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目的:研究纳米ZrO2的添加量对聚甲基丙烯酸甲酯(PMMA)基托树脂机械性能的影响。方法:使用质量分数为1.5%的硅烷偶联剂Z-6030对纳米ZrO2颗粒进行表面处理及FTIR表征分析,未经处理者为对照组。将经过表面处理的纳米ZrO2颗粒按照质量比0.5%、1%、1.5%、2%、2.5%、3%和3.5%的添加量与PMMA混合,合成纳米ZrO2/PMMA复合材料,共7组。制作标准试件,进行表面硬度测试和三点弯曲测试,从每组中随机选取1个试件进行扫描电镜观察。采用SAS 6.12软件包对数据进行ANOVA单因素方差分析。结果:红外光谱图显示,经过表面处理的纳米ZrO2颗粒表面有硅烷偶联剂吸附。纳米ZrO2添加量为1.5%和2%组复合材料的表面硬度最佳(P<0.05)。1.5%组复合材料的挠曲强度最高(P<0.05),SEM观察断裂面为韧性断裂。结论:使用硅烷偶联剂能提高纳米ZrO2与PMMA的结合强度,经过表面处理的纳米ZrO2,能提高复合材料的机械性能,但加入量会影响纳米ZrO2的增强作用。纳米ZrO2以1.5%和2%的添加比例对基托材料的表面硬度增强效果最佳,以1.5%的添加比例对PMMA挠曲强度的增强效果最好。
Objective: To study the effect of nano ZrO2 loading on the mechanical properties of PMMA base resin. Methods: Nano-ZrO2 particles were surface-treated and characterized by FTIR with silane coupling agent Z-6030 (mass fraction of 1.5%). The untreated group was the control group. The nanosized ZrO2 particles were mixed with PMMA at the mass ratio of 0.5%, 1%, 1.5%, 2%, 2.5%, 3% and 3.5% respectively to synthesize nanostructured ZrO2 / PMMA composites. Standard test pieces were prepared, tested for surface hardness and three-point bending test, and one specimen randomly selected from each group was observed by scanning electron microscopy. Data were analyzed by ANOVA one-way ANOVA with SAS 6.12 software package. Results: The infrared spectrum showed that silane coupling agent adsorbed on the surface of nano-ZrO2 particles. The surface hardness of nanocomposites with 1.5% ZrO2 and 2% nanostructures was the best (P <0.05). The flexural strength of 1.5% composite was the highest (P <0.05), and the fracture surface was ductile fracture by SEM. Conclusion: The silane coupling agent can improve the bonding strength between nano-ZrO2 and PMMA. The surface treated nano-ZrO2 can improve the mechanical properties of the composites, but the addition amount will affect the enhancement of nano-ZrO2. The nano-ZrO2 with the addition ratio of 1.5% and 2% has the best effect on the enhancement of the surface hardness of the base material, and the enhancement effect on the flexural strength of PMMA is best with the addition ratio of 1.5%.