Elastic beam

An elastic beam is a slender body that supports bending moments and transverse loads. In the small-slope Euler-Bernoulli approximation its bending moment is $M=EI{y}_{xx}$, and transverse momentum balance gives $\rho A{y}_{tt}+EI{y}_{xxxx}=q$ for a uniform beam under distributed force per length $q$. Here $I$ is the second moment of area; boundary conditions specify the end constraints and applied moments and forces.

The component of the internal resultant force perpendicular to an elastic beam's tangent. For a planar elastic beam, one consistent sign convention has bending moment $M=-B{w}^{\prime \prime }$ and $M^{\prime }=N$, so $N=-B{w}^{\prime \prime \prime }$. Reversing the section or bending moment convention reverses the corresponding signs.

An Euler strut is an initially straight slender elastic column loaded in axial compression. Its lateral stability depends on bending stiffness, length and endpoint constraints. Euler buckling predicts a loss of straight-configuration stability before material failure only in the appropriate slender elastic regime; it is distinct from crushing or plastic yielding of a short column.

For the angular endpoint conditions $\theta (0)=\theta (1)=0$, take $U=(EI/L){\int }_0^1[{\theta }^{\prime 2}/2-{\pi }^2\lambda (1-\cos \theta )]d\xi$. The Poincaré inequality and $1-\cos t\le t^2/2$ give $U\ge 0$ for $0\le \lambda \le 1$. At $\lambda =1$ a nonzero continuous angle still has strictly positive energy because $1-\cos t<t^2/2$ for $t\ne 0$. Thus the straight state remains the unique global minimum at the critical parameter even though the quadratic variation vanishes on the first sine mode.

Along that mode the energy is ${\pi }^{2}EI{A}^4/(64L)+O(A^6)$ at criticality. For $\lambda >1$ a sufficiently small first-mode variation lowers energy, so the straight state loses local minimality. Near the first threshold, expansion of the Euler strut equation gives $A_0(A_0^2-8{\lambda }_1)=0$ when $\lambda =1+ϵ{\lambda }_1$ and $\theta \sim \sqrt{ϵ}{A}_0\sin \pi \xi$. The resulting supercritical pitchfork bifurcation is stabilized by the same positive quartic term. A zero second variation is therefore an inconclusive quadratic test, not proof of failure of minimization.

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.

A beam fixed at one end and free at the other. A transverse end force $F$ gives bending moment $M(x)=F(L-x)$ and, for a rectangular section, maximum bending stress $6FL/(w{h}^2)$ at the fixed end. Longitudinal tension and compression are greatest on opposite outer fibres.

A uniform square section has second moment of area $a^4/12$ and gravitational load per length $\rho g{a}^2$. Euler-Bernoulli bending therefore gives $y^{\prime \prime \prime \prime }=12\rho g/(E{a}^2)$, with zero displacement and slope at the clamp and zero moment and shear at the free end. Integrating gives the displayed tip deflection and $y(x)=\rho g{x}^2(x^2-4lx+6l^2)/(2E{a}^2)$. Small slope and strain are required.