用ANSYS進(jìn)行滯回分析

/PREP7

!定義單元類型，實(shí)常數(shù)，材料特性

ET,1,SHELL143

R,1,12,,,,,

MP,EX,1,196784

MP,NUXY,1,0.3

!雙線性隨動(dòng)強(qiáng)化模型

TB,BKIN,1,1,2,1

TBDATA,,310,600,,,,

!定義關(guān)鍵點(diǎn)、線、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

!定義邊界荷強(qiáng)迫位移,劃分網(wǎng)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,30TIME,1AUTOTS,0NSUBST,10,,,1KBC,0LSWRITE,01,!第2荷載步D,46,ux,-30TIME,3AUTOTS,0NSUBST,20,,,1KBC,0LSWRITE,02,!第3荷載步D,46,ux,30TIME,5AUTOTS,0NSUBST,20,,,1KBC,0LSWRITE,03,!第4荷載步D,46,ux,-30TIME,7AUTOTS,0NSUBST,20,,,1KBC,0LSWRITE,04,

!第1荷載步

D,46,ux,40

TIME,1

AUTOTS,0

NSUBST,10,,,1

KBC,0

LSWRITE,05,

!第2荷載步

D,46,ux,-40

TIME,3

AUTOTS,0

NSUBST,20,,,1

KBC,0

LSWRITE,06,

!第3荷載步

D,46,ux,40

TIME,5

AUTOTS,0

NSUBST,20,,,1

KBC,0

LSWRITE,07,

!第4荷載步

D,46,ux,-40

TIME,7

AUTOTS,0

NSUBST,20,,,1

KBC,0

LSWRITE,08,

!求解

FINISH

/SOLU

LSSOLVE,1,8,1,

!畫出荷載位移曲線

FINISH

/POST26

NSOL,2,46,U,X,

RFORCE,3,46,F,X,

XVAR,2

PLVAR,3,,,,,,,,,,回復(fù)：【分享】用ANSYS進(jìn)行滯回分析

請(qǐng)教這位高手，本人也要做一個(gè)滯回分析，是個(gè)軟鋼圓柱，而我采用的是實(shí)體建模，采用SOLID45單元，雙線性隨動(dòng)強(qiáng)化，可結(jié)果在大位移的情況就是出現(xiàn)蝶形曲線，而實(shí)驗(yàn)的情況則是出現(xiàn)一個(gè)梭形，跟你所畫的圖一樣，我換了其他可用的SOLID單元，但結(jié)果還是一樣。希望你能給我提些建議。

能給個(gè)聯(lián)系方式嗎？

以下是本人的一個(gè)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,

!建立單元及劃分網(wǎng)格

block,,15,,15,,200

esize,5

vmesh,all

!施加底部約束

nsel,s,loc,z,0

d,all,all,0

ALLSEL,ALL

!定義加載的位移數(shù)組

*dim,disp,,6

wlw1=80

disp(1)=0

disp(2)=wlw1

disp(3)=0

disp(4)=-wlw1

disp(5)=0

disp(6)=wlw1

!進(jìn)入solution階段

/solu

nlgeom,on

sstif,off

autots,on

outres,all,all

outpr,all,all

!施加位移荷載

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那么怎么提取滯回?cái)?shù)據(jù)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,,!阻尼器線性系數(shù)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!模態(tài)分析

ANTYPE,2

MODOPT,SUBSP,8

MXPAND,8,,,1

SOLVE

FINI

!得到自振頻率1

*GET,FREQ1,MODE,1,FREQ

/CONFIG,NRES,20210

/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單柱滯回曲線問題（命令流＋圖）建立一個(gè)單柱模型，進(jìn)行位移加載。分別采用了隨動(dòng)強(qiáng)化和等向強(qiáng)化兩種強(qiáng)化準(zhǔn)則。材料的應(yīng)力應(yīng)變曲線（見命令流中）為三折線，均有明顯的下降段，但是計(jì)算后的柱頂位移－柱底剪力滯回曲線上沒有發(fā)現(xiàn)結(jié)構(gòu)有明顯的剛度退化現(xiàn)象，反而呈現(xiàn)一種理想彈塑性的滯回曲線樣式，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

!隨動(dòng)強(qiáng)化

TB,KINH,1,1,3,PLASTIC

TBTEMP,0

TBPT,,6.6e-3,37.23e6

TBPT,,0.034,28.034e6

TBPT,,0.051,0

!等向強(qiáng)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

采用隨動(dòng)強(qiáng)化時(shí)的滯回曲線如下圖：

先把命令流貼一下：

/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,,300,12021,,,,

/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)

其中定義施加往復(fù)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：

我在分析一懸臂板的滯回曲線時(shí)，底邊固定，頂部節(jié)點(diǎn)X方向的自由度耦合(編號(hào)為1)，頂部節(jié)點(diǎn)采用位移加載，命令流如下。請(qǐng)問我如何得到頂部節(jié)點(diǎn)的力(即頂部所有節(jié)點(diǎn)的合力)-水平位移(頂部所有節(jié)點(diǎn)耦合后節(jié)點(diǎn)位移相同)關(guān)系，請(qǐng)指教。fini/clear/PREP7

ET,1,SHELL63R,1,3,,,,,,MP,EX,1,2.06E+005

MP,PRXY,1,0.3

TB,BKIN,1,1,2,1

TBDATA,,235,3000,,,,k,1k,2,50k,3,50,200k,4,0,200a,1,2,3,4TYPE,

1

MAT,

1REAL,

1ESIZE,5,0,MSHAPE,0,2D

MSHKEY,1AMESH,1NSEL,S,LOC,Y,200CP,1,UX,ALLSAVEFINISH

施加的位移為：0,5,0,-5,0,10,0,-10,0,15,0,-15,0提供一個(gè)本人做支撐位移控制低周往復(fù)荷載下命令流，希望能對(duì)你有幫助/solu！求解選項(xiàng)設(shè)置ANTYPE,STATIC

nlgeom,on

pred,off

nropt,full,,on

sstif,