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A Maxwell material is a model material obtained by connecting a perfectly elastic solid and a Newtonian liquid in series.

Let $\sigma$ be the stress, $\gamma$ the shear strain, $G$ the shear modulus and $\eta$ the viscosity.

For an elastic solid, we have

$$\sigma (t) = \gamma (t) \ G$$

and for a Newtonian liquid,

$$\sigma (t) = \dot \gamma (t) \ \eta$$

In a Maxwell material, the total shear strain is the sum of the strains of the single elements ($e$=elastic, $N$=Newtonian):

$$\gamma (t)= \gamma_e (t)+ \gamma_N (t) \tag{1}\label{1}$$

and each elements bears the same stress:

$$\sigma (t) = G_M \ \gamma_E (t) = \eta_M \ \dot \gamma_N (t) \tag{2}\label{2}$$

Let us apply a step strain,

$$\gamma(t) = \begin{cases} 0 & t<0\\ \gamma & t \geq 0\\ \end{cases} $$

From \ref{1} and \ref{2} we get

$$\tau \ \dot \gamma_N(t) = \gamma - \gamma_N(t)$$

where we have defined the relaxation time $\tau=\eta_M/G_M$. Using the initial condition $\gamma_N(0)=0$, we get, for $t \geq 0$,

$$\gamma - \gamma_N(t) = \gamma_E(t) = \gamma \exp(-t/\tau)$$

What I don't understand is: why is it reasonable to assume that $\gamma_N(0)=0$? How is this initial condition justified physically?


Addendum

I am starting to think that the problem is actually ill-defined at $t=0$ if we use the above equations, and that the initial condition must actually be intended as

$$\gamma_N (\epsilon) = 0 \ \ \forall \epsilon >0 \tag{3}\label{3}$$

Indeed, using \ref{1} and \ref{2} and introducing Heaviside's function $\theta(t)$ so that we can write $\gamma(t) = \gamma \ \theta(t)$, we have

$$\sigma(t) = \eta_M \frac{d}{dt}[\gamma \ \theta(t) - \gamma_E(t)] = \eta_M[ \gamma \ \delta(t) - \dot \gamma_E (t)]$$

having introduced the Dirac delta $\delta(t)$. So it looks like if we want \ref{1} and \ref{2} to be valid for any $t$, a Dirac delta, and thus an infinite stress, appears at $t=0$, which is unphysical.

On the other hand, if we take \ref{3} as initial condition or interpret \ref{1} and \ref{2} as valid only at $t>0$, then we may still be able to save the reasoning. Of course then one may argue that in reality there will never be a perfect step strain...

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The idealized viscous liquid (represented by the lumped-component damper/dashpot in the Maxwell model) has infinite stiffness for instantaneous movements. You state this yourself in your constitutive law relating $\sigma_N(t)$ and $\dot\gamma_N(t)$; a step increase in $\gamma_N$ results in $\sigma_N\to\infty$.

This result is nonsensical because the load on two components in series must be equal, but the spring cannot sustain an infinite load; the other constitutive law implies that the maximum load that can be applied by the spring on the damper attached to one of its ends is $\gamma (0) G$.

Therefore, the spring must take up the entire initial displacement in the case of a step strain, and so $\gamma_N(0)=0$.

(Another interesting case is the configuration of a spring and damper attached in parallel. In that case, the assumption of a step increase in strain becomes completely unviable!)

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  • $\begingroup$ Ok, now I understand intuitively, but not formally. Since the elements have to bear the same load ($\sigma_e=\sigma_N=\sigma$), if $\sigma_N$ explodes then also $\sigma_e$ explodes. Wouldn't this imply that also $\gamma_e(0)=0$? But we know that this is not the case. Or maybe the assumption that $\sigma_e=\sigma_N=\sigma$ fails at $t=0$? $\endgroup$ – valerio Mar 20 '18 at 17:12
  • $\begingroup$ Great question! I've edited my answer to address this point. $\endgroup$ – Chemomechanics Mar 20 '18 at 18:52
  • $\begingroup$ It takes an infinite shear stress to deform the viscous element instantaneously. So, even though both elements bear the same stress initially, the elastic element suffers all the deformation. $\endgroup$ – Chet Miller Mar 21 '18 at 12:35
  • $\begingroup$ I updated the question. I am still not 100% convinced, but my doubts are purely mathematical. I understand the physics intuitively, it's just the math that it's strange. $\endgroup$ – valerio Mar 21 '18 at 23:20

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