I want to know the mode shapes and associated eigenfrequencies and damping ratios of a bar that vibrates in a longitudinal direction and has linear elastic as well as viscous material properties. Consider the bar shown below which has length $L$ and is clamped on the left hand side and free on the right. alt text Its motion is desribed by the longitudinal displacements $u(x,\,t)$. The material is assumed to behave according to the viscoelastic Kelvin–Voigt model, which relates stress and strain at any point of the bar by

$$\sigma = E\,\varepsilon + \eta\,\dot{\varepsilon}$$

where $\varepsilon = \frac{\partial\,u}{\partial\,x}$ is the longitudinal strain, $E$ the elastic modulus of the material and $\eta$ the viscosity. The first thing I did was to try and obtain the equation of motion for this system. The relationship between stress and acceleration is

$$\rho\,\frac{\partial^2\,u}{\partial\,t^2} = \frac{\partial\,\sigma}{\partial\,x}$$

where $\rho$ is the bar's density. Combining this with the material model and multiplying with the cross section area $A$ results in the equation of motion

$$\rho A\,\frac{\partial^2\,u}{\partial\,t^2} = EA\,\frac{\partial^2\,u}{\partial\,x^2} + \eta A\,\frac{\partial^3\,u}{\partial\,x^2\,\partial\,t}$$

This is where I'm stuck because I don't know how to solve this. What I'm interested in is the eigenfrequency $\omega$ and damping ratio $\zeta$ of the first eigenmode (if that's even applicable). Without the damping term the problem reduces to the 1D wave equation which can be solved e.g. by separation of variables, but that doesn't seem to work in this case.

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    $\begingroup$ Probably assuming the form $u(x,t) =X(x) \exp(-i \omega t)$ will simplify the problem. $\endgroup$ Commented Nov 20, 2019 at 16:01

1 Answer 1


There was a mistake in the governing equation of motion, a missing derivative in the damping term. I edited the question. The problem can now be solved by separation of variables.

Step 1: Separation of variables

Assume that $u(x,\,t) = X(x)\,T(t)$. Insert that into the equation of motion and separate terms that depend on $t$ from those that depend on $x$,

\begin{align} \rho A\,X(x)\,\ddot{T}(t) &= EA\,X''(x)\,T(t) + \eta A\,X''(x)\,\dot{T}(t) \\ &\ \ \vdots \\ \frac{X''(x)}{X(x)} &= \frac{\rho A\,\ddot{T}(t)}{EA\,T(t) + \eta A\,\dot{T}(t)} =: -\alpha^2 \end{align}

Apply the usual argument: As one side of the equation only depends on position and the other only on time they have to be constant in order to be equal. We choose that constant to be $-\alpha^2$ and obtain two ordinary differential equations in $x$ and $t$ respectively.

Step 2: Solving the ODE for X

The first equation $X''(x) + \alpha^2 X(x) = 0$ has the general solution $X(x) = C\cdot\sin(\alpha\,x - \varphi).$ The constants $\alpha$ and $\varphi$ are determined by the boundary conditions of the bar. The left hand side is fixed, therefore

\begin{align} u(0,\,t) = 0: \quad\quad\quad X(0)\,T(t) &= 0 \\ C\cdot\sin(-\varphi) &= 0 \\ \varphi &= 0 \end{align}

The case $T(t) = 0$ was ignored here because it is of no interest. The right hand side is free, so the boundary condition is

\begin{align} \sigma(L,\,t) = 0: \quad\quad E\,\varepsilon(L,\,t) + \eta\,\dot{\varepsilon}(L,\,t) &= 0 \\ &\ \ \vdots \\ (EA\,T(t) + \eta A\,\dot{T}(t))\,X'(L) &= 0 \\ X'(L) = C\alpha\cdot\cos(\alpha\,L) &= 0 \\ \\ \alpha = \frac{\pi}{L}k - \frac{\pi}{2L}&,\quad k \in \mathbb{Z} \end{align}

So there is an infinte number of solutions for $\alpha$, each corresponding to a different mode shape $X(x)$. The case $EA\,T(t) + \eta A\,\dot{T}(t) = 0$ was ignored because it implies $\ddot{T}(t) = 0$.

Step 3: Analyzing the ODE for T

The differential equation for $T(t)$ resembles that of a damped harmonic oscillator. The undamped natural frequency $\omega_0$ and the damping ratio $\zeta$ can be obtained by simply comparing the coefficients,

$$\ddot{T}(t) + \underbrace{\alpha^2\frac{\eta A}{\rho A}}_{\Large{2\,\zeta\,\omega_0}}\,\dot{T}(t) + \underbrace{\alpha^2\frac{EA}{\rho A}}_{\Large{\omega_0^2}}\,T(t) = 0$$

Using the values for $\alpha$ from above the final answer to the initial question is

$$\omega_0 = \frac{\pi(2k - 1)}{2L}\cdot\sqrt{\frac{EA}{\rho A}},\quad \zeta = \frac{\pi(2k - 1)}{4L}\cdot\frac{\eta A}{\sqrt{\rho A \cdot EA}},\quad k \in \mathbb{Z}$$

Sanity check 1: The undamped eigenfrequencies are the same as found in the literature for the more common case of a bar without damping.

Sanity check 2: The damping ratio increases with $\eta A$ and $k$ and decreases with $\rho A$, $EA$ and $L$, which makes sense.

Sanity check 3: The units seem to be correct.


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