Module 18 · Lesson 18.4
Real columns, and buckling by energy
Why no column reaches its Euler load, and how to estimate one that has no formula.
Why this matters
Euler's load is exact — for a perfectly straight, perfectly elastic, perfectly concentrically loaded strut. No such column has ever been built. Test a real one and it fails below the Euler load, and the shorter it is the further below. Understanding why is what connects the elegant theory to the design curves in every steel code.
By the end of this lesson you should be able to
- Explain why real columns fall short of both the Euler and the squash load
- Use the Rankine–Gordon formula across the whole slenderness range
- Describe how initial imperfections and residual stresses reduce capacity
- Estimate a critical load by the energy (Rayleigh–Ritz) method
What you should already know
- Euler's critical load and effective length (this module)
- Slenderness ratio (this module)
- Strain energy (Module 15)
Plot failure load against slenderness for a set of real columns and you get a curve that sits below both simple theories.
At very low slenderness the failure load approaches the squash load A σy, but does not quite reach it. At very high slenderness it approaches the Euler load, and again falls a little short. In between — which is where most real columns live — it is well below both, and the two theories together would have you believe the column is far stronger than it is.
Three things account for the shortfall, and none of them is in Euler's derivation:
- 1.Initial crookedness. No column is straight. An initial bow means the axial load has a lever arm from the very first increment, so the column bends immediately rather than at a critical load. There is no sudden bifurcation — deflection grows from zero load.
- 2.Residual stresses. Rolling and welding lock in stresses of the order of 100 N/mm² before any load is applied. Parts of the section reach yield early, those parts stop contributing stiffness, and the effective EI falls.
- 3.Load eccentricity. The load is never applied exactly through the centroid, which adds moment for the same reason as an initial bow.
All three make the column bend earlier than theory allows, and bending is what buckling is.
A useful bridge between the two extremes is the Rankine–Gordon formula, which combines them by adding flexibilities rather than strengths:
What it calculates: a failure load valid at any slenderness
- PR
- Rankine–Gordon failure load (N)
- Pc
- squash (crushing) load, A σy (N)
- Pe
- Euler critical load (N)
This assumes
- An empirical blend, not a derivation from first principles
- The constants in the original form were fitted to test data of its era
In plain terms: The reciprocal form does exactly what is needed at both ends. When the column is very stocky Pe is enormous, 1/Pe vanishes and PR → Pc. When it is very slender Pc is the large one and PR → Pe. In between it is automatically below both, which matches what tests show.
Predict first
A column has a squash load of 900 kN and an Euler load of 900 kN. What does Rankine–Gordon predict?
The last technique is the energy method, and it reaches cases Euler's differential equation cannot.
The idea: at the critical load, the strain energy stored in bending exactly equals the work done by the axial load as the strut shortens along its chord. Setting the two equal:
Pcr ≈ ∫EI(v″)² dx ÷ ∫(v′)² dx
You supply an assumed deflected shape. It need not be the true one — it only has to be admissible, meaning it satisfies the geometric boundary conditions (zero deflection at a pin, zero deflection and slope at a fixed end).
The crucial property is that any admissible shape gives an answer at or above the true critical load. Assuming a shape is equivalent to adding restraint the structure does not have, and restraint can only stiffen it. So the estimate errs on the unsafe side — which is exactly why you would never use a single crude guess for design, and why the method is used with a family of trial shapes, taking the lowest.
Its value is for members Euler cannot handle at all: varying section, varying axial force, elastic restraints along the length.
Worked example
A column of intermediate slenderness
Given
- Pin-ended steel column, length 4.00 m
- A = 3000 mm², I = 4.00 × 10⁶ mm², σy = 355 N/mm², E = 205 000 N/mm²
Find
The squash load, the Euler load, and the Rankine–Gordon estimate.
Practice
A column has A = 3000 mm² and σy = 355 N/mm². What is its squash load, in kN?
Practice
The same column is pin-ended, 4.00 m long, with I = 4.00 × 10⁶ mm⁴ and E = 205 000 N/mm². What is its Euler critical load, in kN?
Practice
Combining those two by Rankine–Gordon, what failure load does the formula predict, in kN?
Worked example
Estimating a buckling load by the energy method
Given
- Pin-ended strut of length L, uniform EI
- Trial shape: the parabola v = a·x(L − x), which is zero at both ends as a pin-ended strut requires
Find
The Rayleigh–Ritz estimate of the critical load, and how far above the exact value it lies.
Practice
A Rayleigh–Ritz estimate using a parabolic trial shape gives Pcr ≈ 12EI/L² for a pin-ended strut. By what percentage does this exceed the exact Euler value of π²EI/L²?
Practice
In the Rayleigh quotient Pcr ≈ ∫EI(v″)² dx ÷ ∫(v′)² dx, what happens to the amplitude of the assumed shape?
Summary
- Real columns fail below both the squash load and the Euler load
- Initial crookedness, residual stresses and eccentricity all reduce capacity
- The shortfall is largest at intermediate slenderness, where most columns sit
- Rankine–Gordon adds flexibilities: 1/PR = 1/Pc + 1/Pe
- The energy method estimates Pcr from an assumed shape and always overestimates it
- Design capacity comes from code buckling curves, not from Euler directly
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