 Solution

Write $B$ for the filament bending modulus and $C$ for the filament torsional stiffness, so that the energy per unit length is $(B{\kappa }^2+C{\Omega }^2)/2$. For one circular filament,where $I_1$ is the second moment of area, $J_1$ the polar moment, $G_s$ the shear modulus, and $\nu$ Poisson's ratio. Numerical twisting constants cannot be fixed by Young's modulus alone without specifying $\nu$.

For a filled sliding bundle, $N\sim (R/r{)}^2$. Filaments share the centerline curvature but slide instead of accumulating the additional axial strains of a bonded cross-section. Their bending energies add: $B\sim N{B}_1\sim E{r}^2{R}^2$. If end constraints impose the same material twist on every filament, their torsional energies also add: $C\sim G_s{r}^2{R}^2$.

For a bonded, one-filament-thick circular tube, the wall thickness $h$ is of order $r$. Its moments are $I\sim \pi R^{3}h$ and $J_p\sim 2\pi R^{3}h$, giving $B\sim Er{R}^3$ and $C\sim G_{s}r{R}^3$. For a bonded filled rod, $I\sim \pi R^4/4$ and $J_p\sim \pi R^4/2$, giving $B\sim E{R}^4$ and $C\sim G_s{R}^4$. Packing fractions and the convention for tube radius affect constants, not these powers.

There is an important qualification to bending and torsion of filament bundles: free axial sliding does not specify whether each filament can also spin freely. If both spin and slip are unconstrained, the common-material-twist calculation is not forced by the geometry. A helix of small twist rate has centerline curvature of order $\rho {\Omega }^2$, so its bending energy starts at ${\Omega }^4$, not ${\Omega }^2$. A nonzero linear collective torsional stiffness then requires additional end, contact, or spin constraints.