= Solution
Use the downward-positive displacement and the <filament bending modulus> $B$ and load density $w$ defined above. Two applications of <integration by parts> give the <first variation>
$$
\delta E=\int_0^L(Bh_{xxxx}-w)\delta h\,dx
+B[h_{xx}\delta h_x-h_{xxx}\delta h]_0^L.
$$
Thus the elastic-plus-gravitational <force> density is $-\delta E/\delta h=-Bh_{xxxx}+w$. In <resistive-force theory>, transverse motion has drag per unit length $-\zeta_\perp h_t$. Neglecting filament and fluid inertia, instantaneous force balance gives the forced <small-slope elastohydrodynamic filament equation>
$$
\boxed{\zeta_\perp h_t=-Bh_{xxxx}+w.}
$$
Clamping fixes position and tangent, while the free end has zero bending moment and zero transverse shear force. Therefore
$$
\boxed{h(0,t)=h_x(0,t)=0,\qquad h_{xx}(L,t)=h_{xxx}(L,t)=0.}
$$
The last two conditions also follow as natural boundary conditions from the variation, since the free-end values of $\delta h$ and $\delta h_x$ are arbitrary. Distributed weight does not add a concentrated force or moment at the tip.
At steady state, $Bh_s''''=w$. Integrating first from the free end gives $h_s'''=(w/B)(x-L)$ and $h_s''=(w/2B)(x-L)^2$. Integrating twice more with the clamp conditions gives the <uniform-load bending of a clamped filament>:
$$
\boxed{h_s(x)=\frac{w}{24B}x^2(x^2-4Lx+6L^2).}
$$
In particular the free-end displacement is $\delta=h_s(L)=wL^4/(8B)$ and the total buoyancy-corrected gravitational load is $F=wL$. Thus the <tip stiffness of a uniformly loaded cantilever> is
$$
\boxed{F=k_{\rm eff}\delta,\qquad k_{\rm eff}=\frac{8B}{L^3}
=\frac{2\pi A a^4}{L^3}.}
$$
The last expression uses $A$ as <Young's modulus>; if $A$ itself is flexural rigidity, the same result reads $k_{\rm eff}=8A/L^3$. This is <Hooke's law> for the specified loading pattern. It is not the stiffness for a point force at the tip, which would be $3B/L^3$; replacing a distributed force by its resultant does not preserve its bending moment distribution.
For <dimensional analysis>, $[A]=$ force/length squared, $[I]=$ length to the fourth power, hence $[B]=$ force times length squared. Consequently $[B/L^3]=$ force/length, as required for a spring constant. Also $[\zeta_\perp]=$ force times time/length squared makes $\zeta_\perp h_t$, $Bh_{xxxx}$ and $w$ all forces per unit length. Neither viscosity nor density enters the stiffness, although viscous drag controls the rate of approach to the steady shape. The small-slope approximation requires $wL^3/B\ll1$, since $h_s'(L)=wL^3/(6B)$.
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