Structural Analysis Fundamentals / Structural instability
Effective length, slenderness and the limits
How end restraint helps, which axis buckles, and where the theory runs out.
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Checks needed for project use
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Read the lesson register and outstanding review notesWhy this matters
The Euler formula is easy. Choosing the effective length and the right axis is where real judgement is needed — and where most of the errors happen.
By the end of this lesson you should be able to
- Choose K for the end conditions
- Calculate slenderness
- Identify the buckling axis
- State the limits of Euler theory
Effective length is the length of an equivalent pin-ended column that would buckle at the same load. Holding the ends more firmly shortens it and raises the capacity. For ideal conditions: pinned at both ends gives K = 1.0; fixed at both ends gives K = 0.5; fixed at one end and free at the other gives K = 2.0; fixed at one end and pinned at the other gives about K = 0.7.
Those are ideal values. Real connections rarely achieve full fixity, so codes give more conservative values to use in practice.
A column will always buckle about the axis that gives it the least resistance — the one with the smaller second moment of area. For a rectangular section that means bending about the axis parallel to the longer side, so the column bows in the direction of its thinner dimension. Checking only the strong axis is a classic and dangerous mistake.
What it calculates: How slender a column is, which decides whether buckling or yielding governs.
- λ
- Slenderness ratio (no units)
- Le
- Effective length (mm)
- r
- Radius of gyration about the weak axis (mm)
- I
- Weak-axis second moment of area (mm⁴)
- A
- Cross-sectional area (mm²)
This assumes
- The weak axis governs
- Uniform section along the length
In plain terms: High λ means slender, so Euler buckling governs. Low λ means stocky, so the material yields first. In between, real columns fail by a mixture of the two, which is what design codes actually model.
Try it
Buckle a column
Euler's load depends on the effective length and the smaller second moment of area. Buckling happens about the weak axis.
mm
End conditions
mm
mm
- Weak-axis I
- 1.250e+7 mm⁴
- Radius of gyration r
- 28.9 mm
- Slenderness λ = Le/r
- 139
- Euler load Pcr
- 1580.7 kN
Doubling the length quarters the Euler load. Fixing the ends shortens the effective length and raises it. Real design must also allow for imperfections and follow a code.
Worked example
Critical load of a pin-ended steel column
Given
- Rectangular section 100 mm × 150 mm
- Length 3.5 m, pinned at both ends
- E = 205 000 N/mm²
Find
The Euler critical load, and whether the theory is appropriate.
Assumptions
- Ideal pinned ends
- Perfectly straight column, axial load
- Elastic behaviour up to buckling
Practice
A pin-ended column has E = 205 000 N/mm², weak-axis I = 8.0 × 10⁶ mm⁴ and a length of 4.0 m. What is its Euler critical load in kN?
Practice
A column 5.0 m long is fixed at both ends. E = 205 000 N/mm² and the weak-axis I = 2.0 × 10⁷ mm⁴. What is its Euler critical load in kN?
Practice
A column has an effective length of 3000 mm and a weak-axis radius of gyration of 25 mm. What is its slenderness ratio?
Practice
A pin-ended column has a critical load of 800 kN. If its length is doubled with everything else unchanged, what is the new critical load?
Summary
- Effective length Lₑ = KL; K = 1.0 pinned, 0.5 fixed–fixed, 2.0 fixed–free, 0.7 fixed–pinned
- Compare EI/Lₑ² about both axes; the smaller I governs only when the other conditions are equal
- λ = Lₑ/r decides whether buckling or yielding governs
- Euler is an ideal upper bound — real design needs code-based checks