混凝土非线性分析

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1、各位同学，这里几乎已经是最完整的命令流了，关于这方面工作，初步建议如下： （1）初步的模拟最好先不要考虑混凝土下降段，不要考虑混凝土的压碎，同时单元尺寸不 要过小，以免影响收敛。（2）反复试算结果表明，混凝土的MISO和KINH模型、DP和MISES准则、剪力传递系 数对模拟结果影响均不大，但在一定程度上影响了收敛性的好坏。（3）当解决了收敛问题之后，最大的难点是下降段，对于一个桥墩（柱）的反复荷载试验 引起承载力下降的原因大体可归结为：塑性铰区混凝土压碎脱落、纵筋屈曲，对于以剪切破 坏为主的试件，还包括剪切滑移的影响。对于以黏结破坏为主的试件，还包括纵筋混凝土 黏结滑移破坏。因此，能否正确模

2、拟这些将是模型正确与否的关键。众所周知，以Iink8单 元模拟钢筋自然无法模拟纵筋屈曲，这样就把一个因素忽略了。（4）作者们最后采用了扩大的破坏面，坦率的讲，这也是为了模拟混凝土的压碎破坏而采 取的没有办法的办法，鉴于作者水平有限，至今也无法对此做出更为合理的解释。破坏面到 底扩大多少，仍无规律可循。且一旦打开压碎开关，会出现诸如难以收敛、 “假压碎”等一系 列问题。所以我们建议大家起初还是先不要考虑压碎为好。定义应力应变曲线1， 定义变量：拾取主菜单：Ma in Menu >Time Hist postproc>Defi ne Variables>在随之弹出的对话 框中点击

3、Add键，定义第一个变量序号为2,选取第一个变量stress,确定与之对应的下一 级选项（如Y-direction SY等）；返回定义变量对话框，再点击add键，定义第二个变量序 号为3,选取第二个变量strain-elastic及以及对赢得下一级选项（如Y-dirn EPEL Y等，在 应力-应变图中，其向量的取向应相同）。同理再定义变量4,选取变量strain-plastic及与 之对应的下一级选项如Y-dirn EPEL Y等），在应力-应变图中，应变是弹性应变和塑性应 变累加的总应变。为使其实现相加，还需进行以下操作：拾取主菜单： Main Menu>Time Hist post

4、proc>math operation>add,定义计算变量序号为5,同时在相应交互框内输 入3和 4。点击确认键，则由变量3， 4代表的应变之和就存在变量5中。2，绘制应力-应变曲线：拾取主菜单：Main Menu>Time Hist postproc>setting>graph.设置 x 轴向变量为单 变量，并将其变量序号定义为5。点击确定键退出退化框。拾取应用菜单:Utility Menu>plot ctrIs>styIes>Graphs>Modify axis.将 x,y 坐标轴分别命名为 Y-strains,Y-stress,拾取

5、 主菜单：Main Menu>Time Hist postproc>graph variables.在对话框上the first variable对应的交互框中输入2。点击确定键，则预想的应力-应变曲线就显示在屏幕上。 ok!试试看！欢迎大家继续批评指正！/prep7et,1,solid65et,2,link8et,3,solid45r,1r,2,236e-6r,3,75.4e-6r,4,151.2e-6!混凝土本构关系mp,ex,1,2.522e10mp,prxy,1,0.2mp,dens,1,2500tb,concr,1,1tbdata,0.5,0.95,5e6,-1tb,mi

6、so,1,1,7tbtemp,0tbpt,0.00068,17.15e6tbpt,0.00136,28.03e6tbpt,0.00204,33.62e6tbpt,0.00272,35.96e6tbpt,0.0034,36.53e6tbpt,0.00408,36.53e6!纵筋的本构关系mp,ex,2,2.0e11mp,prxy,2,0.28tb,bkin,2tbtemp,0tbdata,362e6,2e9！箍筋的本构关系mp,EX,3,2.1E11mp,PRXY,3,.27tb,bkin,3tbtemp,0tbdata,272.5e6,2.1e9!建立模型Iocal,11,1,0,0,0,0,

7、-90,0!建立局部坐标系，采用柱坐标，绕y轴顺时针转90度csys,11*do,j,1,8n,j,0.185,j*360/8,-0.6*enddongen,3,100,1,8,1,0,0,0.3 !复制节点， 3 次，编号增量，源节点18， 0,0,0.3 ngen,4,100,201,208,1,0,0,0.15ngen,7,100,501,508,1,0,0,0.3type,2 real,3 mat,3 *do,i,1,6 *do,j,1,7 e,(i-1)*100+j,(i-1)*100+j+1 *enddo！指定单元属性，建立基础箍筋单元e,(i-1)*100+1,(i-1)*100

8、+8*enddotype,2 ！建立柱上箍筋单元 real,4mat,3*do,i,1,5*do,j,1,7 e,(i+5)*100+j,(i+5)*100+j+1*enddo e,(i+5)*100+1,(i+5)*100+8*enddotype,2real,2mat,2*do,i,1,11*do,j,1,8 e,(i-1)*100+j,i*100+j *enddo*enddo！建立纵向钢筋allsel,all nummrg,all numcmp,all！合并节点，压缩节点wpcsys,1,11cylind,0,0.185,-0.6,0,0,360 cylind,0,0.185,0,0.45

9、,0,360 cylind,0.185,0.3,-0.6,0,0,360 cylind,0,0.185,0.45,2.25,0,360!建立圆柱，内外径i-or，高度z1-z2，起终角度cylind,0,0.185,2.25,2.55,0,360 cylind,0.185,0.20,2.25,2.55,0,360 cylind,0,0.185,2.55,2.565,0,360wprota,90 vsel,s,1,7,1 vsbw,all numcmp,all wprota,90 vsel,s,1,14,1 vsbw,all numcmp,all!绕y轴逆时针转90度!选择体，选择类型，item

10、, comb，最小编号，最大编号，增量 !体相减!绕z轴逆时针转90度vsel,s,1,28,1 vglue,all numcmp,all wprota,-90 wprota,-90 allsel,all vsel,s,1,16,1 vsel,a,21,24,1 vatt,1,1,1 vsel,s,17,20,1 vsel,a,25,28,1 vatt,2,1,3allsel,alllsel,s,1,8,1 lsel,a,11,22,1 lsel,a,25,40,1 lsel,a,43,46,1 lesize,all,2！选择线!补选（a）线！线尺寸为 2lsel,s,47,50,1lsel,

11、a,53,56,1lsel,a,58,73,1lesize,all,2 lsel,s,98,107,1 lsel,a,110,112,2 lesize,all,1 lsel,s,23,24,1lsel,a,9,10,1lesize,all,2 lsel,s,78,82,1lesize,all,3lsel,s,83,87,1lesize,all,6lsel,s,41,42,1lsel,a,88,92,1lesize,all,2lsel,s,93,97,1lesize,all,1 mshkey,1 mshape,0,3d vsel,s,1,28 vmesh,all allsel,all nummr

12、g,all numcmp,all wprota,90 csys,0 allsel,all /solu ！进入加载、求解！选择半径在 x=3 处的节点，并施加约束csys,11 nsel,s,loc,x,0.3 d,all,all allsel,allOUTRES,ALL,2！输出获得的解csys,0pred,on cnvtol,f,0.05,2 autots,1 lnsrch,1 ncnv,2！打开预应力选项!定义收敛条件,使用缺省的 VALUE!打开自动时间步控制!打开线性搜索 !如果不收敛时结束而不退出neqit,50save!每一子步中方程的迭代次数限值time,1nsubst,10 a

13、sel,s,29,32,1 sfa,all,pres,5468699 allsel,allacel,9.8kbc,1lswrite,1！定义第 1 载荷步！在已选择的面上施加压力荷载！重力加速度 y=9.8！阶跃加载方式time,10cp,1,ux,298,300,303d,298,ux,0.015AA* c-4 I H.！定义第 2 荷载步!将节点 298,300,303ux 向位移耦合!再借点298处施加ux位移0.015allsel,allKBC,0allsel,allNSUBST,20,1000000,10lswrite,2time,20！定义第 3 载荷步d,298,ux,-0.01

14、5allsel,allKBC,0allsel,allNSUBST,100,1000000,10lswrite,3time,30！定义第 4 载荷步d,298,ux,0.015allsel,allKBC,0allsel,allNSUBST,100,1000000,10lswrite,4time,40！定义第 5 载荷步d,298,ux,-0.015allsel,allKBC,0allsel,allNSUBST,100,1000000,10lswrite,5 time,50 ！定义第 6 载荷步 d,298,ux,0.015allsel,allKBC,0allsel,allNSUBST,100,1

15、000000,10lswrite,6time,60 ！定义第 7 载荷步 d,298,ux,-0.015allsel,allKBC,0allsel,allNSUBST,100,1000000,10lswrite,7 time,70 ！定义第 8 载荷步 d,298,ux,0.015allsel,allKBC,0allsel,allNSUBST,100,1000000,10 lswrite,8time,80 d,298,ux,-0.015allsel,allKBC,0allsel,allNSUBST,100,1000000,10 lswrite,9time,90d,298,ux,0.015all

16、sel,allKBC,0allsel,allNSUBST,100,1000000,10lswrite,10time,100d,298,ux,-0.015allsel,allKBC,0allsel,allNSUBST,100,1000000,10lswrite,11time,110d,298,ux,0.03allsel,allKBC,0allsel,allNSUBST,200,1000000,10lswrite,12time,120d,298,ux,-0.03allsel,allKBC,0allsel,allNSUBST,200,1000000,10lswrite,13time,130d,298

17、,ux,0.03allsel,allKBC,0allsel,allNSUBST,200,1000000,10lswrite,14time,140d,298,ux,-0.03allsel,allKBC,0allsel,allNSUBST,200,1000000,10 lswrite,15time,150d,298,ux,0.03allsel,allKBC,0allsel,allNSUBST,200,1000000,10lswrite,16time,160d,298,ux,-0.03allsel,allKBC,0allsel,allNSUBST,200,1000000,10lswrite,17ti

18、me,170d,298,ux,0.03allsel,allKBC,0allsel,allNSUBST,200,1000000,10lswrite,18time,180d,298,ux,-0.03allsel,allKBC,0allsel,allNSUBST,200,1000000,10lswrite,19time,190d,298,ux,0.03allsel,allKBC,0allsel,allNSUBST,200,1000000,10lswrite,20time,200d,298,ux,-0.03allsel,allKBC,0allsel,allNSUBST,200,1000000,10 l

19、swrite,21time,210d,298,ux,0.045allsel,allKBC,0allsel,allNSUBST,300,1000000,10lswrite,22lssolve,1,22,1命令流：1、单调水平荷载作用下的命令流：建模：!mono brick wall!solid65-整体式建模!*Finish/clear/units,si/com,Structural/config,nres,400000/prep7/title,mono-brick-wall!*depth_sup=0.24 !弹性支垫 240mm!单元属性!for brick, mortar and level

20、ing coatet,1,solid65!*KEYOPT,1,1,0KEYOPT,1,5,0KEYOPT,1,6,0KEYOPT,1,7,1!for elastic supportet,2,solid45!*砌体材料!MU10,M5 砌体材料属性mp,prxy,1,0.15mp,dens,1,1700mp,ex,1,0.24e10!砌体屈服准则TB,MKIN,1TBTEMP,strain TBdata,1,0.2E-3,1.0E-3,2.0E-3,3.0E-3,4.8E-3TBTEMP, !TBDATA,1,0.48E6,0.82E6,1.32E6,1.5E6,1.32E6 TBDATA,1,

21、0.48E6,1.31E6,2.11E6,2.4E6,2.11E6!/XRANGE,0,0.02 !TBPLOT,MKIN,1 !砌体破坏准则!hntrl=0.13E6 hntrl=0.21E6 tb,concr,1,4, tbdata,0.2,0.9,hntrl,-1 !不考虑压碎!*弹性支撑!弹性支撑暂取钢筋材料属性mp,ex,2,2.0e11 mp,dens,2,7.600 mp,prxy,2,0.30 !real constantsr,1!*modeling!geometry modelblock,0,3.12,0,0.24,0,3.12 block,-0.24,3.12,0,0.24

22、,3.12,3.36 Vsel,s,volu,1,2,1$aslv,s$lsla,s lesize,all,0.24Vsel,s,volu,1$vatt,1,1,1, !brick wallVsel,s,volu,2$vatt,2,1,2, !brick wall vsel,all !必须再次选择所有单元VSWEEP,ALLsavenumcmp,allallsel!*/device,vector,1/eshape,1!*耦合allsel,all!elastic support nodes set esel,s,type,2$nsle,s,all$nsel,r,loc,z,3.12 nsel,u

23、,loc,x,-0.24$cm,slavenode,node!other node set esel,s,type,1$nsle,s,all$cm,masternode,node /input,ucouple,macsave finish耦合文件 ucouple,mac：!dowhile,kk k=nnear(slave_node(i) nsel,s,cp,all kk=nsel(k)+0.001开始 allsel !最好保留这句命令 !*将从属节点编号依次存入数组* cmsel,s,slavenode*get,count1_node,node,0,count *del,slave_node*

24、dim,slave_node,array,count1_node *get,slave_node(1),node,0,num,min *do,i,2,count1_node slave_node(i)=ndnext(slave_node(i-1) *enddo !*将主节点编号依次存入数组 allsel cmsel,s,masternode*get,count2_node,node,0,count *del,master_node *dim,master_node,array,count2_node *get,master_node(1),node,0,num,min *do,i,2,coun

25、t2_node master_node(i)=ndnext(master_node(i-1) *enddo *del,cp_node*dim,cp_node,array,count1_node*do,i,1,count1_nodecp_node(i)=0*enddo allsel cmsel,s,masternode *do,i,1,count1_node kk=1k=1allsel cmsel,s,masternode nsel,u,node,k cm,masternode,node *enddo cp_node(i)=k *enddo !*选择完毕*!*开始耦合* allsel,all/p

26、rep7*do,i,1,count1_nodecp,next,UX,slave_node(i),cp_node(i) cp,next,UY,slave_node(i),cp_node(i) cp,next,UZ,slave_node(i),cp_node(i) *enddo求解方法 NR 法：brick wall NR solution!displacement load!用NR法!*solution/soluantype,0!constraintnsel,s,loc,z,0d,all,all!*esel,s,type,2$nsle,s,all$nsel,r,loc,x,-0.24d,all,

27、uy!*!esel,s,type,2$nsle,s,all$nsel,r,loc,x,-0.24!d,all,uzlumpm,on solcontrol,on pred,on lnsrch,on autots,1 !打开自动时间步控制ncnv,2 !如果不收敛时结束而不退出compression=0.24E7!*load steps!*1st acel,9.8 nsel,s,loc,z,3.36$nsel,r,loc,x,0.001,3.119 esln,s,0,allsfe,all,6,pres,0,0.03*compression allsel,all!*OUTRES,ERASEOUTRE

28、S,ALL,ALLNSUBST,2,200,1 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.01,2,0.05 allsellswrite,1!*2nd stepnsel,s,loc,x,-0.24!esln,s,0,all !sfe,all,4,pres,0,60*0.0593E6 !d,all,ux,2.79E-3 d,all,ux,1.0E-4allsel,all!*OUTRES,ERASEOUTRES,ALL,ALLNSUBST,200,2000,100 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.05,2,0.05 allsellswr

29、ite,2!*3rd stepnsel,s,loc,x,-0.24!esln,s,0,all !sfe,all,4,pres,0,60*0.0593E6 !d,all,ux,2.79E-3 d,all,ux,2.5E-4allsel,all!*OUTRES,ERASE OUTRES,ALL,ALL NSUBST,200,2000,100 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.09,2,0.25 allsellswrite,3!*4th stepnsel,s,loc,x,-0.24!esln,s,0,all !sfe,all,4,pres,0,60*0.0593E6

30、 !d,all,ux,2.79E-3 d,all,ux,4.0E-3allsel,all!*OUTRES,ERASEOUTRES,ALL,ALLNSUBST,200,2000,100 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.1,2,0.5 allsellswrite,4!*allsel lssolve,1,4,1 FINISH也可以用弧长法，不过没有一定经验这比较难控制：!mono brick wall-1 arclength solution!displacement load!*solution/soluantype,0!constraintnsel,s,loc

31、,z,0d,all,all!*esel,s,type,2$nsle,s,all$nsel,r,loc,x,-0.24 d,all,uy!*lumpm,on solcontrol,on ncnv,2 !如果不收敛时结束而不退出 compression=0.24E7!*load steps!*first step- acel,9.8 nsel,s,loc,z,3.36$nsel,r,loc,x,0.001,3.12 esln,s,0,allsfe,all,6,pres,0,0.03*compression allsel,all!*OUTRES,ERASEOUTRES,ALL,ALLarclen,o

32、n,1E-5, !使用弧长法NSUBST,200neqit,15KBC,0nlgeom,1cnvtol,f,0.1,2,0.5allsel,alllswrite,1!*second stepesel,s,type,2$nsle,s,all$nsel,r,loc,x,-0.24 d,all,uz!*nsel,s,loc,x,-0.24!esln,s,0,all!sfe,all,4,pres,0,60*0.0593E6 d,all,ux,2.50E-3!d,all,ux,3.51E-3allsel,all!*OUTRES,ERASEOUTRES,ALL,ALLarclen,on,1E-5, !使用

33、弧长法kbc,0NSUBST,250!arctrm,u,0.0028,313,uxneqit,15nlgeom,1!cnvtol,f,0.005,2,0.01!cnvtol,f,0.05,2,0.5 cnvtol,f,0.1,2,0.5 allsellswrite,2!*third stepesel,s,type,2$nsle,s,all$nsel,r,loc,x,-0.24 d,all,uz!*nsel,s,loc,x,-0.24!esln,s,0,all!sfe,all,4,pres,0,60*0.0593E6!d,all,ux,2.79E-3!d,all,ux,3.18E-3d,all,

34、ux,3.00E-3allsel,all!*OUTRES,ERASEOUTRES,ALL,ALLarclen,on,1E-7, !使用弧长法kbc,0NSUBST,2000!arctrm,u,0.0028,313,uxneqit,15nlgeom,1 !cnvtol,f,0.005,2,0.01 !cnvtol,f,0.05,2,0.5 cnvtol,f,0.5,2,0.8 allsellswrite,3allsel!*fourth stepesel,s,type,2$nsle,s,all$nsel,r,loc,x,-0.24 d,all,uz!*nsel,s,loc,x,-0.24!esln

35、,s,0,all!sfe,all,4,pres,0,60*0.0593E6 d,all,ux,3.51E-3allsel,all!*OUTRES,ERASEOUTRES,ALL,ALLarclen,on,1E-7, !使用弧长法kbc,0NSUBST,300 !arctrm,u,0.0028,313,ux neqit,15nlgeom,1!cnvtol,f,0.005,2,0.01!cnvtol,f,0.05,2,0.5cnvtol,f,0.6,2,0.9allsellswrite,4allsel!*lssolve,1,4,1FINISH后处理：/post26 esel,s,type,1 ep

36、lot lines,5000 nsol,2,326,u,x,ux !nsol,2,313,u,x,ux*do,i,1,14 jj=node(i-1)*0.24,0.24,0) kk=node(i-1)*0.24,0,0)! ll=node(225.5+(i-1)*225.5,3300,0) rforce,3,jj,f,x,fxadd,6,3,6 rforce,4,kk,f,x,fxadd,6,4,6! rforce,5,ll,f,x,fx !add,6,5,6*enddo prod,7,2,1000 prod,8,6,-1/1000 /axlab,x,Displacement(mm) /axl

37、ab,y,F(kN)/XRANGE,0,4 /YRANGE,0,100xvar,7plvar,8!rforce,5,340,f,x,fx!esol,5,87,221,f,x,fx2、循环水平荷载作用下的命令流： 出求解文件不同外，其他都相同； !cyclic NR solution!displacement load!*solution/soluantype,0!constraintnsel,s,loc,z,0d,all,all!*esel,s,type,2$nsle,s,all$nsel,r,loc,x,-0.24d,all,uy!*esel,s,type,2$nsle,s,all$nsel

38、,r,loc,x,3.36d,all,uy!*lumpm,on solcontrol,on pred,on lnsrch,on autots,1 !打开自动时间步控制 ncnv,2 !如果不收敛时结束而不退出 compression=0.24E7!*load steps!*1st step acel,9.8 nsel,s,loc,z,3.36$nsel,r,loc,x,0.001,3.119 esln,s,0,allsfe,all,6,pres,0,0.03*compression allsel,all!*OUTRES,ERASEOUTRES,ALL,ALLNSUBST,200,2000,10

39、neqit,15KBC,0 nlgeom,1 cnvtol,f,0.1,2,0.5 allsel,alllswrite,1!*初裂荷载20%做1 次循环!*用 NR 法!*2nd step esel,s,type,2$nsle,s,all$nsel,r,loc,x,-0.24 d,all,uz!*esel,s,type,2$nsle,s,all$nsel,r,loc,x,3.36 d,all,uz!*nsel,s,loc,x,-0.24!esln,s,0,all!sfe,all,4,pres,0,60*0.0593E6 d,all,ux,1.48E-4 allsel,all!*OUTRES,E

40、RASEOUTRES,ALL,ALLNSUBST,20,2000,10neqit,15NROPT ,FULL,OFFKBC,0 nlgeom,1 cnvtol,f,0.1,2,0.5 allsel lswrite,2!*3rd stepnsel,s,loc,x,-0.24 ddele,all,uxnsel,s,loc,x,3.36d,all,ux,-1.48E-4 allsel,all !*OUTRES,ERASEOUTRES,ALL,ALLNSUBST,40,2000,20neqit,15NROPT ,FULL,OFFKBC,0 nlgeom,1cnvtol,f,0.1,2,0.5allse

41、l lswrite,3 !*4th stepnsel,s,loc,x,3.36 ddele,all,ux nsel,s,loc,x,-0.24 d,all,ux,1.48E-4allsel,all!*OUTRES,ERASE OUTRES,ALL,ALL NSUBST,40,2000,20 neqit,15NROPT ,FULL,OFFKBC,0 nlgeom,1 cnvtol,f,0.1,2,0.5 allsel lswrite,4!*one cycle!*初裂荷载50%做1 次循环!*用 NR 法!*5th step nsel,s,loc,x,-0.24 ddele,all,ux nsel

42、,s,loc,x,-0.24 d,all,ux,2.85E-4 allsel,all!*OUTRES,ERASEOUTRES,ALL,ALLNSUBST,100,2000,10 neqit,15NROPT ,FULL,OFFKBC,0 nlgeom,1 cnvtol,f,0.1,2,0.5 allsel lswrite,5!*6th stepnsel,s,loc,x,-0.24ddele,all,ux nsel,s,loc,x,3.36d,all,ux,-2.85E-4 allsel,all !* OUTRES,ERASE OUTRES,ALL,ALL NSUBST,200,2000,30 n

43、eqit,15NROPT ,FULL,OFFKBC,0 nlgeom,1 cnvtol,f,0.1,2,0.5 allsel lswrite,6!*7th stepnsel,s,loc,x,3.36 ddele,all,uxnsel,s,loc,x,-0.24d,all,ux,2.85E-4 allsel,all !*OUTRES,ERASEOUTRES,ALL,ALLNSUBST,200,2000,30 neqit,15NROPT ,FULL,OFFKBC,0 nlgeom,1 cnvtol,f,0.1,2,0.5 allsel lswrite,7!*two cycle!*初裂荷载80%做1

44、 次循环!*用 NR 法!*8th step nsel,s,loc,x,-0.24 ddele,all,ux nsel,s,loc,x,-0.24 d,all,ux,3.5E-4 allsel,all!*OUTRES,ERASEOUTRES,ALL,ALL!arclen,on,1E-7, !使用弧长法 kbc,0NSUBST,100,2000,20 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.5,2,0.9 allsel lswrite,8!*9th stepnsel,s,loc,x,-0.24 ddele,all,ux nsel,s,loc,x,3.36 d,all,

45、ux,-3.5E-4allsel,all!*OUTRES,ERASEOUTRES,ALL,ALL!arclen,on,1E-7, !使用弧长法 kbc,0NSUBST,200,2000,30 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.5,2,0.9 allsel lswrite,9!*10th step-nsel,s,loc,x,3.36 ddele,all,uxnsel,s,loc,x,-0.24d,all,ux,3.5E-4 allsel,all !*OUTRES,ERASEOUTRES,ALL,ALL!arclen,on,1E-7, !使用弧长法 kbc,0NS

46、UBST,200,2000,30 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.5,2,0.9 allsel lswrite,10!*three cycle!*初裂荷载 100%做 1 次循环!*用 NR 法!*11th step nsel,s,loc,x,-0.24 ddele,all,ux nsel,s,loc,x,-0.24 d,all,ux,4.42E-4 allsel,all!*OUTRES,ERASEOUTRES,ALL,ALL!arclen,on,1E-7, !使用弧长法 kbc,0NSUBST,100,2000,20 neqit,15KBC,0 nlgeo

47、m,1 cnvtol,f,0.5,2,0.9 allsel lswrite,11!*12th stepnsel,s,loc,x,-0.24 ddele,all,uxnsel,s,loc,x,3.36d,all,ux,-4.42E-4 allsel,all !*OUTRES,ERASEOUTRES,ALL,ALL !arclen,on,1E-7, !使用弧长法 kbc,0NSUBST,200,2000,30 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.5,2,0.9 allsel lswrite,12!*13th stepnsel,s,loc,x,3.36 ddele,a

48、ll,uxnsel,s,loc,x,-0.24 d,all,ux,4.42E-4 allsel,all!*OUTRES,ERASE OUTRES,ALL,ALL !arclen,on,1E-7, !使用弧长法 kbc,0NSUBST,200,2000,30 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.5,2,0.9 allsel lswrite,13!*four cycle!*61.4KN 的荷载做 1 次循环!*用 NR 法!*14th step nsel,s,loc,x,-0.24 ddele,all,ux nsel,s,loc,x,-0.24 d,all,ux,6

49、.72E-4 allsel,all!*OUTRES,ERASEOUTRES,ALL,ALL !arclen,on,1E-7, !使用弧长法 kbc,0NSUBST,100,2000,20 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.5,2,0.9 allsel lswrite,14!*15th stepnsel,s,loc,x,-0.24 ddele,all,uxnsel,s,loc,x,3.36 d,all,ux,-6.72E-4 allsel,all!*OUTRES,ERASEOUTRES,ALL,ALL !arclen,on,1E-7, !使用弧长法 kbc,0N

50、SUBST,200,2000,30 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.5,2,0.9 allsel lswrite,15!*16th stepnsel,s,loc,x,3.36 ddele,all,uxnsel,s,loc,x,-0.24 d,all,ux,6.72E-4 allsel,all!*OUTRES,ERASEOUTRES,ALL,ALL!arclen,on,1E-7, !使用弧长法 kbc,0NSUBST,200,2000,30 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.5,2,0.9 allsel lswrite,16!

51、*five cycle!*74.1KN 的荷载做 1 次循环!*用 NR 法!*17th step nsel,s,loc,x,-0.24 ddele,all,ux nsel,s,loc,x,-0.24 d,all,ux,9.01E-4 allsel,all!*OUTRES,ERASEOUTRES,ALL,ALL!arclen,on,1E-7, !使用弧长法 kbc,0NSUBST,100,2000,20 neqit,15KBC,0nlgeom,1cnvtol,f,0.5,2,0.9allsellswrite,17!*18th step-nsel,s,loc,x,-0.24 ddele,all,

52、ux nsel,s,loc,x,3.36 d,all,ux,-9.01E-4allsel,all!*OUTRES,ERASEOUTRES,ALL,ALL!arclen,on,1E-7, !使用弧长法 kbc,0NSUBST,200,2000,30 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.5,2,0.9 allsel lswrite,18!*19th step-nsel,s,loc,x,3.36 ddele,all,uxnsel,s,loc,x,-0.24d,all,ux,9.01E-4 allsel,all !*OUTRES,ERASEOUTRES,ALL,ALL!

53、arclen,on,1E-7, !使用弧长法 kbc,0NSUBST,200,2000,30 neqit,15KBC,0 nlgeom,1 cnvtol,f,0.5,2,0.9 allsel lswrite,19!*six cycleallsellssolve,1,19,1FINISH混凝土也好，砌体也好，开裂之后 ANSYS 的弧长法很难计算，特别麻烦，这时候只要不是 计算下降段最好还是用NR法。个人感觉，无论是NR法，还是Arc法，都不好得到曲线的下降段。NR法得到的往往是伪 下降段，而Arc法得到的下降段很不可靠。能不能分析要从本构关系的角度来说明,最重要的是本构关系1. 对于混凝土结构

54、而言,ansys的破坏准则是William-Warnke五参数破坏准则,在模型中弯曲的子午线由二次抛物线表达式来描述，偏平面中的非圆迹线用椭圆曲线对 0-60的每个部分予以近似。因此，完整破坏面的表示分为两 个部分：第一，对于 0-60，推导偏斜横截面的椭圆表达式；第二，按二次 抛物线来近似拉伸和压缩子午线，然后将这两条子午线由偏曲线为基准面的 椭球面连接起来。这是 ansys 中定义混凝土破坏面的办法.2. 破坏以后,它采用的是弥散裂缝的处理办法,当某一单元内高斯数值积分点上的拉应力（实 际上是一定区域）或应变超过抗拉强度或极限拉应变时，则认为这一高斯积分点垂直于该主 拉应力方向的混凝土开裂

55、，假设开裂的混凝土仍保持连续，即裂纹是以“分布”方式出现的， 即存在无限多平行的、间距很密且垂直于引起主拉应力（或主拉应变）方向的裂缝。裂隙穿 越开裂的混凝土单元，仍然可用处理连续体介质力学的方法来处理。由于不必增加节点和重 新划分单元，很容易由计算自动进行处理.所以采用ansys分析混凝土的滞回是作不同的,最根本的它不能考虑混凝土的刚度的损伤强 度的衰减,以及捏缩效应,除非你自己进行二次开发,对于 solid65 的材料属性进行了改造.3. 附件是混凝土的各种破坏准则,进行分析之前一定要明白各个软件最基本的原理是什么./prep7et,1,65et,2,8 et,3,solid45r,1r,

56、2,236e-6 r,3,75.4e-6 r,4,151.2e-6 mp,ex,1,2.522e10mp,prxy,1,0.2 mp,dens,1,2500 tb,concr,1,1tbdata,0.5,0.95,5e6,-1 tb,miso,1,1,12 tbtemp,0 tbpt,0.00068,17.15e6 tbpt,0.00136,28.03e6 tbpt,0.00204,33.62e6 tbpt,0.00272,35.96e6 tbpt,0.0034,36.53e6 tbpt,0.00408,36.17e6 tbpt,0.00476,35.35e6 tbpt,0.00544,34.

57、32e6 tbpt,0.00612,33.214e6 tbpt,0.0068,32.1e6 tbpt,0.00748,31.02e6 tbpt,0.00816,29.99e6 mp,ex,2,2.0e11mp,prxy,2,0.28tb,bkin,2tbtemp,0 tbdata,362e6,2e9 mp,EX,3,2.1E11mp,PRXY,3,.27tb,bkin,3tbtemp,0tbdata,272.5e6,2.1e9 local,11,1,0,0,0,0,-90,0csys,11*do,j,1,8n,j,0.185,j*360/8,-0.6*enddo ngen,3,100,1,8,

58、1,0,0,0.3ngen,4,100,201,208,1,0,0,0.15ngen,7,100,501,508,1,0,0,0.3 type,2real,3mat,3*do,i,2,6*do,j,1,7 e,(i-1)do,i,1,11*do,j,1,8 e,(i-1)*100+j,i*100+j*enddo*enddo100+j,(i-1)*100+j+1*enddo e,(i-1)*100+1,(i-1)*100+8*enddotype,2real,4mat,3*do,i,1,5*do,j,1,7 e,(i+5)*100+j,(i+5)*100+j+1*enddoe,(i+5)*100+

59、1,(i+5)*100+8*enddo type,2real,2mat,2allsel,all nummrg,all numcmp,all wpcsys,1,11cylind,0,0.185,-0.6,0,0,360 cylind,0,0.185,0,0.45,0,360cylind,0.185,0.3,-0.6,0,0,360cylind,0,0.185,0.45,2.25,0,360cylind,0,0.185,2.25,2.55,0,360cylind,0.185,0.20,2.25,2.55,0,360cylind,0,0.185,2.55,2.565,0,360wprota,90 vsel,s,1,7,1 vsbw,all numcmp,all wprota,90 vsel,s,1,14,1 vsbw,all numcmp,allvsel,s,1,28,1 vglue,all numcmp,all wprota,-90 wprota,-90 allsel,allvsel,s,1,16,1 vsel,a,21,24,1vatt,1,1,1 vsel,s,17,20,1 vsel,a,25,28,1 vatt,2,1,3allsel,alllsel,s,