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I have been working extensively on modeling physical objects using the stress/strain relationships of springs and dampers as described here:

https://en.wikipedia.org/wiki/Viscoelasticity#Constitutive_models_of_linear_viscoelasticity

Basic Equations

The stress (σ) of a simple spring (where E is the elastic modulus) can be described in terms of strain (ε) as:

$$σ = Eε$$

The stress of a dashpot/damper (where η is its viscosity) can be described as:

$$σ = η\dot{ε}$$

One of the simplest combinations of spring and dashpots used in these models is putting both in series like so:

enter image description here

The stress-strain relationship can be derived by assuming:

$$σ_{total} = σ_{E} = σ_{η}$$

$$ε_{total} = ε_{E} + ε_{η}$$

This system of equations can be solved to get the final equation of stress-strain:

$$σ + \frac{η}{E}\dot{σ} = η\dot{ε}$$

The Problem

Currently there is no way to set the coefficients so that this can progressively simplify down to a simple spring.

I would like some method by which if you reduce η to zero (ie. make the dashpot contribute no force) this simplifies to a simple spring. $σ = E{ε}$. Currently due to the assumption that the dashpot and the spring have the same force, if the dasphot has zero force, so too does the spring (the whole equation becomes zero).

Is there any way to write the stress-strain relationship for the dashpot and spring (with them still both in series) so that it can under some condition the equation can decompose to that of simple spring?

Is there no way for the spring to behave like a simple spring still? What must be done to the dashpot to allow this? Do you need a third coefficient and how would that work?

Thanks.

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I don't think this will reduce to a simple spring by letting $\eta$ go to zero. The dashpot allows the spring to relax over time to its zero stress state. Reducing $\eta$ to zero means the spring will relax instantaneously. That is, the dashpot end of the spring is free to move.

I think what you want is to let $\eta$ go to infinity which, in effect, connects the dashpot end of the spring to the other end of the dashpot (which is now a rigid rod).

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  • $\begingroup$ Okay yes that is what I want - to make the dashpot rigid somehow, so it simplifies to a simple spring. But how does setting it to infinity accomplish that? The equation is: $$σ + \frac{η}{E}\dot{σ} = η\dot{ε}$$ If η=infinity, how does that simplify to: $$σ = Eε$$? Thanks $\endgroup$
    – mike
    Apr 12 at 0:42
  • $\begingroup$ @mike, as $\eta\rightarrow\infty$, the differential equation goes to $\frac{\eta}{E}\dot\sigma=\eta\dot\epsilon$. Cancel $\eta$ on both sides and integrate. $\endgroup$ Apr 12 at 0:50

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