1、用 ANSYS 进行滞回分析/PREP7 !定义单元类型，实常数，材料特性 ET,1,SHELL143 R,1,12, , , , , MP,EX,1,196784 MP,NUXY,1,0.3 !双线性随动强化模型 TB,BKIN,1,1,2,1 TBDATA,310,600, !定义关键点、线、面 K,1,54,0,0 K,2,-54,0,0 K,3,54,0,1000 K,4,-54,0,1000 A,1,2,4,3 !定义边界荷强迫位移,划分网格 AESIZE,ALL,27, MSHAPE,0,2D MSHKEY,0 CM,_Y,AREA ASEL, , , , 1 CM,_Y1,ARE
2、A CMSEL,S,_Y AMESH,_Y1 *do,i,1,5 D,i,ALL,0 *enddo OUTPR,BASIC,ALL, OUTRES,ALL,ALL, D,46,ux,30 TIME,1 AUTOTS,0 NSUBST,10, , ,1 KBC,0 LSWRITE,01, !第 2 荷载步 D,46,ux,-30 TIME,3 AUTOTS,0 NSUBST,20, , ,1 KBC,0 LSWRITE,02, !第 3 荷载步 D,46,ux,30 TIME,5 AUTOTS,0 NSUBST,20, , ,1 KBC,0 LSWRITE,03, !第 4 荷载步 D,46,u
3、x,-30 TIME,7 AUTOTS,0 NSUBST,20, , ,1 KBC,0 LSWRITE,04,!第 1 荷载步 D,46,ux,40 TIME,1 AUTOTS,0 NSUBST,10, , ,1 KBC,0 LSWRITE,05, !第 2 荷载步 D,46,ux,-40 TIME,3 AUTOTS,0 NSUBST,20, , ,1 KBC,0 LSWRITE,06, !第 3 荷载步 D,46,ux,40 TIME,5 AUTOTS,0 NSUBST,20, , ,1 KBC,0 LSWRITE,07, !第 4 荷载步 D,46,ux,-40 TIME,7 AUTOTS
4、,0 NSUBST,20, , ,1 KBC,0 LSWRITE,08, !求解 FINISH /SOLU LSSOLVE,1,8,1, !画出荷载位移曲线 FINISH /POST26 NSOL,2,46,U,X, RFORCE,3,46,F,X, XVAR,2 PLVAR,3, , , , , , , , , ,回复： 【分享】 用 ANSYS 进行滞回分析请教这位高手，本人也要做一个滞回分析，是个软钢圆柱，而我采用的是实体建模，采用 SOLID45 单元，双线性随动强化，可结果在大位移的情况就是出现蝶形曲线，而实验的情况则是出现一个梭形，跟你所画的图一样，我换了其他可用的 SOLID 单
5、元，但结果还是一样。希望你能给我提些建议。 能给个联系方式吗？ 以下是本人的一个命令流 /prep7 et,1,solid45 mp,ex,1,2.01e5 mp,prxy,1,0.26 TB,BKIN,1,1,2 TBTEMP,0 TBDATA,185,0, !建立单元及划分网格 block,15,15,200 esize,5 vmesh,all !施加底部约束 nsel,s,loc,z,0 d,all,all,0 ALLSEL,ALL !定义加载的位移数组 *dim,disp,6 wlw1=80 disp(1)=0 disp(2)=wlw1 disp(3)=0 disp(4)=-wlw1
6、disp(5)=0 disp(6)=wlw1 !进入 solution 阶段 /solu nlgeom,on sstif,off autots,on outres,all,all outpr,all,all !施加位移荷载 time,0.0001 nsel,s,loc,z,200 nsubst,1,0,0 d,all,ux,disp(1) ALLSEL,ALL solve *do,i,2,6 time,i nsel,s,loc,z,200 d,all,ux,disp(i) nsubst,40,0,0 allsel,all solve *enddo finish那么怎么提取滞回数据呢/PREP7
7、 ET,1,BEAM3 ET,2,COMBIN14 KEYOPT,2,1,0 KEYOPT,2,2,0 KEYOPT,2,3,2 R,1,0.16,0.00213333,0.4, , , , R,2,0.18,0.0054,0.6,0,0,600, R,3, ,5000000, , !阻尼器线性系数 C1MPTEMP, MPTEMP,1,0 MPDATA,EX,1,3e10 MPDATA,PRXY,1,0.2 MPTEMP, MPTEMP,1,0 MPDATA,DENS,1,2500 K,1, K,2,6, K,3,6, K,4,6,6, KPLOT LSTR,1,3 LSTR,3,4 LST
8、R,2,4 LSEL,s,LINE,1,3,2, LATT,1,1,1, , , , LSEL,S,LINE,2 LATT,1,2,1, , , , LSEL,all, LESIZE,ALL,1, , , , , , ,1 LMESH,ALL, /SHRINK,0 /ESHAPE,1.0 /EFACET,1 /RATIO,1,1,1 /REPLOT TYPE, 2 MAT, 1 REAL, 3 ESYS, 0 E,2,14 D,1,all,14,13, FINISH *SET,NT,1001 *SET,DT,0.02 *DIM,AC,NT *VREAD,AC(1),RECORD,TXT (F8
9、.3) /SOLU !模态分析 ANTYPE,2 MODOPT,SUBSP,8 MXPAND,8, , ,1 SOLVE FINI !得到自振频率 1 *GET,FREQ1,MODE,1,FREQ /CONFIG,NRES,20000 /SOLU ANTYPE,TRANS TRNOPT,FULL ALPHAD,2*DAMPRATIO*FREQ1*2*3.1415926 BETAD,2*DAMPRATIO/(FREQ1*2*3.1415926) *DO,I,1,500 ACEL,AC(I),0,0 TIME,I*0.02 OUTRES,ALL,ALL SOLVE *ENDDO FINISH单柱滞
10、回曲线问题（命令流图）建立一个单柱模型，进行位移加载。分别采用了随动强化和等向强化两种强化准则。材料的应力应变曲线（见命令流中）为三折线，均有明显的下降段，但是计算后的柱顶位移柱底剪力滞回曲线上没有发现结构有明显的刚度退化现象，反而呈现一种理想弹塑性的滞回曲线样式，不得其解！ 命令流如下： fini /clear /prep7 n,1, n,16,1.5 n,17,0,1000 fill,1,16 et,1,beam188 mp,ex,1,3E10 mp,nuxy,1,0.167 mp,dens,1,0 !随动强化 TB,KINH,1,1,3,PLASTICTBTEMP,0TBPT,6.6e-
11、3,37.23e6TBPT,0.034,28.034e6 TBPT,0.051,0!等向强化 !TB,MISO,1,1,3!TBTEMP,0.0!TBPT,DEFI,0.001,3E7 !TBPT,DEFI,6.6e-3,37.23e6!TBPT,DEFI,0.034,28.034e6 r,1, SECTYPE,1, BEAM, RECT, pier, 0SECOFFSET, CENTSECDATA,0.25,0.25,10,10,0,0,0,0,0,0 type,1 mat,1 real,1 *do,ii,1,15 e,ii,ii+1,17 *enddo d,1,all /solu anty
12、pe,static nropt,full outpr,all,all outres,all,all *do,tt,1,20,2 time,tt nsubst,10, d,16,tt*0.01,uy !lswrite,tt solve time,tt+1 nsubst,10, d,16,-1*tt*0.01,uy !lswrite,tt+1 solve *enddo !lssolve,1,20,1 save fini 采用随动强化时的滞回曲线如下图： 先把命令流贴一下： /PREP7 K, ,0,0,0, K, ,0,10,0, K, ,60,0,0, K, ,60,10,0, FLST,2,4
13、,3 FITEM,2,2 FITEM,2,1 FITEM,2,3 FITEM,2,4 A,P51X FLST,2,1,5,ORDE,1 FITEM,2,1 VEXT,P51X, , ,0,0,3, /VIEW, 1 ,1,1,1 /ANG, 1 /REP,FAST SAVE ET,1,SOLID45 MPTEMP, MPTEMP,1,0 MPDATA,EX,1,206000 MPDATA,PRXY,1,0.29 TB,BISO,1,1,2, TBTEMP,0 TBDATA,300,12000, /prep7 MSHAPE,0,3D MSHKEY,1 VMESH,all /SOLU DA,3,A
14、LL, *DIM,dis,TABLE,9,1,TIME, , DIS(1,0) = 0,1,2,3,4,5,6,7,8 DIS(1,1) = 0,3,0,-3,0,4,0,-4,0 D,22, , %DIS% , , , ,UZ, , , , , NSUBST,40,0,0 OUTRES,BASIC,-40 TIME,9 /STATUS,SOLU SOLVE FINISH /post26 NSOL,2,22,U,z,Uz RFORCE,3,22,F,z,Fz PROD,3,3, , , , , ,0.001,1,1, VARNAM,3,LOAD PLTIME,0,0 XVAR,2 SPREAD
15、,0 PLCPLX,0 PLVAR,3, , , , , , , , , , /AXLAB,X,displacement(cm) /AXLAB,Y,load(N) 其中定义施加往复位移的命令： *DIM,dis,TABLE,9,1,TIME, , DIS(1,0) = 0,1,2,3,4,5,6,7,8 DIS(1,1) = 0,3,0,-3,0,4,0,-4,0 D,22, , %DIS% , , , ,UZ, , , , , 各位朋友： 我在分析一悬臂板的滞回曲线时，底边固定，顶部节点 X 方向的自由度耦合(编号为 1)，顶部节点采用位移加载，命令流如下。请问我如何得到顶部节点的力(即顶部
16、所有节点的合力)-水平位移( 顶部所有节点耦合后节点位移相同)关系，请指教。 fini /clear /PREP7 ET,1,SHELL63 R,1,3, , , , , , MP,EX,1,2.06E+005 MP,PRXY,1,0.3 TB,BKIN,1,1,2,1 TBDATA,235,3000, k,1 k,2,50 k,3,50,200 k,4,0,200 a,1,2,3,4 TYPE, 1 MAT, 1 REAL, 1 ESIZE,5,0, MSHAPE,0,2D MSHKEY,1 AMESH,1 NSEL,S,LOC,Y,200 CP,1,UX,ALL SAVE FINISH 施
17、加的位移为： 0,5,0,-5,0,10,0,-10,0,15,0,-15,0 提供一个本人做支撑位移控制低周往复荷载下命令流，希望能对你有帮助 /solu ！求解选项设置 ANTYPE,STATIC nlgeom,on pred,off nropt,full,on sstif,off NSUBST,100,10000,10 cnvtol,u,0.03,0 autots,on OUTRES,ALL,1 kbc,0！斜坡荷载 *dim,disp,number *VREAD,disp(1),disp2,txt,IJK,number (E3.0)！读荷载数组 *DO,I,1,NUMBER time,i DK,s_num+1, ,-235*s_length*disp(i)/6/2.06e5, ,0,UX, , , , , ,！位移加载 ACEL,0,9.8,0, SOLVE *ENDDO save
