Euler's critical load

Remarks

The proof is a short computation. For $P > 0$ write $k^2 = P/(EI)$; the general solution is $w = A \cos kx + B \sin kx$. The condition $w(0) = 0$ gives $A = 0$, and $w(L) = 0$ with $B \ne 0$ forces $\sin kL = 0$, so $kL = n\pi$. For $P \le 0$ the only solution is $w = 0$.

For other end conditions the same analysis gives $P_{cr} = \pi^2 E I / (K L)^2$ with an effective length factor $K$. One has $K = 1/2$ for both ends clamped, $K = 2$ for one end clamped and the other free, and $K \approx 0.699$ for one end clamped and the other pinned. Dividing by the cross-section area $A$ and writing $r = \sqrt{I/A}$ for the radius of gyration gives the critical stress $\sigma_{cr} = \pi^2 E / (KL/r)^2$.

The hypotheses matter. The formula assumes the stress stays below the elastic limit. For short columns $\sigma_{cr}$ would exceed the yield stress and the column fails by yielding first, so Johnson's parabolic formula is used there. It also assumes a perfectly straight column with a centred load; real columns have imperfections and are treated by the Perry–Robertson formula.

Going deeper

For a student

The formula comes from a linear eigenvalue problem: the critical loads are the eigenvalues of $-EI\, d^2/dx^2$ with Dirichlet conditions, and the buckling modes are its eigenfunctions. The linear theory predicts the load at which a bent shape first appears but not the size of the deflection. The nonlinear elastica shows that the bent branch rises slowly above $P_{cr}$. The same eigenvalue method gives critical loads for plates and shells, and it is the base of Lagrange's problem of the strongest column.