摘要
依据BCC结构中的"团簇加连接原子"模型确定Ti-Mo二元团簇式[MoTi14]Mo1为基础成分式,根据合金化组元Sn、Zr和Nb与基体Ti的混合焓实现其在基础成分式中的组元替换,从而形成多元成分式[(Mo,Sn)(Ti,Zr)14]Nb1。利用铜模吸铸快冷技术制备d 6 mm合金棒状样品,并对其进行950℃保温2 h并水淬。组织结构分析和拉伸力学性能结果表明:低模量Sn、Zr和Nb分别取代基础成分式中高模量的Mo时形成的三元BCCβ-Ti合金结构稳定,且具有较低的弹性模量;当Mo0.5Sn0.5占据团簇心部,Nb作为连接原子,Zr替代部分Ti时形成的低Nb含量的β-Ti合金[(Mo0.5Sn0.5)(Ti13Zr)]Nb1(Ti68.10Mo5.25Sn6.50Zr9.98Nb10.17,质量分数,%)具有最低的弹性模量(为43 GPa),且断裂强度σb为569 MPa,应变ε为5.6%。
The basic formula of cluster formula [MoTi14]Mo1 in Ti-Mo binary alloys was determined using the clusterplus-glue-atom-model,then a general cluster formula [(Mo,Sn)(Ti,Zr)14]Nb1 was finally formed according to the mixing enthalpies between alloying elements Sn,Zr,Nb and the matrix Ti.The designed alloys were prepared into d 6 mm rods by copper mould suction cast method,and then solution-treated at 950 ℃ for 2 h followed by water-quenching.The structures of alloys were characterized,and the tensile mechanical properties were measured.The experimental results show that the ternary alloys by Sn,Zr and Nb substituting for Mo in cluster center or shell exhibit a BCC β-Ti structure and have lower elastic modulus.A multi-component alloy with the lowest elastic modulus as well as low Nb content is gained at the composition of [(Mo0.5Sn0.5)(Ti13Zr)]Nb1(Ti68.1Mo5.2Sn6.5Nb10.2Zr10.0,mass fraction,%),where a combination of Mo and Sn forms the cluster center Mo0.5Sn0.5,Nb serves as the glue atom,and partial Zr substitutes for Ti in the cluster shell.Its characteristic mechanical parameters are elastic modulus E of 43 GPa,tensile strength σb of 569 MPa and strain ε of 5.6%,respectively.
出处
《中国有色金属学报》
EI
CAS
CSCD
北大核心
2012年第12期3378-3385,共8页
The Chinese Journal of Nonferrous Metals
基金
国家自然科学基金资助项目(51171035
50901012)
