在转子瞬态分析中设定时间步提前结束的问题
我在做转子瞬态分析的时候，deltim,tinc,tinc/200,tinc*200 其中tinc=4但是运算的时候还没有到4就结束了（3秒就结束），问题就出来了，这样导致我只能查看0秒到3秒的振动情况，3秒到4秒之间 的振动就查看不了问这是怎么回事，应该怎么调参数？是不是和时间步的取值有关？
载荷步终点时间是否设置？对照一下吧，说实话你这个响应曲线也太差了
下面是典型的用模态叠加法进行瞬态动力学分析的输入命令流：!Build the Model
/FILNAM,...! Jobname
/TITLE,...! Title
/PREP7! Enter PREP7
---
---! Generate model
---
FINISH
!Obtain the Modal Solution
/SOLU! Enter SOLUTION
ANTYPE,MODAL! Modal analysis
MODOPT,REDU! Reduced method
M,...! Master DOF
TOTAL,...
D,...! Constraints
SF,...! Element loads
ACEL,...
SAVE
SOLVE
FINISH
!Obtain the Mode Superposition Transient Solution
/SOLU! Re-enter SOLUTION
ANTYPE,TRANS! Transient analysis
TRNOPT,MSUP,...! Mode superposition method
LVSCALE,...! Scale factor for element loads
F,...! Nodal Loads
MDAMP,...! Modal damping ratios
DELTIM,...! Integration time step sizes
LSWRITE! Write first load step (Remember: the first load step
---! is solved statically at time=0.)
---
---! Loads, etc. for 2nd load step
TIME,...! Time at end of second load step
KBC,...! Ramped or stepped loads
OUTRES,...! Results-file data controls
---
LSWRITE! Write 2nd load step (first transient load step)
SAVE
LSSOLVE! Initiate multiple load step solution
FINISH
!Review results of the mode superposition solution
/POST26! Enter POST26
FILE,,RDSP! Results file is Jobname.RDSP
SOLU,...! Store solution summary data
NSOL,...! Store nodal result as a variable
PLVAR,...! Plot variables
PRVAR,...! List variables
FINISH
!Expand the Solution
/SOLU! Re-enter SOLUTION
EXPASS,ON! Expansion pass
NUMEXP,...! No. of solutions to expand; time range
OUTRES,...! Results-file data controls
SOLVE
FINISH
!Review the Results of the Expanded Solution
/POST1
SET,...! Read desired set of results into database
PLDISP,...! Deformed shape
PRRSOL,...! Reaction loads
PLNSOL,...! Contour plot of nodal results
PRERR! Global percent error (a measure of mesh adequacy)
---
---! Other postprocessing as desired
---
FINISH 回复 2 # yejet 的帖子
你好！我加的是离心力，不知道按照你上面的方法怎么来加？下面是我一开始编写的程序，运行之后就出现上面图形的问题，希望能够给予指点~~谢谢
finish
/clear
/prep7
/file,rotordynamic
ro_shaft =0.0095/2
MP,EX,1,2.1E+11
MP,DENS,1,7830
MP,PRXY,1,0.26
!定义单元类型
! ** elements types
et,1,188
sect,1,beam,csolid
secdata, ro_shaft,20
et,2,21 !集中质量单元
r,2,1.4,1.4,1.4,5E-4,5E-4,1E-3
et,5,21 !集中质量单元
r,5,1.15
et,3,14,,1
r,3,90000,48
et,4,14,,2
r,4,90000,48
! ** shaft
type,1
secn,1
mat,1
k,1
k,2,,,0.4
l,1,2
lesize,1,,,10
lmesh,all
! ** disk
type,2
real,2
e,8
type,5
real,5
e,2
! ** bearing
n,60,-0.01,,0.4
type,3
real,3
e,2,60
type,4
real,4
e,2,60
! ** constraints
dk,1,ux,,,,uy
d,all,uz
d,all,rotz
d,60,all
finish
! ** transient tabular force (unbalance)
pi = acos(-1)
spin = 6000*pi/30
tinc = 0.001
tend = 4
spindot = spin/tend
nbp = nint(tend/tinc) + 1
unb =1.4
ro_disk= 5e-5
f0 = unb*ro_disk
*dim,spinTab,table,nbp,,,TIME
*dim,rotTab, table,nbp,,,TIME
*dim,fxTab,table,nbp,,,TIME
*dim,fyTab,table,nbp,,,TIME
*vfill,spinTab(1,0),ramp,0,tinc
*vfill,rotTab(1,0), ramp,0,tinc
*vfill,fxTab(1,0),ramp,0,tinc
*vfill,fyTab(1,0),ramp,0,tinc
tt = 0
*do,iloop,1,nbp
spinVal = spindot*tt
spinTab(iloop,1) = spinVal
spin2 = spinVal**2
rotVal = spindot*tt**2/2
rotTab(iloop,1) = rotVal
sinr = sin(rotVal)
cosr = cos(rotVal)
fxTab(iloop,1)= f0*(-spin2*sinr + spindot*cosr)
fyTab(iloop,1)= f0*( spin2*cosr + spindot*sinr)
tt=tt+tinc
*enddo
fini
! ** transient analysis
/solu
antype,transient
nlgeom,on !! so that the gyroscopic matrix is updated
time,tend
deltim,tinc,tinc/200,tinc*200
kbc,0
coriolis,on,,,on
omega,,,spin
f,8,fx,%fxTab%
f,8,fy,%fyTab%
outres,all,all
solve
fini
回复 2 # yejet 的帖子
说明一下：上面的图是8号点的振幅，也就是盘的振动情况 是用ANSYS做的分析吗？ 回复 5 # hustxyong 的帖子
是的~~~不知道你有没有看出什么问题来呢 woshiaq 发表于 2011-1-13 11:10 static/image/common/back.gif
回复 2 # yejet 的帖子
你好！我加的是离心力，不知道按照你上面的方法怎么来加？下面是我一开始编写的程序 ...
加角速度就行 回复 7 # yejet 的帖子
多谢~~~我试一试
不知道你有没有（或者知道哪里有）类似的例子参考一下
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