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1、学 海 无 涯焊接模拟ansys实例!下面的命令流进行的是一个简单的二维焊接分析, 利用ANSYS单元生死和热-结构耦合分析功能进 !行焊接过程仿真, 计算焊接过程中的温度分布和应力分布以及冷却后的焊缝残余应力。finish/clear/filnam,1-2D element birth and death /title,Weld Analysis by Element Birth and Death/prep7/unit,si !采用国际单位制!*et,1,13,4!13号二维耦合单元, 同时具有温度和位移自由度et,2,13,4!1号材料是钢!2号材料是铝!3号材料是铜!铝是本次分析中的焊
2、料, 它将钢结构部分和铜结构部分焊接起来!下面是在几个温度点下, 各材料的弹性模量mptemp,1,20,500,1000,1500,2000mpdata,ex,1,1,1.93e11,1.50e11,0.70e11,0.10e11,0.01e11mpdata,ex,2,1,1.02e11,0.50e11,0.08e11,0.001e11,0.0001e11mpdata,ex,3,1,1.17e11,0.90e11,0.30e11,0.05e11,0.005e11!假设各材料都是双线性随动硬化弹塑性本构关系!下面是各材料在各温度点下的屈服应力和屈服后的弹性模量tb,bkin,1,5tbtemp
3、,20,1tbdata,1,1200e6,0.193e11tbtemp,500,2tbdata,1, 933e6,0.150e11tbtemp,1000,3tbdata,1, 435e6,0.070e11tbtemp,1500,4tbdata,1, 70e6,0.010e11tbtemp,2000,5tbdata,1, 7e6,0.001e11!tb,bkin,2,5tbtemp,20,1tbdata,1,800e6,0.102e11tbtemp,500,2tbdata,1,400e6,0.050e11tbtemp,1000,3tbdata,1, 70e6,0.008e11tbtemp,150
4、0,4tbdata,1, 1e6,0.0001e11tbtemp,2000,5tbdata,1,0.1e6,0.00001e11!tb,bkin,3,5tbtemp,20,1tbdata,1,900e6,0.117e11tbtemp,500,2tbdata,1,700e6,0.090e11tbtemp,1000,3tbdata,1,230e6,0.030e11tbtemp,1500,4tbdata,1, 40e6,0.005e11tbtemp,2000,5tbdata,1, 4e6,0.0005e11!材料密度(假设为常值)mp,dens,1,8030mp,dens,2,4850mp,dens
5、,3,8900! 热膨胀系数(假设为常值)mp,alpx,1,1.78e-5mp,alpx,2,9.36e-6mp,alpx,3,1.66e-5!泊松比(假设为常值)mp,nuxy,1,0.29mp,nuxy,2,0.30mp,nuxy,3,0.30!热传导系数(假设为常值)mp,kxx,1,16.3mp,kxx,2,7.44mp,kxx,3,393!比热(假设为常值)mp,c,1,502mp,c,2,544mp,c,3,385!热膨胀系数(假设为常值)!由于该13号单元还有磁自由度, 此处假设一磁特性, 但并不影响我们所关心的结果mp,murx,1,1mp,murx,2,1mp,murx,3
6、,1!假设焊料(铝)焊上去后的初始温度是1500mp,reft,1,20mp,reft,2,1500mp,reft,3,20!*!下面建立几何模型csys,0k,1,0,0,0k,2,0.5,0,0k,3,1,0,0!长1米k,4,0,0.3,0!厚度0.3米（二维中叫做宽度）k,5,0.35,0.3,0k,6,0.65,0.3,0k,7,1,0.3,0a,1,2,5,4a,2,6,5a,2,3,7,6!划分网格!esize,0.025type,2mat,2amesh,2 ! esize,0.05 !网格划分出现问题type,1mat,1amesh,1!mat,3amesh,3eplot!/s
7、oluantype,4! 瞬态分析trnopt,full!在模型的左边界加位移约束!nsel,all*get,minx,node,mnloc,xnsel,s,loc,x,minxd,all,ux,0*get,miny,node,mnloc,ynsel,r,loc,y,minyd,all,uy,0!*假设模型的左右边界处温度始终保持在20摄氏度左右*!其他边界条件如对流和辐射等均可施加，此处因为只是示意而已，故只施加恒温边界条件nsel,all*get,minx,node,mnloc,xnsel,s,loc,x,minxd,all,temp,20nsel,all*get,maxx,node,mx
8、loc,xnsel,s,loc,x,maxxd,all,temp,20!由于第2个面是焊接所在区域，因此首先将该区域的单元“杀死”!nna=2esel,all*get,emax,elem,num,maxasel,s,area,nnaesla*get,nse,elem,count*dim,ne,nse*dim,nex,nse*dim,ney,nse*dim,neorder,nsemine=0!* !下面的do循环用于将焊料区的单元按其形心y坐标排序!以便后面模拟焊料由下向上逐步“生长”过程*do,i1,1,nseesel,u,elem,mine*get,nse1,elem,countii=0*d
9、o,i,1,emax*if,esel(i),eq,1,thenii=ii+1ne(ii)=i*endif*enddo*do,i,1,nse1*get,ney(i),elem,ne(i),cent,y*get,nex(i),elem,ne(i),cent,x*enddominy=1e20minx=1e20*do,i,1,nse1*if,ney(i),lt,miny,thenminy=ney(i)minx=nex(i)mine=ne(i)*else*if,ney(i),eq,miny,then*if,nex(i),lt,minx,thenminy=ney(i)minx=nex(i)mine=ne(
10、i)*endif*endif*endif*enddoneorder(i1)=mine*enddo!* max_tem=1500 !按照前面假设，焊料的初始温度为1500dt1=1e-3 !用于建立初始条件的一个很小的时间段dt=5 !焊接一个单元所需的时间t=0!起始时间esel,alleplot/auto,1/replot*do,i,1,nseekill,neorder(i)esel,s,liveeplot*enddoallsel,alloutres,all,allic,all,temp,20kbc,1timint,0,structtimint,1,thermtimint,0,magtint
11、p,0.005,1,0.5,0.2!nsub1=2nsub2=40 !* *do,i,1,nseealive,neorder(i)esel,s,liveeplotesel,all!*下面的求解用于建立温度的初始条件*t=t+dt1time,tnsubst,1*do,j,1,4d,nelem(neorder(i),j),temp,max_tem*enddosolve!*下面的求解用于保证初始的升温速度为零*t=t+dt1time,tsolve!*下面的步骤用于求解温度分布* *do,j,1,4ddele,nelem(neorder(i),j),temp*enddot=t+dt-2*dt1time,tnsubst,nsub1solve*enddot=t+50000 !*下面的步骤用于冷却过程求
