Zirconia-ceramic versus metal-ceramic: thermal expansion mismatch and residual stresses

Reference	Presenter	Authors (Institution)	Abstract
09-013	Alice Natsuko Jikihara	Jikihara, A.N.(Faculdade de Odontologia da USP);	The objective of this study was to test the hypothesis that the thermal expansion mismatches ($∆\alpha$) of veneered-zirconia systems could be responsible for high thermal residual stresses, which could justify the greater frequency of chipping. A finite element analysis was performed on flat specimens with 0.7 mm-thickness of framework and 1.5 mm-thickness of porcelain veneer. The models simulated eight $∆\mathrm{\alpha }$ conditions, resulting from the combination of two framework materials (zirconia and metal) and six veneering porcelains (diferentes CTE values). Four of the models presented $∆\alpha$ considered thermally compatible for metal (0.4 and 0.9 ppm°C-1) and for zirconia (1.3 and 1.7 ppm°C-1), two presented the same $∆\alpha$ for metal and zirconia (1ppm°C-1) and two presented negative $∆\alpha$ (-2.6 and -3.1 ppm°C-1), simulating zirconia framework combined with porcelain for metal. The simulation was consisted of two steps: (1) heat conduction analysis, in which a slow cooling of the specimen between 600 ºC and 25ºC was represented, generating the result of temperature variation along the cooling for each point of the model, and (2) mechanical analysis, in which residual thermal stresses were analyzed from the temperature variations obtained in the first analysis. The distributions of stress were analyzed. When zirconia frameworks were combined with metal-compatible- porcelains (cases of negative $∆\mathrm{\alpha }$), the residual stresses values were even higher, with inversion of the distribution pattern. When the $∆\mathrm{\alpha }$ of metal-based and zirconia-based models were similar, there was no difference in thermal residual stresses pattern or magnitude. A possible solution to avoid chipping on veneered-zirconia prosthesis would be to enhance the thermal contraction adjustment of the materials, so that the $∆\mathrm{\alpha }$ becomes zero.
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