When a Column Buckles

Squeeze a long ruler from both ends and it stays straight, then suddenly bows out. Mathematically, a bent shape is only possible at special loads, the eigenvalues of a boundary value problem. The smallest one is the buckling load.

How it works

Boundary value problems and eigenvalues

In one sentence

The boundary value problem $y'' + \lambda y = 0$ , $y(0) = y(L) = 0$ has nonzero solutions only for the eigenvalues $\lambda_n = (n\pi/L)^2$ , with eigenfunctions $y_n = \sin(n\pi x/L)$ .

General formula

{\htmlData{sym=1}{E}}{\htmlData{sym=2}{I}}\,{\htmlData{sym=4}{y}}'' + {\htmlData{sym=0}{P}}\,{\htmlData{sym=4}{y}} = 0, \;\; {\htmlData{sym=4}{y}}(0) = {\htmlData{sym=4}{y}}({\htmlData{sym=3}{L}}) = 0 \;\Rightarrow\; \text{bent shapes exist only for } {\htmlData{sym=5}{P_n}} = \frac{n^2\pi^2{\htmlData{sym=1}{E}}{\htmlData{sym=2}{I}}}{{\htmlData{sym=3}{L}}^2}

A sine wave must start and end at zero, so a whole number of half-waves has to fit in the column: only certain wavelengths, and so only certain loads, work.

What the symbols mean

$P$
Load
How hard the ends are pushed together.
$E$
Young's modulus
The material's stiffness.
$I$
Second moment of area
The cross-section's resistance to bending.
$L$
Length
Distance between the pinned ends.
$y$
Sideways bend
How far the column bows out at each point.
$P_n$
Critical loads
The eigenvalues: the only loads that allow a bent shape.

A column pushed with load $P$ bends sideways only if a nonzero $y$ satisfies $EIy'' + Py = 0$ with both ends pinned.
The solutions are $\sin kx$ and $\cos kx$ with $k = \sqrt{P/EI}$ ; $y(0) = 0$ leaves $A\sin kx$ .
$y(L) = 0$ needs $\sin kL = 0$ , so $kL = n\pi$ : only special loads, the eigenvalues, allow a bent shape.
The smallest, $P_1 = \pi^2EI/L^2$ , is Euler's buckling load: beyond it the straight column is unstable.

Learn the math

Boundary Value ProblemsDiff Eq · Lesson 41, Boundary Value Problems and Fourier Series Eigenvalues and EigenfunctionsDiff Eq · Lesson 42, Boundary Value Problems and Fourier Series