Euler Buckling Load
Critical axial load at which a column buckles.
Formula
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Example
Longer columns buckle at lower loads
How it works
This formula gives the load at which a long, slender compression member becomes unstable and suddenly bows sideways instead of shortening further. It comes from elastic stability theory: a perfectly straight column under axial compression has a critical point where a tiny lateral deflection can grow because the compressive load creates a bending moment that the member can no longer resist. The stiffness E I and the boundary-condition factor K together determine how much load the member can carry before buckling.
When to use it
Use this when checking slender columns, struts, or compression members where failure is expected by buckling rather than crushing. It is especially relevant for steel, timber, or aluminum members in frames, bracing systems, and temporary supports.
Worked example, step by step
Given: E = 200 GPa = 200 × 10^9 Pa I = 8.0 × 10^-6 m^4 L = 3.0 m K = 1.0 (pinned-pinned ends) Step 1: Write Euler's buckling formula P_cr = (π^2 E I) / (K L)^2 Step 2: Substitute the values P_cr = (π^2 × 200 × 10^9 × 8.0 × 10^-6) / (1.0 × 3.0)^2 Step 3: Compute the denominator (KL)^2 = 3.0^2 = 9.0 Step 4: Compute the numerator π^2 × 200 × 10^9 × 8.0 × 10^-6 = 9.8696 × 1.6 × 10^6 = 1.57914 × 10^7 Step 5: Divide P_cr = (1.57914 × 10^7) / 9.0 P_cr ≈ 1.75 × 10^6 N Final answer: P_cr ≈ 1.75 MN
Common mistakes to avoid
- Using the actual member length instead of the effective length K L
- Mixing up units for I, E, or L so the load comes out wrong
- Applying Euler buckling to a stocky column where material yielding governs before buckling
Did You Know?
Euler derived this buckling formula in 1757.
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