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The part of an elastic rod's elastic energy that penalizes departure of its bending strain of a rod $u$ from the intrinsic value $\hat{u}$. Here $B$ is the bending stiffness and $S$ is reference arc length. For an inextensible rod, the same formula can be written using spatial curvature and current arc length.

For planar director kinematics of a rod measured in reference arc length $S$, define the signed bending strain by $d{\mathbf{d}}_3/dS=u{\mathbf{d}}_1$. If $s$ is current arc length and $\alpha =ds/dS>0$ is the stretch ratio, the signed spatial curvature is $u/\alpha$: the chain rule gives $d{\mathbf{d}}_3/ds=(u/\alpha ){\mathbf{d}}_1$. Thus bending strain and spatial curvature coincide for an inextensible rod, but differ in general. An intrinsic value $\hat{u}$ is the preferred reference-length bending strain in an energy density $B(u-\hat{u}{)}^2/2$ per unit reference length.

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 inextensible rod preserves material arc length, so its stretch ratio is one. In an unshearable elastic rod, $d\mathbf{r}/dS={\mathbf{d}}_3$ for reference arc length $S$.

The right-handed material frame of a rod has ${\mathbf{d}}_2\times {\mathbf{d}}_1=-{\mathbf{d}}_3$ and ${\mathbf{d}}_2\times {\mathbf{d}}_3={\mathbf{d}}_1$. Hence the director kinematics of a rod reduce toSolving these constant-coefficient ordinary differential equations with the given initial frame givesThe inextensible rod has ${\mathbf{r}}^{\prime }={\mathbf{d}}_3$. An integration using $\mathbf{r}(0)=0$ yields, for $\hat{u}\ne 0$,Indeed,so the centreline is a circular arc of radius $1/|\hat{u}|$ in the plane perpendicular to ${\mathbf{e}}_2$. The continuous $\hat{u}\to 0$ limit is ${\mathbf{d}}_1={\mathbf{e}}_1$, ${\mathbf{d}}_3={\mathbf{e}}_3$, and $\mathbf{r}=S{\mathbf{e}}_3$.

Because the two elastic rods each have thickness $a$, their centrelines are separated by $a$ along the common director ${\mathbf{d}}_1$. The bonded geometry isDifferentiate with respect to reference arc length. The first rod is an inextensible rod, and the printed director kinematics of a rod giveFor the physically admissible orientation-preserving configuration $1-au>0$, the rods have the same unit tangent ${\mathbf{d}}_3^{+}={\mathbf{d}}_3$ and the same planar material frame of a rod. ThereforeIn particular, the second rod is shortened on the inside of a bend with $u>0$.

The bending strain of a rod is defined here by differentiation with respect to reference arc length. Since the reference labels match,The distinction from spatial curvature matters: the second centreline hasThus $u/(1-au)$ is its signed spatial curvature, rather than the $u^{+}$ appearing in the given elastic energy.