kushou 发表于 2007-8-23 15:03

帮看看我的命令流有何错误？

我用三维实体模型分析一个转子－轴承系统的模态，转子是中空的，内径一致，带一个盘，总算不对，请问我的命令流中哪里错了？谢谢各位
/PREP7
!*
ET,1,MESH200
!*
ET,2,SOLID45
!*
ET,3,COMBIN14   
!*
ET,4,COMBIN14   
!*
KEYOPT,1,1,6
KEYOPT,1,2,0
!*
KEYOPT,3,1,0
KEYOPT,3,2,2
KEYOPT,3,3,0
!*
KEYOPT,4,1,0
KEYOPT,4,2,3
KEYOPT,4,3,0
!*
R,1,10000000, , ,         !弹簧系数
!*
*DIM,SPIN,,4                                                    ! SPIN VELOCITY (RPM)
SPIN(1) = 0.
SPIN(2) = 20000.
SPIN(3) = 50000.
SPIN(4) = 90000.
SPIN(5) = 70000.
SPIN(6) = 90000.
SPIN(7) = 110000.
SPIN(8) = 130000.
!*
MP,EX,1,2.06e11         !弹模
MP,DENS,1,7900      !密度
mp,prxy,1,0.275          !泊松比
K,1,0,0.006,,   
K,2,0,0.0175,,
K,3,0.0555,0.0175,,
K,4,0.0555,0.0198,,
K,5,0.1557,0.0198,,
K,6,0.1557,0.0509,,
K,7,0.1657,0.0509,,
K,8,0.1657,0.025,,
K,9,0.1989,0.025,,
K,10,0.1989,0.033,,
K,11,0.5039,0.033,,
K,12,0.5039,0.0198,,   
K,13,0.5969,0.0198,,   
K,14,0.5969,0.0175,,   
K,15,0.6369,0.0175,,   
K,16,0.6369,0.006,,
K,17,0.0555,0.006,,   
K,18,0.1557,0.006,,
K,19,0.1657,0.006,,
K,20,0.1989,0.006,,
K,21,0.5039,0.006,,
K,22,0.5969,0.006,,
!*
LSTR,1,2
LSTR,2,3
LSTR,3,4
LSTR,4,5
LSTR,5,6
LSTR,6,7
LSTR,7,8
LSTR,8,9
LSTR,9,10
LSTR,10,11
LSTR,11,12
LSTR,12,13
LSTR,13,14
LSTR,14,15
LSTR,15,16
LSTR,16,1
!*
AL,ALL         !创建面
!*
LSTR,3,17
LSTR,5,18
LSTR,8,19
LSTR,9,20
LSTR,12,21
LSTR,14,22
!*
FLST,3,6,4,ORDE,2   
FITEM,3,17
FITEM,3,-22
ASBL,1,P51X            !用线分割面
ESIZE,0.006,0,
LCCAT,3,17
LCCAT,5,18
LCCAT,7,19
LCCAT,9,20
LCCAT,11,21
LCCAT,13,22
!*
TYPE,1
MSHAPE,0,2D
MSHKEY,1
AMESH,ALL
!*
LSEL,S,LCCA   !选择连接生成的线
LDELE,ALL       !删除连接线
ALLSEL,ALL      !选择所有图元
!*
K,40,,,,
K,41,100,,,
!*
TYPE,2
VROTAT,ALL, , , , , ,40, 41,360, ,
VGLUE,ALL
VSWEEP,ALL
!*
N,40001,0.090865, -0.0198, 0
N,40002,0.56202, -0.0198, 0
N,40003,0.090865, 0, 0.0198
N,40004,0.56202, 0, 0.0198
TYPE,3            !创建轴承单元
REAL,1
E,3262,40001
E,3442,40002
TYPE,4
REAL,1
E,2002,40003
E,2182,40004
!*
FINI
/SOLU
D,ALL,UX                                                ! NO TRACTION & NO TORSION
D,40001,ALL
D,40002,ALL
D,40003,ALL
D,40004,ALL
RATIO = 4*ATAN(1)/30
ANTYPE,MODAL
CORIOLIS,ON,,,ON                                       ! CORIOLIS ON IN A STATIONARY REFERENCE FRAME
NBF = 10
MODOPT,QRDAMP,NBF,,,ON
/OUT,SCRATCH.txt
*DO,I,1,8
   OMEGA,SPIN(I)*RATIO
   MXPAND,NBF
   SOLVE
*ENDDO
FINI
/POST1
PRCAMP,,1.,RPM                                          ! PRINT CAMPBELL VALUES FOR SLOPE=1, UNIT= RPM
PLCAMP,,1.,RPM

kushou 发表于 2007-8-23 15:52

*DIM,SPIN,,4 中应该定义为8，我发错了

kushou 发表于 2007-8-25 10:23

没人能帮帮小弟吗？

rodge 发表于 2007-8-26 10:44

不是没人帮，最好是将问题描述清楚，算不对，是怎么样的算不对？结果不可信？还是其它的？
或者你认为真实的结果与算出来的结果差别在哪？
像这样，没说清楚，就放上一堆命令流，说让人看看哪里有问题，这是很让人反感的，至少我会这么觉得。
让人有针对性去帮你解决问题，也可以节省不少时间啊，如果还是没人回答，那就是都不太懂这方面的，或者懂这方面的都没上线。
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