🏛️ Euler Buckling Load Calculator

Calculate the critical buckling load of a column using Euler's formula.

Inputs

Young's Modulus (E)

Area Moment of Inertia (I)

Unsupported Length (L)

Effective Length Factor (K)

Results

Critical Buckling Load (P_cr)0.000e+0 lbf

Buckling Mode Shape

Calculator Description

Buckling is the instability in which a slender column under compression suddenly bows sideways and collapses before the material reaches its compressive strength. Euler’s critical load formula predicts the minimum compressive load at which an ideal elastic column begins to buckle.

What this calculator finds: the critical buckling load P_cr

P_cr is the maximum axial compressive load a column can carry without buckling. Beyond this load, the column bows and collapses regardless of its material strength. For slender members, therefore, this buckling load — not the compressive strength — governs the design.

Why it matters

Verifying the safety of compression members such as columns, struts, screws, and piston rods
Comparing how changes in cross-section, material, or length affect stability
Accounting for end-support conditions (fixed/pinned) via the effective length

Formula

Euler's Critical Load Formula

The critical load is set by the material stiffness (E), the cross-section shape (I), and the effective length (K·L).

P_{cr} = \dfrac{\pi^2 E I}{(K L)^2}

P_cr — Critical buckling load [N (lbf)]
E — Young's modulus [Pa (psi)]
I — Area moment of inertia (use the minimum) [m⁴ (in⁴)]
L — Unsupported length of column [m (in)]
K — Column effective-length factor (pinned–pinned = 1.0, fixed–fixed = 0.5, fixed–free = 2.0)

How the formula works

The critical load is inversely proportional to the square of length — doubling the length cuts the buckling load to one-quarter.
A larger moment of inertia I (material placed farther from the center) increases the critical load.
A smaller effective-length factor K (stiffer end fixity) makes the column more resistant to buckling.

Worked example

For a steel column (E = 200 GPa), I = 5×10⁻⁶ m⁴, L = 3 m, pinned–pinned (K = 1.0): P_cr = π² × 200×10⁹ × 5×10⁻⁶ / (1.0 × 3)² = π² × 1,000,000 / 9 ≈ 1.097×10⁶ N, i.e. about 1,097 kN.

Useful Tips

Practical tips

A column buckles about its weakest axis, so always use the minimum moment of inertia (I_min).
In practice, apply a safety factor (e.g. 2–3) to P_cr to set the allowable load.
Adding an intermediate support to shorten the unsupported length L raises the buckling load quadratically.

Limitations & cautions

Euler’s formula is valid only for slender elastic columns; short, stocky columns yield before buckling and need the Johnson formula instead.
The material must remain within its proportional (elastic) limit; the formula fails once stress exceeds the yield strength.
Initial crookedness, eccentric loading, or residual stress lowers the real buckling load below the theoretical value.