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An elastic beam is a slender body that supports bending moments and transverse loads. In the small-slope Euler-Bernoulli approximation its bending moment is $M=EI{y}_{xx}$, and transverse momentum balance gives $\rho A{y}_{tt}+EI{y}_{xxxx}=q$ for a uniform beam under distributed force per length $q$. Here $I$ is the second moment of area; boundary conditions specify the end constraints and applied moments and forces.

The printed diagram and the word “parallel” describe a Kelvin-Voigt model, not a series Maxwell fluid. In the depicted model both elements have displacement $x$ and their forces add:For $x=a{e}^{i\omega t}$ and the convention $G(\omega )=F_{\mathrm{amplitude}}/x_{\mathrm{amplitude}}$,The storage modulus and loss modulus here are element stiffnesses with units $\mathrm{N}{\mathrm{m}}^{-1}$; conversion to stress/strain moduli requires a geometric factor. If “response” instead denotes displacement per force, it is the compliance $1/(k+i\omega \mu )$. Specifying which ratio is used removes that convention ambiguity. The mean dissipated power is $\mu {\omega }^2|a{|}^2/2$, confirming the positive loss sign for $e^{i\omega t}$.

For completeness, the actual series Maxwell fluid analogue has a common force and additive spring/dashpot extensions. Thus $\dot{x}=\dot{F}/k+F/\mu$. With $\tau =\mu /k$,These different functions cannot both describe the printed parallel arrangement. The distinction also changes the elastic beam's response to a constant load.

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.