 × 3 mm tube — 706.86 mm² of steel either way, so the same weight to carry and the same price to buy — and the second moment goes up 12.52 times, taking the buckling load from 8.72 kN to 109.18 kN. Twelve times the column for none of the money, by moving the metal outwards rather than upgrading it. That is the whole trade, and it is why the answer to a buckling problem is never a better alloy.

$P_{E}_{250} = P_{E}_{690} = 8.72 kN \Rightarrow \frac{I _{78 \times 3}}{I _{30}} = 12.52$

Learning path

Where the load goes

References (2)

Euler's own, and it is an appendix rather than the book. The critical load appears in the additamentum De curvis elasticis, which is where the elastica and this page's formula both start: L. Euler, Methodus inveniendi lineas curvas maximi minimive proprietate gaudentes, Additamentum I: De curvis elasticis. Lausanne and Geneva, 1744.
Where the five materials' stiffness and strength figures would be checked. THEY WERE NOT TAKEN FROM HERE — see the note: A standard materials-property handbook, e.g. the CRC Handbook of Chemistry and Physics or a structural-steel design code such as EN 1993-1-1 for steel and EN 338 for graded timber.

By the 42Math team · Reviewed against published sources. Corrections welcome.