声振论坛

 找回密码
 我要加入

QQ登录

只需一步,快速开始

查看: 2130|回复: 1

[综合讨论] 关于ansys中转子启动的瞬态响应问题

[复制链接]
发表于 2010-8-28 18:22 | 显示全部楼层 |阅读模式

马上注册,结交更多好友,享用更多功能,让你轻松玩转社区。

您需要 登录 才可以下载或查看,没有账号?我要加入

x


ansys转子启动的瞬态响应在ansys 12.0帮助文件中,有一个转子启动瞬态响应的例子,
主要想请教各位大侠的是:例子中转子的不平衡质量是怎么被加到转轴上的,相关理论依据是什么?(下边命令流中用【】选中的即是不平衡质量力的命令流)

PS:本人感觉这里不平衡质量的等效方法和谐响应中的等效方法好像不一样。

问题描述如下:
The model is a simply supported shaft. A rigid disk is located at 1/3 of its length. A bearing is located at 2/3 of its length. The rotational velocity varies with a constant slope from zero at t = 0 to 5000 rpm at t = 4 s.   
The unbalance mass (0.1g) is located on the disk at a distance of 0.15 m from the center line of the shaft.
命令流如下:
/batch,list
/title, Simply Supported Shaft with Rigid Disk and Bearing
/config,nres,10000
/prep7
! ** parameters
length = 0.4
ro_shaft = 0.01
ro_disk = 0.15
md = 16.47
id = 9.427e-2
ip = 0.1861
kxx = 2.0e+5
kyy = 5.0e+5
beta = 2.e-4
! ** material = steel
mp,ex,1,2.0e+11
mp,nuxy,1,.3
mp,dens,1,7800
! ** elements types
et,1,188
sect,1,beam,csolid
secdata,ro_shaft,20
et,2,21
r,2,md,md,md,id,id,ip
et,3,14,,1
r,3,kxx,betta*kxx
et,4,14,,2
r,4,kyy,beta*kyy
! ** shaft
type,1
secn,1
mat,1
k,1
k,2,,,length
l,1,2
lesize,1,,,9
lmesh,all
! ** disk
type,2
real,2
e,5
! ** bearing
n,21,-0.05,,2*length/3
type,3
real,3
e,8,21
type,4
real,4
e,8,21
! ** constraints
dk,1,ux,,,,uy      
dk,2,ux,,,,uy      
d,all,uz            
d,all,rotz
d,21,all
finish
【! ** transient tabular force (unbalance)
pi = acos(-1)
spin = 5000*pi/30
tinc = 0.5e-3
tend = 4
spindot = spin/tend
nbp = nint(tend/tinc) + 1
unb = 1.e-4
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/10,tinc*10
kbc,0
coriolis,on,,,on
omega,,,spin
f,5,fx,%fxTab%
f,5,fy,%fyTab%
outres,all,all
solve
fini

本帖被以下淘专辑推荐:

回复
分享到:

使用道具 举报

发表于 2013-1-8 16:29 | 显示全部楼层
在两个垂直方向上施加的惯性力而已,这个帮助里的模型时梁单元和集中质量单元建立的,如果实体模型的话不会这么复杂,关于实体模型是如何施加,我也没整明白,期待高手出现啊
您需要登录后才可以回帖 登录 | 我要加入

本版积分规则

QQ|小黑屋|Archiver|手机版|联系我们|声振论坛

GMT+8, 2024-5-4 12:31 , Processed in 0.085224 second(s), 19 queries , Gzip On.

Powered by Discuz! X3.4

Copyright © 2001-2021, Tencent Cloud.

快速回复 返回顶部 返回列表