Uniform-load bending of a clamped filament ($h_s(x)$)

Let positive transverse displacement point along a uniform load per unit length $w$. The filament bending modulus $B$ gives energy ${\int }_0^L[B(h_{xx}{)}^2/2-wh]dx$. The variational derivative is $B{h}_{xxxx}-w$, so resistive-force theory gives ${\zeta }_{\perp }{h}_t=-B{h}_{xxxx}+w$. Clamping imposes $h(0)=h_x(0)=0$, while a free tip imposes $h_{xx}(L)=h_{xxx}(L)=0$. The steady solution isIts tip displacement is $w{L}^4/(8B)$. For a submerged cylindrical filament under Newtonian gravity, buoyancy gives $w=({\rho }_f-\rho )\pi a^{2}g$. This linear approximation requires small slope and negligible inertia.

For a uniform-load bending of a clamped filament, total load $F=wL$ and tip displacement $\delta =w{L}^4/(8B)$ obey Hooke's law in the form $F=k_{\mathrm{eff}}\delta$, with $k_{\mathrm{eff}}=8B/L^3$. This stiffness depends on the specified distributed loading. A concentrated tip force instead gives $3B/L^3$, so equal total forces with different distributions are not interchangeable.