Euler Buckling Load Calculator
Find the compressive load at which a slender column buckles, from its length, section, material and end conditions.
What it does
Stand a steel pole in the ground and push down on it. It feels enormously strong, and for a while it is. Then it bows sideways almost at once and snaps. Nothing about the material failed. The pole simply became too slender to stay straight, and it buckled.
Leonhard Euler worked this out in 1744, and the formula he wrote is still the starting point for every column check. The critical load is pi squared times the modulus of elasticity times the second moment of area, divided by the square of the effective length. The engineering mechanics text used here sets it out with the four standard end conditions, each with its own effective length factor K, and notes that the effective length accounts for boundary conditions other than a simple pin on both ends.
Three things decide the answer. The modulus of elasticity says how stiff the material is. The second moment of area says how hard the section is to bend, and it goes with the fourth power of the depth, so a section twice as deep is sixteen times as good at resisting buckling. The effective length says how the ends behave.
This page covers the four usual cases: pinned at both ends with K equal to 1, fixed at both ends with K equal to 0.5, fixed at one end and pinned at the other with K close to 0.7, and a cantilever with K equal to 2. Give it a length, a section and a material, and it returns the critical load, the critical stress, the slenderness ratio and a plain reading of whether the column is long enough for the formula to be honest.
The limits are clear. The formula assumes a perfectly straight, perfectly centred, perfectly elastic column, and a real one is never any of those. It also gives the load at which buckling starts, not the load the column can safely carry, and a design check has to add the safety factors, the imperfections and the strength limit of the material.
How to use
- Choose metric or imperial for the length, the section and the answer.
- Pick the material. The list carries the usual modulus of elasticity for each one.
- Pick the end conditions, which set the effective length factor K.
- Pick the cross section and enter its size: a diameter for a solid round, a side for a square, or a width and a depth for a rectangle.
- Read the critical load and the critical stress, and check the critical stress against the yield strength of the material you chose.
- Look at the slenderness ratio. Above about 100 the Euler number is the right answer, below that the material strength decides and the calculated load is too generous.
Formula
- Pcr
- Load at which the column starts to bow sideways, in newtons or pounds force.
- E
- Modulus of elasticity of the material, in pascals.
- I
- Smallest second moment of area of the section, in metres to the fourth or inches to the fourth.
- L
- Unsupported length of the column between its end restraints.
- K
- Effective length factor: 1 pinned, 0.5 fixed, 0.699 fixed and pinned, 2 cantilever.
The critical stress is the critical load divided by the cross section area, and it can also be written as pi squared times E divided by the square of the slenderness ratio, where the slenderness ratio is the effective length K L over the radius of gyration r, and r is the square root of I over A. For a solid round section the second moment of area is pi d to the fourth over 64, for a square it is the fourth power of the side over 12, and for a rectangle it is the width times the cube of the depth over 12.
Worked example
Example: a 50 mm steel column 2 m long, pinned at both ends
Defaults: metric units, structural steel, pinned at both ends, a solid round section of 50 mm and a length of 2 m.
- Structural steel has a modulus of elasticity of 200 GPa, which is 200000 MPa.
- The area of the round section is pi x 50 squared over 4 = 1963.5 mm2.
- The second moment of area is pi x 50 to the fourth over 64 = 306,796.1576 mm4.
- The radius of gyration is the square root of I over A, which is 12.50 mm, so the slenderness ratio is K L over r = 2000 / 12.5 = 160.0.
- With pinned ends K is 1, so Pcr = pi squared x 200000 x 306,796.1576 over 2000 squared = 151,398 N, which is 151.40 kN.
- The critical stress is the load over the area: 151,398 / 1963.5 = 77.11 MPa.
- The tool shows 151.40 kN, 77.11 MPa, 1,963.5 mm2, 306,796.1576 mm4, 160.0 and 12.50 mm, and it calls this a slender column where elastic buckling governs.
Reference table
| Length | Critical load | Critical stress | Slenderness ratio |
|---|---|---|---|
| 1.5 m | 269.15 kN | 137.08 MPa | 120.0 |
| 2 m | 151.40 kN | 77.11 MPa | 160.0 |
| 2.5 m | 96.89 kN | 49.35 MPa | 200.0 |
| 3 m | 67.29 kN | 34.27 MPa | 240.0 |
| 4 m | 37.85 kN | 19.28 MPa | 320.0 |
Limits
- The formula assumes a perfectly straight, perfectly centred column that stays elastic, and real columns carry out of straightness from the moment they are made.
- It gives the load at which buckling starts, not the safe working load, so it does not include safety factors or codes of practice.
- It ignores local buckling of the plates in a thin section, which can happen before the whole column bows.
- A short, stuby column is governed by crushing or yielding of the material, and the Euler number for it is far too large.
Common mistakes
- Using the depth of the section for a rectangle in the wrong place. The second moment of area needs the depth to the third power, so the cube of the side facing the buckling direction is what matters.
- Forgetting the effective length factor. A column fixed at both ends carries four times the load of the same column pinned at both ends.
- Treating the Euler load as the safe load. It is the point where the column starts to move sideways, and the code load is a fraction of it.
- Using a modulus of elasticity for the wrong material. Aluminium at 70 GPa gives less than half the load of steel at 200 GPa for the same section.
- Checking only the critical load and never the critical stress. A section can have plenty of buckling capacity and still yield before it buckles.
Try it yourself
Quick checks. Answers sit tucked under each question.
What is the critical load of the same column if it is fixed at both ends?
605.59 kN, which is four times the 151.40 kN of the pinned column, because the effective length halves and the load goes with the square of that length. The slenderness ratio drops to 80.0, so the column is now in the range where material strength has to be checked too.
The column is made twice as long. What happens to the load?
It falls to a quarter. At 4 m the critical load is 37.85 kN, against 151.40 kN at 2 m, and the slenderness ratio climbs from 160.0 to 320.0, which is a warning that the column is getting harder to keep straight.
What if the same column is made of aluminium instead of steel?
The critical load falls to 52.99 kN and the critical stress to 26.99 MPa. The section is unchanged, so the area and the second moment of area stay the same, and the whole difference comes from the modulus of elasticity: 70 GPa for aluminium against 200 GPa for steel.
Questions
What is Euler buckling?
It is the sideways bending failure of a slender compression member. The load reaches a critical value and the column deflects more and more until it breaks, even though the material is still well below its strength.
What is the effective length factor K?
A number that folds the end conditions into one formula. The text behind this page lists pinned at both ends as 1, fixed at both ends as 0.5, fixed and pinned as 0.699 and a cantilever as 2, and the effective length is K times the real length.
Why does the second moment of area matter so much?
It enters as the fourth power of the depth for a rectangle, so doubling the depth of a column raises its buckling load by a factor of sixteen. That is why deep sections are used for columns and why a wide flat plate is a poor column.
How do I know if the Euler formula applies?
Look at the slenderness ratio. Above roughly 100 the column is long and the critical stress falls below the proportional limit, so the elastic formula holds. Below that, the column is stocky and the material strength decides.
Does the critical load depend on the material strength?
Not on the strength itself. The Euler load depends on the modulus of elasticity, which says how stiff the material is. Yield strength matters for short columns and for the safety check that comes after the buckling load is known.
What about a column that is braced part way up?
Then the length between the braces is the length to use, and each length gets its own end conditions. A brace near the middle roughly doubles the load the column can take.
Sources
- Engineering LibreTexts: buckling of columns and plates, with the effective length factors
- Engineering mechanics, strength of materials, chapter 15: columns
- Engineering LibreTexts: chapter 8 on beam bending, buckling and torsion