用ANSYS进行滞回分析.docx
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用ANSYS进行滞回分析
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,AREA
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,ux,-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,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
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
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C1
MPTEMP,,,,,,,,
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
LSTR,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
(F8.3)
/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,PLASTIC
TBTEMP,0
TBPT,,6.6e-3,37.23e6
TBPT,,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,0
SECOFFSET,CENT
SECDATA,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
antype,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,2,4,3
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,ALL,
*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,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