Worked problem

Column Buckling: Why End Conditions Matter

Compare the Euler buckling load of one slender steel column pinned at both ends and fixed at both ends, and why Euler is not a full column check.

Civil & Structural · Last reviewed October 17, 2026

Given

A steel column 5.0 m long with area A = 1,800 mm², minimum-axis second moment of area I = 2.0 × 10⁶ mm⁴, E = 200 GPa and yield stress Fy = 250 MPa. Compare two effective-length assumptions: pinned at both ends (K = 1.0) and fixed at both ends (theoretical K = 0.5).

Find

The ideal Euler buckling load for each assumption, and whether Euler is the controlling check.

Solution

P(cr) = π² × E × I ÷ (K × L)²

Pinned-pinned, K = 1.0: P(cr) = 9.8696 × 200,000 × 2.0 × 10⁶ ÷ 5,000² = 157.9 kN.
Fixed-fixed, K = 0.5: effective length 2,500 mm, so P(cr) = 9.8696 × 200,000 × 2.0 × 10⁶ ÷ 2,500² = 631.7 kN, four times larger.
Radius of gyration r = √(I ÷ A) = √(2.0 × 10⁶ ÷ 1,800) = 33.3 mm. Slenderness KL ÷ r = 150 (pinned) and 75 (fixed).
Euler stress Fe = π² × E ÷ (KL ÷ r)²: 87.7 MPa for pinned, 350.9 MPa for fixed.
The yield load is A × Fy = 1,800 × 250 = 450 kN.

Answer

End condition	Euler load	Slenderness	Compare with yield (450 kN)
Pinned-pinned, K = 1.0	157.9 kN	150	Euler is below yield; Euler governs
Fixed-fixed, K = 0.5	631.7 kN	75	Euler is above yield; Euler does not apply

Why this is not a complete column check

Euler assumes a perfectly straight, elastic column. It applies only when the Euler stress is below roughly half the yield stress, which here means a slenderness above about 133. The fixed-fixed case is stockier than that: the 632 kN result is wrong, and the column would crush or buckle inelastically below 450 kN.
Real columns have initial crookedness, residual stress, load eccentricity and imperfect end restraint. Design codes use reduced curves for these effects.
Fully fixed ends are rare. Codes recommend a larger K than 0.5 for a column that is not perfectly restrained, for example 0.65 for fixed-fixed with realistic restraint, which gives 373.8 kN by the Euler formula. At a slenderness of 97.5 that is still too stocky for Euler to be reliable, so the real strength is lower.
Local buckling of the cross section, lateral-torsional behavior and connections are separate checks.

Common mistakes

Using Euler for a short, stocky column. Check the slenderness first.
Using the strong-axis I instead of the weak-axis I. A column buckles about the axis with the smaller I.
Using the ideal K = 0.5 or 2.0 for a connection that is only partly fixed.
Treating the Euler load as an allowable load. It is a theoretical limit, and design values include safety factors.

Related tools and labs

Calculators, labs and simulators are part of the All-Access Pass.

Questions

What does K represent?

The effective-length factor. It shortens or lengthens the real length to the equivalent pinned-pinned length for buckling. A cantilever column (fixed-free) has K = 2.0, so the load is a quarter of the pinned value.

Why does end fixity quadruple the load?

Halving the effective length multiplies the Euler load by four, because the load varies with the inverse square of the effective length.

What do I use for a design check?

The column curves in the governing steel or timber code, which combine Euler buckling and yielding with imperfection effects.

Educational material. Examples use simplified, hypothetical inputs and are not a design, a construction document or professional engineering advice. Confirm requirements with the codes and standards adopted where the work is built and with a licensed engineer.

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