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The second moment of area about an axis weights each element of cross-sectional area by its squared perpendicular distance from that axis. The ellipse extends farther along ${\mathbf{d}}_2$ than along ${\mathbf{d}}_1$, so its material lies, on balance, farther from the ${\mathbf{d}}_1$ axis. ThereforeWith zero intrinsic curvature of an elastic filament, the respective bending moments required by the constitutive equation are $E{I}_1\kappa {\mathbf{d}}_1$ and $E{I}_2\kappa {\mathbf{d}}_2$. Thus their magnitudes satisfyfor $\kappa \ne 0$: bending about the minor axis is harder. Here $E$ is Young's modulus, and $E{I}_i$ is the corresponding bending stiffness.

Zero applied end force and the force balance ${\mathbf{n}}^{\prime }=0$ imply $\mathbf{n}(S)=0$ everywhere. The moment balance then reduces to ${\mathbf{m}}^{\prime }=0$.

Choose an untwisted straight configuration: all three directors are constant, ${\mathbf{r}}^{\prime }={\mathbf{d}}_3$, and $\mathbf{u}=0$. The constitutive equation giveswhich is constant in space and therefore satisfies the moment balance. The required end bending moment is consequentlyIts sign opposes the positive intrinsic curvature of an elastic filament. If the initial position and frame from part (ii) are retained, this solution is simply $\mathbf{r}(S)=S{\mathbf{e}}_3$ with ${\mathbf{d}}_i={\mathbf{e}}_i$; the clamp supplies the opposite reaction moment at the other end.

The first term is the bending energy of rod 1, whose preferred bending strain of a rod is zero. The second is the bending energy of rod 2 relative to its preferred intrinsic curvature of an elastic filament $\hat{u}$. The third is its stretching energy, with stretching stiffness $k{B}_2$ and stress-free stretch ratio ${\alpha }^{+}=1$. Each term is an energy per unit reference arc length before integration.

Substitute $u^{+}=u$ and ${\alpha }^{+}-1=-au$. The resulting quadratic function of $u$ has densityThere are no derivatives of $u$ or imposed end constraints in this minimum potential energy principle in elasticity. Sincethe unique pointwise minimizer is also the constant global minimizer:For example, completing the square giveswhich also proves uniqueness, up to equality almost everywhere, without merely assuming a uniform curvature.

With $k$, $a$ and $\hat{u}$ fixed, the two limits areIn the first limit, rod 1 dominates the bending energy and suppresses the bend. In the second, rod 1 offers negligible bending resistance, but following rod 2's preferred bending strain of a rod still shortens rod 2 through the bonded geometry. The competition between its bending energy and stretching energy leaves $u_{\ast }<\hat{u}$ when $\hat{u}>0$ and $k{a}^2>0$. Only when that stretching penalty also vanishes does $u_{\ast }\to \hat{u}$.

These formulae describe an admissible bonded elastic rod provided $1-a{u}_{\ast }>0$. If parameters instead force $a{u}_{\ast }\ge 1$, the unconstrained quadratic minimizer is outside the kinematic regime of the model; the given energy does not itself enforce positive stretch ratio.

Introduce a constant filament bending modulus $B$, which sets the mechanical force scale. The filament has intrinsic curvature of an elastic filament ${\kappa }_0(s)$, and its actual signed curvature in a small-slope Monge representation is $\kappa \simeq {\zeta }_{xx}$. To quadratic order one may identify material arc length $s$ with $x$ in ${\kappa }_0$, whileAt fixed positive extension force, the mechanical potential is bending energy minus $FL$. Replacing $L$ by $\mathcal{L}-(\mathcal{L}-L)$ leaves, up to a force-dependent constant,Its first variation isThe higher-order Euler-Lagrange equation is thereforeFor a Fourier transform relation, take a periodic bulk segment, or a long-filament bulk approximation with negligible endpoint contributions; these are necessary boundary conditions because the question specifies no filament end conditions. Using $f(x)=\int \hat{f}(q)e^{iqx}dq/(2\pi )$, every nonzero wavenumber obeysA zero displacement mode is an arbitrary translation. In a periodic representation a constant intrinsic-curvature mode contributes a constant bending cost but no bulk slope, so it does not contribute to the following deficit. The Parseval identity givesFor clarity about finite normalization, on a periodic interval of length $L$ define ${\kappa }_n=L^{-1}{\int }_0^L{\kappa }_0(x)e^{-i{q}_{n}x}dx$, $q_n=2\pi n/L$. Then the equivalent exact quadratic-order formula isThis is force-extension of an intrinsically curved filament, determined by its particular fixed shape rather than by a thermal orientation entropy.

For an ensemble with quenched intrinsic curvature, first minimize each realization and then average the last expression, replacing $|{\kappa }_n{|}^2$ by its ensemble mean. For stationary bulk disorder, let $S_{\kappa }(q)$ be the power spectrum, normalized so that its correlation is $⟨{\kappa }_0(x){\kappa }_0(x+r)⟩=\int S_{\kappa }(q)e^{iqr}dq/(2\pi )$. The measurable mean deficit per unit projected length isIf the curvature is smooth enough that the second spectral moment is finite, one can take the large-force limit under this integral, obtaining the smooth-curvature high-force extension lawFor a stationary ensemble the integral is $L⟨({\kappa }_{0,x}{)}^2⟩$. Equivalently, the bulk slope is ${\zeta }_x\simeq -(B/F){\kappa }_{0,x}$; a plot of the mean deficit against $F^{-2}$ measures the mean squared intrinsic-curvature gradient, if $B$ is known. The $F^{-2}$ law is a smooth-curvature bulk result, not a universal law for every random filament. For example, ideal white intrinsic-curvature disorder with $S_{\kappa }(q)=S_0$ has an infinite second spectral moment; retaining the full integral instead gives $⟨\mathcal{L}-L⟩/L=S_0\sqrt{B}/(8\sqrt{F})$ in that ideal continuum model.

Endpoint conditions also matter. If each end is moment-free and has no transverse applied force, the natural conditions from the variation areat each end. These can produce a free-end curvature boundary layer of width $\ell =\sqrt{B/F}$. For a constant intrinsic filament curvature $\kappa$, integration of the higher-order Euler-Lagrange equation with zero transverse force gives $B({\zeta }_x{)}_{xx}-F{\zeta }_x=0$ and henceThe last formula follows by integrating ${\sinh }^2[(x-L/2)/\ell ]$, and tends to ${\kappa }^2{\ell }^3/2\propto F^{-3/2}$, despite the vanishing bulk curvature gradient. Thus a finite free-ended measurement must include or suppress endpoint bending before interpreting a bulk $F^{-2}$ fit.

For the freely jointed chain, rigid links have independent orientations, so the restoring force is entropic. Let $T$ now denote absolute temperature, and $\beta =1/(k_{B}T)$ with Boltzmann constant $k_B$. In the usual three-dimensional chain, a segment at polar angle $\theta$ has force energy $-Fb\cos \theta$. Its angular canonical partition function and the full chain partition function areDifferentiating $\log Z_N$ with respect to $\beta F$ yields the exact mean projected displacementHere $\mathcal{L}$ is the Langevin function; this is force-extension of a three-dimensional freely jointed chain, in the fixed-force canonical ensemble. At high extension, $\coth \xi =1+2e^{-2\xi }+\cdots$, soThus the smooth quenched-curvature bulk deficit falls as $F^{-2}$, whereas the three-dimensional freely jointed chain deficit falls as $F^{-1}$ and depends explicitly on temperature. The distinction is mechanical straightening of a fixed intrinsic shape versus orientational entropy; thermal bending fluctuations of a stiff filament have not been included in the first calculation.

If the comparison chain is also restricted to two dimensions, it is a planar freely jointed chain rather than the three-dimensional model. In this case,The last identity follows directly by differentiating the integral defining $I_0$; $I_0$ and $I_1$ are Modified Bessel functions of the first kind. At large $\xi$ the angular weight is proportional to $e^{-\xi {\theta }^2/2}$, giving $⟨\cos \theta ⟩\simeq 1-⟨{\theta }^2⟩/2=1-1/(2\xi )$. Consequently $Nb-⟨L⟩\sim N{k}_{B}T/(2F)$: the same inverse-force exponent, with half the three-dimensional coefficient. Both versions assume inextensible links and forces below the regime where the links themselves stretch.

Quenched intrinsic curvature of an elastic filament is fixed during a mechanical experiment, although different filaments can have different random intrinsic shapes. Ensemble averaging is then over these shapes after mechanical minimization. It differs from averaging over time-dependent thermal bending fluctuations in a worm-like chain.