Euler buckling load
Both ends pinned. Constant EI.
Method
P = π²EI/L². Worked case: E 210 kN/mm², I 1×10⁻⁵ m⁴, 3 m is 2,303 kN.
Effective length, imperfections and BS EN 1993 buckling curves are not applied.
What to write beside the result
A review can only check this figure if the inputs sit next to it. Write the value, the unit and the assumption the form used. A diameter in millimetres and a diameter in metres are not the same reading. A force in kilonewtons divided as if it were newtons is a thousand-fold error.
The result is the arithmetic on this page. It is not a utilisation, not a selected product, and not a clause from a standard. If the number is going into a shared container, name the page, the date and the inputs. The file-name checker does not read the calculation.
Logic: how this figure is built.
How to read an Euler load
This page answers the elastic critical load for a pin-ended strut: P = π²EI/L². It is an upper bound from the ideal Euler model. It is not the design resistance of a steel member.
E 210 kN/mm², I 1×10⁻⁵ m⁴ and a length of 3 m give 2,303 kN. The length is the buckling length of the pin-ended model. A fixed base, a partial restraint or a different axis changes that length before it changes the formula. The page does not apply an effective-length factor. If the strut is not pin-ended, change the length you type, and say that you did.
I is the second moment about the axis that buckles. A rectangular bar has two. The smaller I governs. A circular bar has the same I about every diameter. Using the strong-axis I for a weak-axis buckle overstates the load.
Eurocode 3 reduces the Euler load by a buckling curve that accounts for imperfections and yielding. That reduction is not applied here. Use the figure as the elastic critical load you would hand to that check, not as the load the member may carry.
- Related: section properties
- structural logic
Last checked 3 October 2026. See the disclaimer.