Issue 33

A. Winkler et alii, Frattura ed Integrità Strutturale, 33 (2015) 262-288; DOI: 10.3221/IGF-ESIS.33.32 282  The fatigue life of plastics is strongly temperature dependent. This is because the primary cause of failure is ‘pull- out’ of the intertwined molecules. This process is facilitated by an increased temperature, since the pull-out resistance reduces significantly.  Plastics have relatively high dissipation rates, which are themselves strongly temperature dependent due to changes in viscosity. This fact causes isothermal measurements of fatigue life to be more complicated than for metals.  Due to the severe temperature dependency of plastics, fatigue life analysis needs to take this into account. This will cause additional analysis costs, because: o We need to compute a temperature distribution, which is related to the magnitude and speed of loading. The computation of such a temperature distribution means that simply using a single linear analysis with a stress correction, one way or another, will be insufficient. o Further, not only will the FEA analysis need to be more involved, but the fatigue calculation will also need to become more complex. Merely using the highest temperature arising from the computed temperature distribution will yield strongly conservative results, and thus the temperature changes need to be taken into account in a fatigue calculation. A N IDEAL ANALYSIS WORLD he following section is concerned with outlining what we would perceive as being an “ideal” analysis world, where we are free to assemble the individual building blocks as we need them and as they present themselves in terms of fitness, without any preconceived boundaries being applied. Our goal is to present a concept which is not only realistic in terms of physics, but also feasible in a contemporary sense. Exemplary SN analysis: We will begin with conducting a simple SN-analysis. As an example, we have chosen a polyacetal cantilever spring with dimensions provided in Fig. 25. The dimensions and material data have been supplied by DuPont [44]. L = 15.0 mm h = 2.54 mm b = 10.0 mm The spring will repeatedly be subjected to a deflection y of 0.75 mm. We wish to determine how many repeats the spring can withstand before it fractures. We first determine the load W required to cause the deflection L/h = 15.0/2.54 = 6.0 y/L = 0.75/15.2  0.05 Using the spring rate chart (see Fig. 26), we find the spring rate (W c /y) by locating the intersection of the L/h and y/L lines. Thus, Wc/y  490 lbs/in. = 85.8 N/mm. The load is therefore calculated as W c ⋅ y = 85.8 ⋅ 0.75  64.3 N. The chart was based on a spring width of 25.4 mm, meaning that we need to scale this result accordingly. Our load therefore equates to 64.3 * 10/25.4 = 25.3 N. Figure 25 : Dimension sketch for a cantilever spring. T