Column Buckling Calculator (Euler and Johnson)
Critical buckling load and critical stress for a compression member, using the Euler formula for slender columns and the Johnson parabola for stocky ones that would buckle inelastically. The calculator reports the slenderness ratio, the effective length for each of the four end conditions, which of the two formulas actually governs, and the factor of safety on the applied axial load.
How to use this calculator
- Enter the unbraced length of the column and choose its end condition, from fixed-fixed through pinned-pinned to the free-ended cantilever.
- Enter Young's modulus, the second moment of area about the weak axis and the cross sectional area of the section.
- Enter the yield strength, which sets the slenderness at which the elastic Euler formula stops applying.
- Enter the axial load the member actually carries to get the factor of safety, and check which formula governs.
Frequently asked questions
What is the difference between Euler and Johnson buckling?
Euler's formula, P = pi^2 E I / (K L)^2, assumes the material stays elastic right up to the moment it buckles, so it only holds for slender members. Below a slenderness ratio of pi x sqrt(2 E / sy) the column would reach a critical stress above half its yield strength, so it starts to yield before it buckles and Euler overstates the capacity. The Johnson parabola, sigma = sy - (sy^2 / 4 pi^2 E) x (K L / r)^2, takes over in that range and the two curves meet exactly at half the yield strength.
Which K factor should I use for the effective length?
The theoretical values are 0.5 for fixed at both ends, 0.7 for fixed at one end and pinned at the other, 1.0 for pinned at both ends and 2.0 for fixed at one end and free at the other. Real structures are never perfectly fixed or perfectly pinned, so design codes push these values up, often to 0.65, 0.8, 1.0 and 2.1, and a joint that looks pinned may behave partly fixed.
Why does the buckling load depend on the second moment of area rather than the area?
Because buckling is a stiffness problem, not a strength one. Two columns with the same cross sectional area but different shapes carry very different loads, since a tube resists bending far better than a solid bar of the same weight. That is also why the calculation must use the axis with the smaller I, because the member always buckles about its weak axis first.
How generous is the factor of safety on a column?
Three is the usual floor, and codes often go higher. The Euler load assumes the member is perfectly straight, the load is exactly axial, the material is homogeneous and the supports are exactly as modelled. None of those hold in a real structure, and a column that buckles loses its capacity almost completely rather than redistributing load the way a beam does.