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Let $y(x,t)$ be downward deflection, $0\le x\le \ell$, with $y(0,t)=y_x(0,t)=0$. Take a uniform circular elastic beam, $I=\pi r^4/4$, small slopes, and linear material response. Write $E$ for its axial elastic modulus and ${\eta }_s$ for its axial viscous coefficient; these are material quantities, not the solvent viscosity. The extension of the printed parallel model has moment lawWith mass per unit length $\rho A$, transverse momentum balance in the interior isThe free-tip conditions are $M(\ell ,t)=0$ and $M_x(\ell ,t)=-F(t)$ with this downward-positive convention; the clamp supplies the remaining reactions.

Neglecting inertia gives $M=F(t)(\ell -x)$. Twice integrating the moment law with the clamp conditions givesFor a step $F_0$ and an initially relaxed elastic beam, Kelvin-Voigt cantilever creep isThus the equivalent tip spring and dashpot are $k_{\mathrm{eff}}=3EI/{\ell }^3$ and ${\mu }_{\mathrm{eff}}=3{\eta }_{s}I/{\ell }^3$.

If the intended material is instead a true series Maxwell fluid, its moment satisfies $\dot{M}+M/\tau =EI{y}_{xxt}$. Together with momentum balance $\rho A{y}_{tt}+M_{xx}=0$, this gives the full interior equation $\rho A(y_{ttt}+y_{tt}/\tau )+EI{y}_{xxxxt}=0$, with the moment law retained for boundary and initial conditions. In the quasistatic limit,This Maxwell cantilever creep has an initial elastic jump followed by linear creep, not a finite plateau. The jump is the ideal massless step-load limit; inertia resolves the transient. A two-element series model has no nonzero relaxed modulus, so “Maxwell solid” is not an additional specification of a solid with finite long-time stiffness. Such a solid would need an additional spring. The small-deflection approximation requires $\delta \ll \ell$ and eventually fails for unbounded creep.