= Solution
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:
$$
\boxed{\sigma(z)=-\lambda g(L-z)}.
$$
For a small transverse displacement $X(z)$, the quadratic bending and gravitational energies are
$$
\mathcal E[X]
=\frac A2\int_0^L(X'')^2\,dz
-\frac{\lambda g}{2}\int_0^L(L-z)(X')^2\,dz.
$$
The <Euler-Lagrange equation> for this functional is
$$
AX''''+\bigl[\lambda g(L-z)X'\bigr]'=0,
$$
which is exactly
$$
\boxed{AX_{zzzz}-(\sigma X_z)_z=0}.
$$
Clamping at the bottom fixes displacement and slope:
$$
X(0)=0,\qquad X'(0)=0.
$$
At the free upper end, bending moment and transverse force vanish:
$$
AX''(L)=0,\qquad AX'''(L)-\sigma(L)X'(L)=0.
$$
Since $\sigma(L)=0$, the four boundary conditions are
$$
\boxed{X(0)=X'(0)=0,\qquad X''(L)=X'''(L)=0}.
$$
Set $u=X'$. Integrating the field equation once and using the free-end shear condition gives
$$
Au''-\sigma u=0,
$$
or, with $s=L-z$,
$$
u_{ss}+\frac{\lambda g}{A}s\,u=0.
$$
Introduce the dimensionless similarity coordinate
$$
\eta=\frac23\left(\frac{\lambda g}{A}s^3\right)^{1/2}
$$
and write $u=\eta^{1/3}F(\eta)$. Direct substitution reduces the equation to
$$
\eta^2F''+\eta F'
+\left(\eta^2-\frac19\right)F=0.
$$
This is the <Bessel differential equation> of order $1/3$, so
$$
\boxed{
u=\eta^{1/3}\left[aJ_{-1/3}(\eta)+bJ_{1/3}(\eta)\right]}.
$$
The free-moment condition is $u'(L)=0$. As $\eta\to0$,
$$
\eta^{1/3}J_{-1/3}(\eta)\sim\text{constant},
\qquad
\eta^{1/3}J_{1/3}(\eta)\sim\eta^{2/3}\propto s.
$$
The second term has nonzero limiting $z$ derivative, so the free-end condition forces $b=0$. The free-shear condition $u''(L)=0$ then follows from the differential equation. At the clamp, $u(0)=0$, giving
$$
J_{-1/3}(\eta_0)=0,\qquad
\eta_0=\frac23\left(\frac{\lambda gL^3}{A}\right)^{1/2}.
$$
Let $j_{-1/3,1}$ be the smallest positive zero of this <Bessel function>. The first <self-buckling of a vertical rod>[self-buckling threshold] is
$$
\boxed{
\frac23\left(\frac{\lambda gL^3}{A}\right)^{1/2}
=j_{-1/3,1}},
$$
or equivalently
$$
\boxed{\frac{\lambda gL^3}{A}
=\frac94j_{-1/3,1}^2}.
$$
Back to article page
