Critical load Pcr
-Allowable load
-Pcr / NSlenderness K L / r
-Safety factor at P
-Pcr / applied loadCritical stress
-Pcr / AAll results
| Area A | - |
| Smallest second moment of area I | - |
| Radius of gyration r (smallest) | - |
| Effective length K L | - |
| Euler critical load | - |
| Johnson critical load (valid when K L / r < Cc) | - |
| Transition slenderness Cc | - |
| Squash load A · Sy | - |
| Longest length for the applied load at factor N | - |
Method and formulas
A straight, concentrically loaded column fails by buckling long before the material crushes if it is slender. These are the standard textbook relations (Euler 1744, Johnson), so you can check every result by hand.
| Quantity | Formula |
|---|---|
| Radius of gyration | r = √(I / A), using the smallest I of the section |
| Slenderness ratio | λ = K L / r |
| Euler critical load | Pcr = π² E I / (K L)² = π² E A / λ² |
| Transition slenderness | Cc = √(2 π² E / Sy) |
| Johnson critical load (λ < Cc) | Pcr = A [Sy − Sy² λ² / (4 π² E)] |
| Allowable load | Pall = Pcr / N |
| Section properties | round: I = π d4/64; rectangle: I = b h³/12 (smaller of the two axes); tubes by subtraction of the inner void |
The effective length factor K accounts for the end conditions: 1.0 pinned-pinned, 0.7 fixed-pinned, 0.5 fixed-fixed and 2.0 fixed-free are the theoretical values. Real supports are not perfectly rigid, so design practice often uses larger values for the fixed cases; choose Custom K to enter one. The tool uses the smaller of the two principal second moments, which is conservative if the end conditions differ about each axis. Elastic modulus and yield strength are typical values for the material in the list: choose Custom and enter certified values for critical work.
Worked example: 1 in steel rod, 40 in long, pinned ends
Carbon steel E = 29.7×106 psi, Sy = 53 700 psi. d = 1 in gives A = 0.7854 in², I = π × 14/64 = 0.04909 in4, r = 0.25 in. With K = 1 and L = 40 in, λ = 160. Cc = √(2π² × 29.7×106 / 53 700) = 104.5, so λ > Cc and Euler applies: Pcr = π² × 29.7×106 × 0.04909 / 40² = 9 003 lbf. With a safety factor of 3 the allowable load is about 3 000 lbf.
Questions
How do you calculate the critical buckling load of a column?
For a long slender column, Euler's formula gives the critical load Pcr = π² E I / (K L)², where E is the elastic modulus, I the smallest second moment of area of the cross-section, L the length and K the effective length factor for the end conditions. Example: a 1 in diameter steel rod (E = 29.7×10⁶ psi, I = 0.0491 in⁴) with pinned ends and L = 40 in has Pcr = π² × 29.7×10⁶ × 0.0491 / 40² = 9 000 lbf.
What is the effective length factor K?
K converts the real column into an equivalent pinned-pinned column. The theoretical values are 1.0 for pinned-pinned, 0.5 for fixed-fixed, 0.7 for fixed-pinned and 2.0 for fixed-free (cantilever). Because real supports are never perfectly rigid, design codes recommend larger values for the fixed cases, for example 0.65, 0.80 and 2.1. Choose Custom K to enter your own.
What is the slenderness ratio and when does Euler apply?
The slenderness ratio is K L / r, where r = √(I/A) is the radius of gyration. Euler's formula only holds while the buckling stress stays below the proportional limit. The transition occurs at Cc = √(2π² E / Sy). Above Cc the column is long and Euler applies; below it the column is intermediate and the Johnson parabola gives a lower, more realistic critical load.
What is the Johnson formula?
For intermediate columns (K L / r below Cc), Johnson's parabolic formula is Pcr = A [Sy − Sy² (K L / r)² / (4 π² E)]. It joins Euler's curve tangentially at Cc and tends to the squash load A·Sy for very short columns.
Which axis does a column buckle about?
A column buckles about the axis with the smallest second moment of area, unless it is braced differently in each direction. This calculator uses the smallest I of the section, which is conservative when the two axes have different end conditions; check each axis separately if they differ.
What safety factor should I use for a column?
Buckling is sudden and sensitive to imperfections such as initial bow, load eccentricity and out-of-straightness, so factors of 2 to 3 on the critical load are common for machine design, and structural codes use their own resistance factors. The calculator shows both the allowable load for your chosen factor and the actual factor for an applied load.
What does this calculator not cover?
It covers a straight, concentrically loaded prismatic column of uniform material and ignores eccentric load, initial crookedness, local buckling of thin walls, lateral-torsional buckling, residual stress and creep. It is an engineering aid; verify critical designs and use the applicable design code for structural work.
Engineering aid for guidance. It assumes a straight, concentrically loaded, prismatic column and does not cover eccentric load, initial crookedness, local or lateral-torsional buckling, residual stress, creep or code resistance factors. Moduli and strengths are typical handbook values: use certified data. Verify critical designs; you remain responsible for the design. Nothing you enter leaves your browser except through a share link you copy yourself.