欢迎登录材料期刊网

材料期刊网

高级检索

运用分子动力学方法,采用EAM势函数对纳米单晶铜杆拉伸过程进行分析.选择一个无孔的纳米单晶铜杆作为模型进行拉伸实验,分析内部拉伸应力分布情况;在该模型中间挖出一个孔洞进行拉伸实验,分析孔边拉伸应力分布情况,用工程应力的方法计算应力、应力集中系数.改变模型尺寸、孔洞半径,对多个模型孔边应力集中系数进行计算,总结孔边应力集中的特点.

参考文献

[1] 王玉,刘更,朱世俊.带孔纳米单晶铜悬臂梁弯曲的分子动力学模拟[J].机械科学与技术,2006(09):1045-1048.
[2] Zhao, KJ;Chen, CQ;Shen, YP;Lu, TJ .Molecular dynamics study on the nano-void growth in face-centered cubic single crystal copper[J].Computational Materials Science,2009(3):749-754.
[3] 刘光勇.有孔纳米单晶铜薄膜拉伸断裂特性的分子动力学模拟[J].机械强度,2004(z1):84-86.
[4] Yang Xinhua;Zhou Tao;Chen Chuanyao .Effective elastic modulus and atomic stress concentration of single crystal nano-plate with void[J].Computational Materials Science,2006,40:51.
[5] Irving J H;Kirkwood J G .The statistical mechanical theory of transport processes Ⅳ The equations of hydrodynamics[J].Chemical Physics,1950,8(06):817.
[6] Cormier J.;Delph TJ.;Rickman JM. .Stress calculation in atomistic simulations of perfect and imperfect solids[J].Journal of Applied Physics,2001(1):99-104.
[7] Lutsko J F .Stress and elastic constants in anisotropic soli-ds:Molecular dynamics techniques[J].Journal of Applied Physics,1988,64(03):1152.
[8] Tsai D H .The virial theorem and stress calculation in molecular dynamics[J].Journal of Chemical Physics,1979,70(03):1375.
[9] Cheung K S;Yip S .Atomic-level stress in an inhomogenous system[J].Journal of Applied Physics,1991,70(10):5688.
[10] Cormier J.;Delph TJ.;Rickman JM. .Stress calculation in atomistic simulations of perfect and imperfect solids[J].Journal of Applied Physics,2001(1):99-104.
[11] Zhou M .A new look at the atomic level virial stress on continuum-molecular system equivalence[J].Royal Soc A,2003,459(08):2347.
[12] Kin S Cheumg;Sidney Yip .Atomic-level stress in an inhomogeneous system[J].Journal of Applied Physics,1991,70(10):5688.
[13] Wu HA .Molecular dynamics study on mechanics of metal nanowire[J].Mechanics research communications,2006(1):9-16.
[14] 陈明,张文飞.纳米尺度固体中应力计算方法研究[J].兵器材料科学与工程,2009(04):87-90.
上一张 下一张
上一张 下一张
计量
  • 下载量()
  • 访问量()
文章评分
  • 您的评分:
  • 1
    0%
  • 2
    0%
  • 3
    0%
  • 4
    0%
  • 5
    0%