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[结构分析] 帮看看我的命令流有何错误?

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发表于 2007-8-23 15:03 | 显示全部楼层 |阅读模式

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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
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 楼主| 发表于 2007-8-23 15:52 | 显示全部楼层
*DIM,SPIN,,4 中应该定义为8,我发错了
 楼主| 发表于 2007-8-25 10:23 | 显示全部楼层
没人能帮帮小弟吗?
发表于 2007-8-26 10:44 | 显示全部楼层
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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