Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 71 / 3

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.

The compressive tip force is limited by Euler buckling of an elastic filament. With unsupported length $\ell$ and filament bending modulus $B=EI$,For a hollow microtubule, use $I=\pi (r_o^4-r_i^4)/4$ rather than the solid-cylinder value. For a clamped seed and a moment-free tip free to slide along the wall, linearized angle balance is $B{\theta }^{\prime \prime }+F\theta =0$, with $\theta (0)=0$ and ${\theta }^{\prime }(\ell )=0$. Its first nonzero solution has $\sqrt{F/B}\ell =\pi /2$, hence $F_c={\pi }^{2}B/(4{\ell }^2)$. A different tip constraint changes this coefficient.

Increasing the unsupported length decreases that threshold. For growth against a wall at fixed projected gap $d$, polymerization-driven cantilever postbuckling gives a more direct quantitative statement. The inextensible elastica obeys $B{\theta }^{\prime \prime }+F\sin \theta =0$. Multiplying by ${\theta }^{\prime }$ and using the zero tip moment givesPut $k=\sin ({\theta }_t/2)$ and substitute $\sin (\theta /2)=k\sin \varphi$. Integrating arc length and its horizontal projection giveswhere the complete elliptic integrals are $K(k)={\int }_0^{\pi /2}(1-k^2{\sin }^2\varphi {)}^{-1/2}d\varphi$ and $E(k)={\int }_0^{\pi /2}(1-k^2{\sin }^2\varphi {)}^{1/2}d\varphi$. Using $K=(\pi /2)(1+k^2/4+\cdots )$ and $E=(\pi /2)(1-k^2/4+\cdots )$ gives $\ell /d=1+k^2+O(k^4)$, and thereforeThus the initial postbuckling force decreases as more contour is inserted. This result describes the weakly buckled branch before overhang, further wall contact, or altered tip constraints invalidate the model. It is a mechanical force law; a chemical force limit from polymerization can impose a smaller force.