Euler's critical load ($P_{\mathrm{cr}}={\pi }^{2}E{I}_A/(KL{)}^2$)

A straight slender elastic beam under axial compression loses stability when bending can decrease its total potential energy. With pinned ends a distance $L$ apart, its quadratic potential is $\frac{1}{2}{\int }_0^L[E{I}_A(y^{\prime \prime }{)}^2-P(y^{\prime }{)}^2]dx$. Stationarity gives $E{I}_A{y}^{\prime \prime \prime \prime }+P{y}^{\prime \prime }=0$, with $y=y^{\prime \prime }=0$ at each end. The first nontrivial mode is $\sin (\pi x/L)$, at $P_{\mathrm{cr}}={\pi }^{2}E{I}_A/L^2$. More generally $P_{\mathrm{cr}}={\pi }^{2}E{I}_A/(KL{)}^2$, with an effective-length factor $K$ determined by endpoint constraints: $K=1$ for pinned-pinned, $2$ for fixed-free and $1/2$ for fixed-fixed. Real imperfections remove the perfectly sharp onset, and short columns can yield before this elastic instability.

For a pin-ended slender beam with unstressed profile $y_0$, bending moment depends on curvature change: $M=EI(y-y_0)$. An axial dead load gives $EI(y-y_0)=-Fy$ in the small-slope model.

With $y_0=Y\sin (\pi x/L)$ and $F_c={\pi }^{2}EI/L^2$, the equilibrium branch connected to zero load has $y=Y\sin (\pi x/L)/(1-F/F_c)$. The pole as $F$ approaches $F_c$ marks breakdown of the small-deflection model, not a physically infinite beam deflection.