Buckling Calculator

Find the Euler critical buckling load of a column.

Critical buckling load (N) 109,662.27

Formula: P_cr = π²·E·I ÷ (K·L)²

Step-by-step with your numbers:
1. Values used:
2. Young's modulus = 200 GPa
3. Area moment of inertia = 500,000 mm⁴
4. Column length = 3
5. Effective length factor = 1
6.
7. Critical buckling load = 109,662.27N
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Euler's formula predicts when a slender column suddenly buckles under compression.

How the Math Works

The Euler buckling formula calculates the critical load at which a column will suddenly bend or buckle under compression. The formula P_cr = π²·E·I ÷ (K·L)² combines four key parameters: E (modulus of elasticity) measures the material's stiffness, I (moment of inertia) represents the column's resistance to bending based on its cross-sectional shape, L is the unsupported length of the column, and K is the effective length factor that accounts for how the column ends are restrained. The π² term arises from solving the differential equation of a bent column, and the squared denominator emphasizes that buckling load decreases dramatically with longer columns.

Practical Applications

To use this calculator for a structural column, first determine the material's modulus of elasticity (E) from engineering tables - for example, steel has E = 200 GPa. Calculate the moment of inertia (I) based on your cross-section: for a rectangular column with width b and height h, I = bh³/12. Determine the effective length factor (K) based on end conditions: K=1.0 for pinned-pinned, K=0.5 for fixed-fixed, K=2.0 for fixed-free. Input these values with the column length (L) to find the maximum load your column can safely support before buckling occurs.

Day-to-Day Use

This calculation helps ensure the safety of countless structures in your daily life, from the steel beams in your office building to the bridge you drive across. When furniture makers select wooden table legs, they use these principles to prevent wobbly or collapsed furniture. Even bicycle frame designers rely on buckling calculations to ensure frames are lightweight yet strong enough to support your weight without failing. Understanding buckling helps explain why utility poles have specific bracing and why tall buildings require internal supports at regular intervals throughout their height.

Worked example

Steel, I = 5×10⁵ mm⁴, 3 m, K = 1 → about 110 kN.

FAQ

What is the K factor?

It accounts for end conditions: 1 for pinned ends, 0.5 for fixed-fixed, 2 for cantilever.