Module 18 · Lesson 18.1
Deriving the Euler buckling load
Why a strut fails suddenly at a particular load — an eigenvalue problem, solved from scratch.
Why this matters
Euler's formula is usually presented as something to memorise. Derived properly it is far more interesting: it shows that buckling is not a strength problem at all, but a question of whether a bent shape can exist in equilibrium.
By the end of this lesson you should be able to
- Set up the differential equation for a slightly bent axially loaded strut
- Recognise it as an eigenvalue problem
- Show why only particular loads permit a bent shape
- Generalise to other end conditions using effective length
From first principles
Euler's critical load for a pin-ended strut
We want to show: that a perfect pin-ended strut can only remain bent at P = n²π²EI/L², the lowest of which is the critical load.
Push down on a long ruler. Nothing happens, nothing happens — and then at some load it springs sideways. Nothing about the material changed at that instant, so this is not a strength failure. The trick is to stop asking 'when does it break?' and start asking a different question: is a slightly bent shape possible in equilibrium? For small loads the answer is no — the bending stiffness pushes it straight again. At one particular load a bent shape becomes possible, and beyond it the strut has no reason to stay straight. Finding that load is the whole problem.
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.
Check yourself
Why is buckling described as an eigenvalue problem?
Practice
A pin-ended strut is 4.0 m long, with E = 205 000 N/mm² and I = 1.2 × 10⁷ mm⁴. What is its Euler critical load, in kN?
Practice
The derivation gives P = n²π²EI/L² for n = 1, 2, 3… What is the critical load of the second mode of the same strut, in kN?
Practice
The same strut is instead built in at the base and completely free at the top, so its effective length is twice the actual length. What is the critical load now, in kN?
Summary
- Assume a bent shape, find the moment it creates, and substitute into the beam equation
- EI v″ + Pv = 0 has solutions only when sin kL = 0
- Rejecting the trivial straight solution gives P = n²π²EI/L²
- The first mode is the critical load; effective length generalises it to other supports
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