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Background

For a simple system where you have a mass attached to a spring and damper in parallel:

enter image description here

We can calculate the critical damping from the equation of motion:

$$mx_{tt} + cx_t + kx = 0$$

$$ms^2 + cs + k = 0$$

$$s= \frac{-c ± \sqrt{c^2-4mk}}{2m}$$

There are then three conditions:

  1. $c^2 <4mk$ (under damping)
  2. $c^2 >4mk$ (over damping)
  3. $c^2 =4mk$ (critical damping)

The damping ratio is then expressed by $\frac{c}{\sqrt{4mk}}$.

Question

I'm wondering if or how this can be extended to more complex systems.

Let's say you add a second spring to the parallel system:

enter image description here

This is a viscoelastic model where the constitutive relationship is in terms of stress (σ) and strain (Ɛ):

$$σ = E_1Ɛ + \frac{η(E_1+E_2)}{E_2}\dot{Ɛ} - \frac{η}{E_2}\dot{σ}$$

Is it possible to calculate the critical damping or damping ratio in the same manner? If so how would it work?

What about with two springs and two dampers like this?

enter image description here

$$σ = (η_1+η_2)\dot{ϵ} + \frac{η_1η_2(E_1+E_2)}{E_1E_2} - (\frac{η_1}{E_1} + \frac{η_2}{E_2})\dot{σ} - \frac{η_1η_2}{E_1E_2}\ddot{σ}$$

Is it possible to do the same and if so how?

My Guess

My guess is strain is equivalent to $x$ and stress is equal to force, so we can reword the 3 element system:

$$0 = E_1Ɛ + \frac{η(E_1+E_2)}{E_2}\dot{Ɛ} - \frac{η}{E_2}\dot{σ} - σ$$

$$0 = E_1x + \frac{η(E_1+E_2)}{E_2}\dot{x} - m\frac{η}{E_2}\dddot{x} - m\ddot{x}$$

$$0 = E_1 + \frac{η(E_1+E_2)}{E_2}s - ms^2 - m\frac{η}{E_2}s^3$$

Which according to Wolfram gives the following roots:

$$s = \frac{-E_1E_2}{η(E_1+E_2)}$$

$$s = ±\sqrt{\frac{E_1}{m}}$$

But that is not useful for solving any critical damping scenario that I can figure out.

Is there some other approach or what did I screw up?

Any help or guidance is appreciated.

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enter image description here

Look at this example:

put the coordinates and the forces for the free body diagram , put dummy mass where you don't have real masse. (here on $~x_3~,x_4$), write the equations of motion.

\begin{align*} & m_{{1}}{\ddot{x}}_{{1}}+D_{{1}} \left( {\it \dot{x}}_{{1}}-{\it \dot{x}}_{{3}} \right) +D_{{2}} \left( {\it \dot{x}}_{{1}}-{\it \dot{x}}_{{4}} \right)=0 \\& m_{{2}}{\ddot{x}}_{{2}}+C_{{1}} \left( x_{{2}}-x_{{3}} \right) +C_{{2}} \left( x_{{2}}-x_{{4}} \right) =0\\ &m_{{d}}{\ddot{x}}_{{3}}+C_{{1}} \left( x_ {{2}}-x_{{3}} \right) -D_{{1}} \left( {\it \dot{x}}_{{1}}-{\it \dot{x}}_{{3}} \right) =0\\ & m_{{d}}{\ddot{x}}_{{4}}+C_{{2}} \left( x_ {{2}}-x_{{4}} \right) -D_{{2}} \left( {\it \dot{x}}_{{1}}-{\it \dot{x}}_{{4}} \right)=0 \end{align*} Where $~C_i~$ are the spring constants $~[\frac{N}{m}]~$ and $~D_i~$ are the dumper constants $~[\frac{N\,s}{m}]$

with $~m_d=0$ you obtain two second order differential equations $~\ddot x_1,~\ddot x_2$ and two first order differential equations $~\dot x_3,~\dot x_4$

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  • $\begingroup$ Thank you very much Eli. That's very helpful. I've been staring at your equations of motion for the past few hours, while trying to figure out what to do next. This is not as simple for me as the single mass-spring system. Critical damping is the value of the dampers that will most quickly and smoothly extinguish oscillations in the springs. At a given $m_1$, $m_2$, $c_1$, and $c_2$, how would I next figure out what $d_1$ and $d_2$ for that would be? (If needed we can set one mass (eg. $m_1$) as a fixed point I think for my purposes.) Thanks again for your help. $\endgroup$
    – mike
    Jan 17 at 20:54
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Just like you can replace resistors in electric circuits connect in series or parallel with a single equivalent resistor, we can do the same thing with springs and dampers, being able to replace them with a single spring and damper in parallel.

Let's consider the following system:

enter image description here

Note the nodes between each spring and damper has its displacement labelled. We can break this system up into five free-body diagrams:

enter image description here

Note how the force is the same through each branch, due to there being no mass in each branch itself. We can now express equations for each free-body diagram.

For now, let's consider the two free-body diagrams in the upper branch. They obey the equations

$$f_1 = k_1 x_1$$

and

$$f_1 = c_1 (\dot{x}-\dot{x}_1)$$

respectively.

Now, let's take advantage of the linearity of the system, and assume all displacements and forces are varying sinusoidally at the frequency $\omega$:

$$f_1(t) = \hat{f}_1 \, \mathrm{e}^{\mathrm{i}\omega t}, \qquad x_1(t) = \hat{x}_1 \, \mathrm{e}^{\mathrm{i}\omega t}, \qquad \mathrm{etc...}$$

Substituting these expressions into the two equations yields us

$$\hat{f}_1 = k_1 \hat{x}_1$$

$$\hat{f}_1 = \mathrm{i}\omega c_1 (\hat{x}-\hat{x}_1)$$

(equivalently, we could have taken the Fourier transform of the equations) Note how the second of the equations has the same $f=kx$ form of a spring. That is, a damper is equivalent to a spring with an imaginary, frequency-dependent stiffness $\mathrm{i} \omega c_1$.

And, we can combine the spring and damper in the branch as if they were two springs. By eliminating $\hat{x_1}$, which wasn't a displacement we really cared about to begin with, we get the equation

$$\hat{f}_1 = \left(\frac{1}{\mathrm{i}\omega c_1} + \frac{1}{k_1}\right)^{-1} \hat{x}$$

or, more compactly,

$$\hat{f}_1 = k_1^*(\omega) \hat{x}$$

That is, we has successfully replaced the spring and damper in one branch with a single complex-stiffness spring, where $k_1^*(\omega)$ is the complex stiffness. Note how the rule for combining two springs in series is equivalent to the rule of combining two resistors in parallel.

We can repeat this for the lower branch to get

$$\hat{f}_2 = k_2^*(\omega) \hat{x}$$

We have now reduced our system to having two complex springs in parallel. Parallel spring can be combined into a single spring, whose stiffness is the sum of the individual springs. Therefore, we have now reduced our system to the following:

enter image description here

where the equivalent complex stiffness $k^*(\omega)$ is

$$k^*(\omega) = \frac{k_1 \mathrm{i}\omega c_1}{k_1+ \mathrm{i}\omega c_1} + \frac{k_2 \mathrm{i}\omega c_2}{k_2+ \mathrm{i}\omega c_2}$$

Our equivalent system, recalling that the occurring motion is sinusoidal at frequency $\omega$, obeys the following equation:

$$-\omega^2 m \hat{x} + k^*(\omega)\hat{x} = \hat{f}$$

If we were to do things in Laplace domain instead of Fourier domain, the system equation would look like ($\mathrm{i}\omega \rightarrow s$)

$$\left(s^2 m + \frac{k_1 s c_1}{k_1+ sc_1} + \frac{k_2 s c_2}{k_2+ sc_2}\right) X(s) = F(s)$$

Now, this is where I'm less certain, but my guess (comparing to the what you did for the simpler case) would be to consider the equation

$$s^2 m + \frac{k_1 s c_1}{k_1+ sc_1} + \frac{k_2 s c_2}{k_2+ sc_2} = 0$$

which can be re-expressed as a cubic equation. (This equation can be reduced to the simpler case by setting $k_1 = k$, $k_2 = \infty$, $c_1 = \infty$, $c_2 = c$.)

A cubic equation may either have three real roots, or one real and two complex roots, depending on the discriminant of the cubic (see here), which seems comparable to the distinction of underdamped and overdamped systems for the simpler case. I believe the value of this discriminant is akin to whether the system is under-/critically/over-damped.

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  • $\begingroup$ That's very interesting and informative Involute. Thank you. I tried solving that final equation for $s$ with Wolfram and it mostly returned unusable results, but gave one good root before "computation time exceeded": wolframalpha.com/input/… $s = \frac{-\sqrt{-k_2 m (4 c_2^2 - k_2 m)} - k_2 m}{2 c_2 m}$ if $c_1 = 0$. Interestingly then, I think that solves the 3 element equation (where one damper is eliminated). $\endgroup$
    – mike
    Jan 18 at 0:31
  • $\begingroup$ I'm not really sure what to do with this next though. Or if there is a good root for the 4 element model. I am also curious - you said in your reply that your equation would reduce to the simple damped spring if $k_1=k, k_2=∞, c_1=∞, c_2=c$. How so? What's $s$ times infinity? Wouldn't the proper way to reduce it be to say $k_2=c_1=0$? ie. Designate that $k_2$ and $c_1$ contribute no force are thus nonexistent? $\endgroup$
    – mike
    Jan 18 at 0:36
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    $\begingroup$ If $k_2$ and $c_1$ are set to zero, you effectively end up cutting the mass loose. You want to set them to infinity so that those elements can no longer extend. To evaluate the expressions with infinity, consider the case where $k_2$ and $c_1$ are finite but tend towards infinity. For instance, note that in the denominator $k_1+sc_1$, the $sc_1$ becomes larger and larger compared to $k_1$, and so the $k_1$ term in that denominator can be neglected. So, $k_1 s c_1 /(k_1 + sc_1) \rightarrow k_1 s c_1/(sc_1) = k_1$. $\endgroup$
    – Involute
    Jan 19 at 20:30

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