【作者】 张强；

【导师】 吴欣；

【作者基本信息】 哈尔滨工业大学 ， 材料学， 2010， 硕士

【摘要】 金属钛是二十一世纪最重要的结构材料之一,由于其优良的性能而越来越广泛的应用于各领域,金属钛的塑性变形机制一直是人们研究的热点,本文采用分子动力学的方法研究金属钛单晶的塑性变形机制。本文利用分子动力学研究方法,采用钛的分析型嵌入式原子势,建立了不同加载条件下金属钛的剪切、拉伸、压缩变形的分子动力学模型,研究了金属钛在不同加载条件下的塑性变形行为。比较了截面尺寸、温度、变形速度对金属钛拉伸性能的影响,得出以下主要研究结论:利用分析型的原子间作用势形式得到了金属钛的嵌入式原子势(EAM),通过计算金属钛的熔点、空位形成能验证了金属钛嵌入式原子势的准确性。计算得到的金属钛的熔点为1920K,空位形成能为1.50eV。通过与实验值进行比较,得出建立的嵌入式原子势准确性较高,可以用于塑性变形模拟。建立了沿(0001)晶面[2110]方向与沿(1010)晶面[2110]剪切的分子动力学模型,研究了加载方向对剪切变形的影响。结果表明:在300K,0.01nm/ps加载速度下,沿(0001)晶面[2110]方向进行剪切起始塑性变形应力、应变分别为0.27GPa、0.05;沿(1010)晶面[2110]方向进行剪切起始塑性变形应力、应变分别为0.57GPa、0.18。基面的剪切变形比柱面的更容易。建立了沿[0001]晶向和[0110]方向拉伸模型,结果表明:两种方向的拉伸均包含弹性变形阶段、均匀塑性变形阶段、颈缩阶段、断裂阶段。沿[0001]方向拉伸时,滑移系少,取向偏离软取向。变形时屈服强度为3.55GPa,屈服应变为0.095,断裂时的应变为0.55。沿[0110]方向拉伸时滑移系多,取向接近软取向。变形的屈服强度为2.41 GPa,屈服应变为0.067,断裂时的应变达到1.28,沿[0110]的拉伸变形更容易进行,表现出的塑性更好。建立了沿[0110]压缩模型,研究了压缩变形过程,结果表明:压缩变形的应力-应变规律经历了弹性变形阶段、屈服阶段、应力上升阶段。沿[0110]压缩,滑移系多,取向接近软取向。变形时的屈服强度为2.42GPa,屈服应变为0.04,可以看出与沿[0110]拉伸屈服应力基本相同。更多还原

【Abstract】 The metal titanium is one of the most important structural materials in 21st century, due to its excellent performance and has been widely used in various fields. The plastic deformation mechanism of titanium has been a focus for researchers. In this paper, through the method of molecular dynamics, the plastic deformation mechanism of single crystal titanium has been investigated.With an embedded atomic potential (EAM) and molecular dynamic (MD)method, deformation model of shear, tensile and compression have been established and plastic deformation behaviour have been investigated on titanium through different loading style. The influence of temperature, size and tensile speed on the properties of titanium has been analyzed. The main conclusions are as follow: An embedded atomic potential (EAM) was selected and used to simulated the melting point, the formation energy of vacancy of titanium to check its accuracy and efficacy. The calculated melt point temperature and the formation energy of a vacancy are 1920K and 1.469 eV, respectively. These values are very close to the experimental findings, providing strong evidence that this EAM could be used to simulate the plastic deformations of titanium.MD model of shear deformation along (0001)[2110] and (1010)[2110] have been established and the influence of shearing direction on deformation behaviour have been compared. The result shows that the initial stress and strain of plastic deformation are 0.27GPa, 0.05 and 0.57GPa, 0.18 respectively for shearing along (0001)[2110] and (1010)[2110], with deforming speed 0.01nm/ps, 300K. Shear deformation along basal plane is easier than prismatic plane.Model of tensile deformation with tensile direction along [0001] and [ 0110] have been established. The results show that tensile process of both two directions contains four stages, ie, elastic deformation, uniform plastic deformation, necking and fracture. The slip system is few and the orientation deviates from the soft orientation when tension applied along [0001]. The yielding stress and strain are 3.55GPa and 0.095 respectively while the fracture strain is 0.55. For [ 0110] tension, the slip system is more and the orientation is close to soft orientation. The yielding stress and strain are 2.41GPa and 0.067 respectively while the fracture strain is 1.28. The plastic deformation along [0110] tension is easier as well as the plasticity is better.Model of compression deformation with compress applied along [0110] has been established and the process of deformation has been investigated. The result shows that curve of stress vs strain implies three deforming stages, ie, elastic, yielding, and hardening. The slip system is close to soft orientation. The yielding stress and strain are 2.42GPa, 0.04 respectively, with the same yielding stress as [0110] tension.更多还原