Issue 33

A. Winkler et alii, Frattura ed Integrità Strutturale, 33 (2015) 262-288; DOI: 10.3221/IGF-ESIS.33.32 279 The first together with the fourth condition gives us      e v . The second condition yields 0 1      , where 0 0    E . Finally, the third conditions gives us 1 1       e v E . Here the dot above the viscous strain refers to the rate of the viscous strain. We now assume that we are going to subject the network to an oscillating load sin     a t . We can now rearrange all the equations to give us an ordinary differential equation to solve:   0 1 1 0 1 sin         v a v E E E t E E We solve this equation to find the strain at a given frequency. The result is that our strain will be described accordingly       1 2 sin cos         a C t C t Here   1  C and   2  C are the in and out of phase compliance. The above allows us at this stage to deduce some interesting behaviour. First note that when 0   , the out of phase compliance is 0, thus we see pure in phase behaviour. The in phase compliance then simplifies to 0 1 / E . A rotational frequency of 0 essentially means that we are performing the oscillations extremely slowly, and we will only see static behaviour. Looking at our network we can observe, that if the damper is given sufficient time to relax, the only stress arising will be that of the elastic layer. For the case of   , the out of phase compliance is again 0, as the denominator will dominate the numerator for large values of  . Conversely, the in phase compliance will drop to   0 1 1 /  E E , since essentially the fraction involving  in our result will slowly approach 1. This is in accordance with what we would expect. An infinite rotational frequency means that we are pulling on the network so fast, that the damper does not have time to relax, essentially rendering it infinitely stiff. This means that we are applying the load in the viscous layer to the spring only. The result is the observation of a stiffness which is the sum of the stiffness in the elastic layer and the viscous layer, which is 0 1  E E . The in phase compliance is then merely the inverse of that as given previously. Heat generated from the Simplified Model Let us now direct our attention the matter of interest, which is the temperature effect. As we already stated, the temperature rise is caused by the dissipation of energy. Let us therefore study how much energy is dissipated per unit time over a single cycle of the load. We can express this rather simply as 0 1       T J dt T Where 2 /    T . We then find that   2 2 1 2 2 2 2 0 1 0 1 4         a E J E E E E Which is caused by the out of phase motion. It should now be recognisable that a viscous material such as a plastic has quite severe amounts of dissipation. We can also see that every doubling of the stress amplitude at a given frequency quadruples the amount of energy being dissipated when the viscosity would remain constant. Let us now look at the one- dimensional energy balance equation [42]. This reads      p d dq c r dt dx In this equation  is the temperature, q is the heat flux (that is the amount of heat transported per second over a given area), r is a heat generating source term and  and p c are the density and specific heat as usual. In order to come up with a representable solution we need to simplify this equation substantially even though we are already operating in the uniaxial domain. The heat flux is typically given through the Fourier law, but that would present us with a non-local problem, which does not have a trivial solution. We should bear this in mind, as it is one of the issues that will come back to haunt us when we want to perform physically realistic analyses pertaining to fatigue of plastics.