Euler buckling of an elastic filament

A pinned filament of length $L$ and bending modulus $A$ first buckles under compressive force $F_c={\pi }^{2}A/L^2$.

A clamped filament touching a laterally sliding, moment-free wall has fixed projected gap $d$. For an inextensible small-amplitude elastica, parameterize its tip angle by $k=\sin ({\theta }_{\mathrm{tip}}/2)$. Then $\ell =\sqrt{B/F}K(k)$ and $d=\sqrt{B/F}[2E(k)-K(k)]$, where $K,E$ are complete elliptic integrals. Their small-$k$ expansions give the displayed decrease of wall force with excess contour length. It concerns the weakly buckled branch, before overhang or additional wall contact changes the boundary conditions.

Thermal fluctuations smooth the zero-temperature pitchfork in filament buckling: the mean compression remains analytic and nonzero below the mechanical threshold, with rounding controlled by $L/L_p$.

A vertical rod can buckle under its own distributed weight. For a rod clamped below and free above, the threshold is set by a Bessel-function eigenvalue rather than the sinusoidal mode of a rod under constant end load.

For a uniform elastic filament clamped below and free above, its own weight gives filament tension $T(z)=-\lambda g(L-z)$. With filament bending modulus $B=EI$, the static slope $u$ satisfies $B{u}^{\prime \prime }+\lambda g(L-z)u=0$, $u(0)=0$, and $u^{\prime }(L)=0$. Setting $\eta =(2/3)\sqrt{\lambda g/B}(L-z{)}^{3/2}$ gives $u={\eta }^{1/3}{J}_{-1/3}(\eta )$ up to amplitude; the $J_{1/3}$ branch violates the zero-moment condition. The first positive zero $j_{-1/3,1}$ therefore yields $\lambda g{L}_c^3/B\simeq 7.83735$. This is a self-buckling of a vertical rod threshold, not a constant-end-load Euler threshold.