Column Buckling

Calculate the Euler critical buckling load, effective length, radius of gyration, slenderness ratio, and critical stress for axially loaded columns with pinned, fixed, and free end conditions, compared against yield.


Euler Buckling Theory

End Conditions

Column Properties

MPa
mm⁴
mm
mm²
MPa

Results

1.10 × 10⁶N
Critical load Pcr
1096.6kN
Pcr
1.00
K factor
6000mm
Effective length K·L
57.74mm
Radius of gyration r
103.9
Slenderness K·L/r
182.77MPa
Critical stress σcr
74.6
Critical slenderness
Elastic buckling governs (σcr < Fy): the column buckles before yielding.

About Column Buckling Calculator

The column buckling calculator computes the Euler elastic critical buckling load Pcr for a straight, axially loaded, prismatic column. It applies the effective length factor K for four idealised end conditions: pinned-pinned (K = 1.0), fixed-fixed (K = 0.5), fixed-pinned (K = 0.7), and fixed-free, the cantilever case (K = 2.0).

Enter the modulus of elasticity E, the least second moment of area I, the unbraced length L, and the cross-sectional area A in a consistent SI unit set (newtons, millimetres, megapascals). The tool returns the critical load, the effective length, the radius of gyration, the slenderness ratio, and the critical buckling stress, and when a yield strength is supplied it reports whether elastic buckling or yielding governs.

How It Works

  1. Select the column end condition, which sets the theoretical effective length factor K.
  2. Enter E (MPa), the least I (mm^4), the unbraced length L (mm), and the area A (mm^2); optionally enter the yield strength Fy (MPa).
  3. The calculator computes the effective length K*L, then Pcr = pi^2 E I / (K L)^2 and the critical stress Pcr / A.
  4. It derives the radius of gyration r = sqrt(I/A) and the slenderness ratio K*L / r, and compares the critical stress with the yield strength to decide whether buckling or yielding governs.

Worked Example

A pinned-pinned steel column (K = 1.0) with E = 200,000 MPa, I = 2.0e7 mm^4, length L = 6000 mm, area A = 6000 mm^2, and Fy = 355 MPa. The effective length is K L = 6000 mm. Pcr = pi^2 * 200000 * 2.0e7 / 6000^2 = 1.097e6 N (1097 kN). The radius of gyration r = sqrt(2.0e7 / 6000) = 57.7 mm, so the slenderness ratio K L / r = 6000 / 57.7 = 103.9. The critical stress is 1.097e6 / 6000 = 182.8 MPa, which is below Fy = 355 MPa, so elastic buckling governs.

Formulas

Euler critical buckling load
Pcr = pi^2 * E * I / (K * L)^2
Effective length factors (theoretical)
K = 1.0 pinned-pinned | 0.5 fixed-fixed | 0.7 fixed-pinned | 2.0 fixed-free
Radius of gyration
r = sqrt(I / A)
Slenderness ratio
lambda = K * L / r
Critical buckling stress
sigma_cr = Pcr / A = pi^2 * E / lambda^2

Standards & References

  • Euler buckling theory
  • Gere & Timoshenko, Mechanics of Materials
  • Theory of Elastic Stability (Timoshenko & Gere)

Frequently Asked Questions

What is the effective length factor K?

K converts the actual unbraced length into the effective length K*L over which the column buckles in a half sine wave. Theoretical values are 1.0 for pinned-pinned, 0.5 for fixed-fixed, 0.7 for fixed-pinned, and 2.0 for fixed-free.

When does the Euler formula stop being valid?

The Euler formula assumes elastic buckling, so it is valid only when the critical stress stays below the yield (or proportional-limit) stress. For stocky columns below the critical slenderness sqrt(pi^2 E / Fy), yielding or inelastic buckling governs instead.

Which moment of inertia should I enter?

Enter the least (minimum) second moment of area, because the column buckles about its weakest axis unless it is braced against that direction. Using the larger I would overestimate the buckling capacity.

Does the calculator include imperfections or a safety factor?

No. It reports the theoretical elastic critical load only. Real designs apply code-based buckling curves, imperfection factors, and partial safety or resistance factors to obtain a reduced design capacity.