Euler elastica ($EI{\theta }^{\prime \prime }+F\sin \theta =0$)

The planar equilibrium of an inextensible slender beam can have large tangent angle while retaining small elastic strain across its thickness. For constant horizontal compressive force $F$, moment balance and curvature ${\theta }^{\prime }=d\theta /ds$ give the displayed nonlinear equation. Multiplication by ${\theta }^{\prime }$ gives the first integral $EI{\theta }^{\prime 2}/2-F\cos \theta =C$. A tensile load has $F<0$ in this convention.

Specifying endpoint tangent angles of an inextensible rod does not by itself specify its endpoint positions. A planar Euler elastica satisfies $x^{\prime }=\cos \theta$, $y^{\prime }=\sin \theta$, so equal endpoint heights require the displayed integral to vanish. For constant compressive resultant $(-P,0)$, the angular equation is $EI{\theta }^{\prime \prime }+P\sin \theta =0$, giving $P\Delta y=-EI[{\theta }^{\prime }(L)-{\theta }^{\prime }(0)]$. Thus a nontrivial branch containing an odd number of pendulum half-periods cannot have equal endpoint heights and the same zero endpoint angles with zero transverse resultant. At small amplitude the tangent-only eigenfunctions are $\sin (n\pi s/L)$, with $P_n=EI(n\pi /L{)}^2$; equal endpoint heights retain even $n$ and give the first fixed-clamp load $4{\pi }^{2}EI/L^2$. Other clamp modes can require a nonzero transverse reaction. Axial shortening is also needed when the reference endpoint separation equals the rod length.

An ideal zero-thickness planar loop under tension $T$ is a separatrix of the elastica equation. Its tangent turns from zero to $2\pi$, the same physical orientation at both ends. With $\alpha =\sqrt{T/(EI)}$, integration gives $x=s-2\tanh (\alpha s)/\alpha$ and $y=2\mathrm{sech}(\alpha s)/\alpha$. Its height is $2\sqrt{EI/T}$. Self-contact and finite beam thickness are absent from this ideal crossing solution.