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After cooling takes place,the product is formed. One cycle of the product takes about35 s including 20 s of cooling time.The material used for producing warpage testing speci-men was ABS and the injection temperature, time and pres-sure were 210 ◦C, 3 s and 60MPa respectively. The materialselected for the mould was AISI 1050 carbon steel.Properties of these materials were important in determin-ing temperature distribution in the mould carried out usingfinite element analysis. Table 2 shows the properties for ABSand AISI 1050 carbon steel.The critical part of analysis for mould is on the cavity andcore plate because these are the place where the product isformed. Therefore, thermal analysis to study the temperatureFig. 5. Model for thermal analysis.distribution and temperature at through different times areperformed using commercial finite element analysis softwarecalled LUSAS Analyst, Version 13.5. A two-dimensional(2D) thermal analysis is carried out for to study the effectof thermal residual stress on the mould at different regions.Due to symmetry, the thermal analysis was performed bymodeling only the top half of the vertical cross section orside view of both the cavity and core plate that were clampedtogether during injection. Fig. 5 shows the model of thermalanalysis analyzed with irregular meshing.Modeling for themodel also involves assigning propertiesand process or cycle time to themodel. This allowed the finiteelement solver to analyze the mould modeled and plot timeresponse graphs to show temperature variation over a certainduration and at different regions.For the product analysis, a two dimensional tensile stressanalysis was carried using LUSAS Analyst, Version 13.5.Basically the product was loaded in tension on one end whilethe other end is clamped. Load increments were applied untilthe model reaches plasticity. Fig. 6 shows loaded model ofthe analysis.3.3. Result and discussion for mould and productanalysisFor mould analysis, the thermal distribution at differenttime intervals was observed. Fig. 7 shows the 2D analysis Fig. 6. Loaded model for analysis of product.contour plots of thermal or heat distribution at different timeintervals in one complete cycle of plastic injection molding.For the 2D analysis of the mould, time response graphsare plotted to analyze the effect of thermal residual stress onthe products. Fig. 8 shows nodes selected for plotting timeresponse graphs.Figs. 9–17 show temperature distribution curves for dif-ferent nodes as indicated in Fig. 8.From the temperature distribution graphs plotted inFigs. 9–17, it is clear that every node selected for the graphplotted experiencing increased in temperature, i.e. from theambient temperature to a certain temperature higher thanthe ambient temperature and then remained constant at thistemperature for a certain period of time. This increase in tem-perature was caused by the injection of molten plastic intothe cavity of the product.After a certain period of time, the temperature is thenfurther increased to achieve the highest temperature andremained constant at that temperature. Increase in temper-ature was due to packing stages that involved high pressure, which caused the temperature to increase. This temperatureremains constant until the cooling stage starts, which causesreduction in mould temperature to a lower value and remainsat this value. The graphs plotted were not smooth due to theabsence of function of inputting filling rate of the moltenplastic as well as the cooling rate of the coolant. The graphsplotted only show maximum value of temperature that canbe achieved in the cycle.
The most critical stage in the thermal residual stress anal-ysis is during the cooling stage. This is because the cooling ther decreasing in temperature and the region away from thecooling channel experienced less cooling effect. More cool-ing effect with quite fast cooling rate means more shrinkageis occurring at the region. However, the farthest region, Node284 experiencemore cooling although far away fromcoolingchannel due to heat loss to environment.As a result, the cooling channel located at the center of theproduct cavity caused the temperature difference around themiddle of the part higher than other locations. Compressivestress was developed at the middle area of the part due tomore shrinkage and caused warpage due to uneven shrinkagethat happened. However, the temperature differences aftercooling for different nodes are small and the warpage effectis not very significant. It is important for a designer to design amould that has less thermal residual stress effectwith efficientcooling system.For the product analysis, fromthe steps being carried out toanalyze the plastic injection product, the stress distributionon product at different load factor is observed in the twodimensional analysis. Figs. 18–21 show the contour plots ofequivalent stress at different load increments.A critical point, Node 127, where the product experiencesmaximum tensile stress was selected for analysis. The stressversus strain curve and the load case versus stress curves atthis point were plotted in Figs. 22 and 23.Fromthe load case versus stress curves at this point plottedin Fig. 23, it is clear that the product experiencing increasedin tensile load until it reached the load factor of 23, whichis 1150N. This means that the product can withstand tensileload until 1150N. Load higher than this value causes failureto the product.Based on Fig. 23, the failure is likely to occur atthe region near to the fixed end of the product with maximumstress of 3.27×107 Pa.The product stress analysis reveals very limited informa-tion since the product produced was for warpage testingpurposes and had no relation with tensile loading analy-sis. In future, however, it is suggested that the product ser-vice condition should be determined so that further analysismay be carried out for other behaviors under various otherloading. Fig. 21. Equivalent stress plot at load increment 23.Fig. 22.
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