如何在ANSYS APDL VM17案例中将节点1的UY1位移导出为文本文件

问题说明 #

需要在ANSYS验证手册VM17提供的APDL代码中，将节点1的UY位移导出为文本文件。当前代码仅生成UY位移曲线图，无法找到对应文本文件，需获取曲线数值或导出为文本文件，同时获取结果数值表格。

原APDL代码 #

C*** USING SHELL63 ELEMENTS

ANTYPE,STATIC ! STATIC ANALYSIS

ET,1,SHELL181   !  Changed

R,1,6.350 ! SHELL THICKNESS

MP,EX,1,3102.75

MP,NUXY,1,0.3

:COM CREATE FINITE ELEMENT MODEL

R1 = 2000   !2540 ! SHELL MID-SURFACE RADIUS

L = 200     !254 ! HALF THE LENGTH

PI = 4*ATAN(1) ! VALUE OF PI COMPUTED

THETA = 0.1*180/PI ! 0.1 RADIANS CONVERTED TO DEGREES

CSYS,1 ! CYLINDRICAL CO-ORDINATE SYSTEM

N,1,R1,90 ! NODES 1 AND 2 ARE CREATED AT POINTS

N,2,R1,90,L ! A AND B RESPECTIVELY.

K,1,R1,90

K,2,R1,(90-THETA)

K,3,R1,90,L

K,4,R1,(90-THETA),L

ESIZE,,2 

! TWO DIVISION ALONG THE REGION BOUNDARY

A,1,3,4,2

AMESH,1

NUMMRG,NODE

FINISH

*CREATE,SOLVIT,MAC

/PREP7

:COM APPLY BOUNDARY CONDITIONS

NSEL,S,LOC,Z,0

DSYM,SYMM,Z

NSEL,S,LOC,Y,90

DSYM,SYMM,X

NSEL,S,LOC,Y,(90-THETA)

D,ALL,UX,,,,,UY,UZ

NSEL,ALL

FINISH

:COM SOLUTION PHASE

:COM SINCE THE SOLUTION OUTPUT IS SUBSTANTIAL IT IS DIVERTED TO A

:COM SCRATCH FILE

/OUTPUT,SCRATCH

/SOLUTION

NLGEOM,ON ! LARGE DEFLECTION TURNED ON

OUTRES,,1 ! WRITE SOLUTION ON RESULTS FILE FOR EVERY SUBSTEP

F,1,FY,-250 ! 1/4 TH OF THE TOTAL LOAD APPLIED DUE TO SYMMETRY

NSUBST,30 ! BEGIN WITH 30 SUBSTEPS

ARCLEN,ON,5 ! ARC-LENGTH SOLUTION TECHNIQUE TURNED ON WITH

! MAX. ARC-LENGTH KEPT AT 5 TO COMPUTE AND STORE

! SUFFICIENT INTERMEDIATE SOLUTION INFORMATION

SOLVE

FINISH

/OUTPUT

:COM POSTPROCESSING PHASE

/POST26

NSOL,2,1,U,Y ! STORE UY DISPLACEMENT OF NODE 1

NSOL,3,2,U,Y ! STORE UY DISPLACEMENT OF NODE 2

PROD,4,1,,,LOAD,,,4*250 ! TOTAL LOAD IS 4*250 DUE TO QUARTER SYMMETRY

PROD,5,2,,,,,,-1 ! CHANGE SIGNS OF THE DISPLACEMENT VALUES

PROD,6,3,,,,,,-1

*GET,UY1,VARI,2,EXTREM,VMIN

*GET,UY2,VARI,3,EXTREM,VMIN

PRVAR,2,3,4 ! PRINT STORED INFORMATION

/AXLAB,X, DEFLECTION (MM)

/AXLAB,Y, TOTAL LOAD (N)

/GRID,1

/XRANGE,0,35

/YRANGE,-500,1050

XVAR,5

PLVAR,4 ! PLOT LOAD WITH RESPECT TO -UY OF NODE 1

/NOERASE

!XVAR,6     !Chane

!PLVAR,4 ! PLOT LOAD WITH RESPECT TO -UY OF NODE 2  !Change

/ERASE

*DIM,LABEL,CHAR,2,2

*DIM,VALUE,,2,3

LABEL(1,1) = 'UY @A ','UY @B '

LABEL(1,2) = 'mm ','mm '

*VFILL,VALUE(1,1),DATA,-30,-26

*VFILL,VALUE(1,2),DATA,UY1,UY2

*VFILL,VALUE(1,3),DATA,ABS(UY1/30) ,ABS(UY2/26 )

FINISH

*END

SOLVIT

SAVE,TABLE_1+

解决方案 #

1. 用*VWRITE导出位移与载荷数据到文本文件 #

在/POST26后处理阶段，插入以下代码到PRVAR,2,3,4之后，可将节点1的UY位移、总载荷按子步导出为结构化文本：

! 打开文本文件存储数据
*CFOPEN,NODE1_UY_RESULTS,TXT
! 写入表头
*VWRITE
(1X,'Substep','UY_Node1(mm)','Total_Load(N)')
! 按格式写入子步号、节点1的UY位移、总载荷
*VWRITE,VARI,2,4
(I10,2F12.6)
! 关闭文件
*CFCLOS

2. 重定向PRVAR输出到文本文件 #

原代码中PRVAR,2,3,4会在命令窗口打印数据，通过输出重定向可直接保存到文件：

/POST26
! 将后续输出重定向到指定文本文件
/OUTPUT,UY_TABLE,TXT
PRVAR,2,3,4
! 恢复输出到命令窗口
/OUTPUT

3. 求解阶段直接输出节点位移 #

在/SOLUTION阶段添加OUTPR命令，让ANSYS直接输出每个子步的节点位移到默认输出文件（Jobname.out）：

/SOLUTION
! 输出每个子步的节点位移
OUTPR,NSOL,ALL
! 若仅需节点1的位移，可先筛选节点
NSEL,S,NODE,,1
OUTPR,NSOL,ALL
NSEL,ALL

修改后的完整APDL代码示例 #

以下是整合了*VWRITE导出方法的完整代码：

C*** USING SHELL63 ELEMENTS

ANTYPE,STATIC ! STATIC ANALYSIS

ET,1,SHELL181   !  Changed

R,1,6.350 ! SHELL THICKNESS

MP,EX,1,3102.75

MP,NUXY,1,0.3

:COM CREATE FINITE ELEMENT MODEL

R1 = 2000   !2540 ! SHELL MID-SURFACE RADIUS

L = 200     !254 ! HALF THE LENGTH

PI = 4*ATAN(1) ! VALUE OF PI COMPUTED

THETA = 0.1*180/PI ! 0.1 RADIANS CONVERTED TO DEGREES

CSYS,1 ! CYLINDRICAL CO-ORDINATE SYSTEM

N,1,R1,90 ! NODES 1 AND 2 ARE CREATED AT POINTS

N,2,R1,90,L ! A AND B RESPECTIVELY.

K,1,R1,90

K,2,R1,(90-THETA)

K,3,R1,90,L

K,4,R1,(90-THETA),L

ESIZE,,2 

! TWO DIVISION ALONG THE REGION BOUNDARY

A,1,3,4,2

AMESH,1

NUMMRG,NODE

FINISH

*CREATE,SOLVIT,MAC

/PREP7

:COM APPLY BOUNDARY CONDITIONS

NSEL,S,LOC,Z,0

DSYM,SYMM,Z

NSEL,S,LOC,Y,90

DSYM,SYMM,X

NSEL,S,LOC,Y,(90-THETA)

D,ALL,UX,,,,,UY,UZ

NSEL,ALL

FINISH

:COM SOLUTION PHASE

:COM SINCE THE SOLUTION OUTPUT IS SUBSTANTIAL IT IS DIVERTED TO A

:COM SCRATCH FILE

/OUTPUT,SCRATCH

/SOLUTION

NLGEOM,ON ! LARGE DEFLECTION TURNED ON

OUTRES,,1 ! WRITE SOLUTION ON RESULTS FILE FOR EVERY SUBSTEP

F,1,FY,-250 ! 1/4 TH OF THE TOTAL LOAD APPLIED DUE TO SYMMETRY

NSUBST,30 ! BEGIN WITH 30 SUBSTEPS

ARCLEN,ON,5 ! ARC-LENGTH SOLUTION TECHNIQUE TURNED ON WITH

! MAX. ARC-LENGTH KEPT AT 5 TO COMPUTE AND STORE

! SUFFICIENT INTERMEDIATE SOLUTION INFORMATION

SOLVE

FINISH

/OUTPUT

:COM POSTPROCESSING PHASE

/POST26

NSOL,2,1,U,Y ! STORE UY DISPLACEMENT OF NODE 1

NSOL,3,2,U,Y ! STORE UY DISPLACEMENT OF NODE 2

PROD,4,1,,,LOAD,,,4*250 ! TOTAL LOAD IS 4*250 DUE TO QUARTER SYMMETRY

PROD,5,2,,,,,,-1 ! CHANGE SIGNS OF THE DISPLACEMENT VALUES

PROD,6,3,,,,,,-1

*GET,UY1,VARI,2,EXTREM,VMIN

*GET,UY2,VARI,3,EXTREM,VMIN

PRVAR,2,3,4 ! PRINT STORED INFORMATION

! --- 添加数据导出代码 ---
*CFOPEN,NODE1_UY_RESULTS,TXT
*VWRITE
(1X,'Substep','UY_Node1(mm)','Total_Load(N)')
*VWRITE,VARI,2,4
(I10,2F12.6)
*CFCLOS
! --- 导出代码结束 ---

/AXLAB,X, DEFLECTION (MM)

/AXLAB,Y, TOTAL LOAD (N)

/GRID,1

/XRANGE,0,35

/YRANGE,-500,1050

XVAR,5

PLVAR,4 ! PLOT LOAD WITH RESPECT TO -UY OF NODE 1

/NOERASE

!XVAR,6     !Chane

!PLVAR,4 ! PLOT LOAD WITH RESPECT TO -UY OF NODE 2  !Change

/ERASE

*DIM,LABEL,CHAR,2,2

*DIM,VALUE,,2,3

LABEL(1,1) = 'UY @A ','UY @B '

LABEL(1,2) = 'mm ','mm '

*VFILL,VALUE(1,1),DATA,-30,-26

*VFILL,VALUE(1,2),DATA,UY1,UY2

*VFILL,VALUE(1,3),DATA,ABS(UY1/30) ,ABS(UY2/26 )

FINISH

*END

SOLVIT

SAVE,TABLE_1+

内容的提问来源于stack exchange，提问作者Muhammad Bilal