Module 18 · Lesson 18.3
Effective length, slenderness and the limits
How end restraint helps, which axis buckles, and where the theory runs out.
Why 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
- Buckling occurs about the axis with the smaller I
- λ = Lₑ/r decides whether buckling or yielding governs
- Euler is an ideal upper bound — real design needs code-based checks
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint