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Interpret $\varphi$ as the small transverse displacement of the elastic beam. The first term is the positive kinetic energy density $\rho {\varphi }_t^2/2$, with $\rho$ the mass per unit length. The Lagrangian density is kinetic minus potential energy, so the potential density encoded by the other two terms isFor a compressed rod, transverse deflection releases compression energy: the small-slope geometric change in length is proportional to ${\varphi }_x^2/2$, so the applied compressive load lowers the potential by $\alpha {\varphi }_x^2/2$. This explains the positive sign of the $\alpha$ term in the Lagrangian. Tension would give the opposite sign and a restoring gradient term.

Bending costs elastic energy. At small slopes the curvature is ${\varphi }_{xx}$, so positive bending stiffness $\beta$ gives potential $\beta {\varphi }_{xx}^2/2$, and hence a negative term in the Lagrangian. For a uniform slender beam $\beta$ is its flexural rigidity. The competition between destabilizing compression and stabilizing bending is responsible for Euler buckling of an elastic filament.

A uniform elastic beam with endpoint conditions $y=y_{xx}=0$ has normal modes $\sin (n\pi x/L)$ for positive integers $n$. Substitute $y=X{e}^{-i\omega t}$ into $EI{y}_{xxxx}+{\rho }_{\ell }{y}_{tt}=0$ and impose the four endpoint conditions to obtain the displayed spectrum. Here ${\rho }_{\ell }$ is mass per unit length, $I$ is a second moment of area, and $EI$ is bending stiffness. frequencies proportional to $n^2$ distinguish flexural waves from the nondispersive ideal stretched string.

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 elastic beam above a smooth obstacle cannot penetrate it, and a nonadhesive obstacle can only push. The displayed constraints express this when $R$ is an upward distributed normal contact force; concentrated point forces must also be nonnegative. In the small-slope Euler-Bernoulli beam equation,Integrating across a contact point with no applied couple gives jump conditions $[w]=[w^{\prime }]=[B{w}^{\prime \prime }]=0$ and $[B{w}^{\prime \prime \prime }]=R_i$, provided the tension is continuous. For a flat obstacle, no distributed load and zero tension, an open contact interval has $R=0$; its entire normal contact force can therefore be concentrated at its endpoints.