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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.

Take $z$ upward along the straight rod. The part above height $z$ has weight $\lambda g(L-z)$, so, with tensile force positive, the internal tension is compressive:For a small transverse displacement $X(z)$, the quadratic bending and gravitational energies areThe Euler-Lagrange equation for this functional iswhich is exactly

Clamping at the bottom fixes displacement and slope:At the free upper end, bending moment and transverse force vanish:Since $\sigma (L)=0$, the four boundary conditions are

Set $u=X^{\prime }$. Integrating the field equation once and using the free-end shear condition givesor, with $s=L-z$,Introduce the dimensionless similarity coordinateand write $u={\eta }^{1/3}F(\eta )$. Direct substitution reduces the equation toThis is the Bessel differential equation of order $1/3$, so

The free-moment condition is $u^{\prime }(L)=0$. As $\eta \to 0$,The second term has nonzero limiting $z$ derivative, so the free-end condition forces $b=0$. The free-shear condition $u^{\prime \prime }(L)=0$ then follows from the differential equation. At the clamp, $u(0)=0$, givingLet $j_{-1/3,1}$ be the smallest positive zero of this Bessel function. The first self-buckling threshold isor equivalently