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第七章 固体材料的塑性变形Chapter 7 Plastic Deformation作业1:Identify the dislocation, in terms of its Burgers vector (Using vector notation) that can cross slip between the (111) and plane of an fcc crystal.Solution: 作业2:There are three slip systems on an fcc octahedral plane. Assume a 2 MPa tensile stress is applied along the 100 direction of a gold crystal, whose critical resolred shear stress is 0.91MPa. Demonstrate quantitatively that measurable slip will not occur on any of the three slip systems in the (111) plane as a result of this applied stress.Solution: The three slip systems in the (111) plane are (111) , (111) , (111) . Because 100, that is =, so ( resolred shear stress in (111) is 0.Another two So:Measurable slip will not occur on any of the three slip systems in the (111) plane.作业3.:An FCC crystal yields under a normal stress of 2 MPa applied in the 23 direction. The slip plane is (111) and the slip direction is 01. Determine cr for this crystal. (10 points)Solution:To solve this problem, we must find both cos and cos. This can be done suing the vector dot product:Cos=Cos=Solving equation for cR and substituting the data given in the problem statement yields:cR=(2Mpa)(0.617)(0.756)=0.933Mpa作业4:假定某面心立方晶体可以开动的滑移系为011。试回答:(1)给出引起滑移的单位位错得相反矢量,并说明之。(2)如果滑移是由纯刃位错引起的,试指出位错线的方向。(3)指出上述情况下,滑移时位错线的运动方向。(4)假定在该滑移系上作用一大小为:的切应力,试计算单位刃型位错线的受力的大小和方向。(设晶格常数为a=0.2nm)解:(1)为滑移方向上的一个原子间距(大小)。方向为滑移方向,故(2)位错线一定在滑移面上,(与法线相垂直),且垂直于,则位错线方向与垂直。011 (3)滑移时位错线运动方向即为方向(4) 单位刃型位错受力大小,方向方向位错线 方向。作业5:若铜单晶可以在所有滑移系滑移,并且其表面平行于(001), 试画出表面出现的滑移线痕迹,并确定滑移线之间的夹角。若铜单晶的表面平行于(111), 情况又如何?解答: 表面平行于(001)面 滑移线痕迹,夹角9060表面平行于(111)面 夹角60作业6:铜单晶拉伸时,若力轴为001,临界分切应力值,问多大的拉伸应力能使晶体开始滑动?解:fcc,四个滑移面与(001)夹角相同,即每四个滑移面上的三条边中有一条与力轴垂直,分切应力为0,而另外两条边共计8条边等级。(即8个等级滑移系),其与力轴夹角为45,即,则使晶体开始滑移的条件是,故作业7: 拉伸铜单晶体时,若拉伸力轴的方向为001,。求(111)面上柏氏矢量的螺型位错线上所受的力()。解:设外加拉应力在(111)滑移面上沿晶向的分切应力为,则式中,为001与111夹角,为001与夹角,则若螺型位错线上受的力为Fd,则
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