modal analysis of a gear

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循环对称结构分析

问题描述

finish
/clear
/filname,modal2
/title,modal analysis of a gear

/prep7
et,1,shell63
et,2,solid45
mp,ex,1,2e8
mp,nuxy,1,0.3
mp,dens,1,7.8e-6

r1=30
r2=r1+10
r3=r1+30
r4=r1+50
r5=r1+70
h2=6
h1=h2+2
h3=h2+2
ang=15
ang1=3
ang2=6

k,1,r1
k,2,r4
k,3,r4,,h3
k,4,r3,,h3
k,5,r3,,h2
k,6,r2,,h2
k,7,r2,,h1
k,8,r1,,h1
a,1,2,3,4,5,6,7,8      !生成齿轮径向的横截面
k,10
k,11,,,1
vrot,1,,,,,,10,11,ang  !vrotat命令,将截面绕一轴旋转成体。na1为1表示area1,pax1与pax2为10与11关键点,两个关键点定义所绕的轴,arc为15(角度),通过右手法则判断旋向

csys,1     !柱坐标系
k,50,r5,ang2
k,51,r5,ang-ang2
k,52,(r4+r3)/2,ang-ang1
k,53,(r4+r3)/2,ang1
nums,area,20  !numstr设定area从20开始编号,即生成面时编号为20
a,50,51,52,53

lsel,s,loc,x,r1  !选择柱坐标系半径为r1处的所有线
lsel,r,loc,y     !继续选择其中位于0度处的线
*get,line_num,line,,num,max  !返回之前选中的line编号。这种得到线编号的方法值得学习
vdrag,20,,,,,,line_num  !vdrag相当于沿着轴将面sweep,20是面号
vadd,all    !将所选的体合为一个整体

asel,s,loc,x,r3,r5
asel,r,loc,z,h3
aadd,all     !将所选的面合为一个整体
asel,s,loc,z 
aadd,all
alls   !allsel

smrt,4   !自动(智能)单元大小为4
asel,s,loc,y  
amesh,all

ngen,2,100,all,,,,ang     !生成节点,数目是当前节点的2倍,编号为当前节点编号加100,all表示当前所有节点,ang为旋转的角度
agen,2,all,,,,ang  !生成第二个面,就在刚刚生成的节点处
alls
numm,node   !nummrg,将重复的合并
numm,kp

type,2
mshap,1,3d
vmesh,all

acle,all   !删除选中的面单元与节点

nsel,s,loc,y
cm,low,node         !define low component
nrot,all            !旋转节点坐标系
nsel,s,loc,y,ang
cm,high,node        !define high component
nrot,all            
alls
save,,db

cycgen       !创建第二个扇区
nsel,s,loc,x,r1
d,all,all
alls
finish

/solu
antype,2
modopt,lanb,5
mxpand,5

save,,db
cycsol,0,2,24,'low'  !定义节径与扇区角
finish

/post1
/expand,24,modal,2,,ang  !15*24是360度,每个扇区15度,扩展24个就是360度
/replot
/view,,1,1,1
/ang,1
/rep
set,list,2   
set,first    
pldisp,0     
anmode,10,0.5e-1  
set,next     
anmode,10,0.5e-1
set,next
pldisp,0
anmode,10,0.5e-1
set,next
pldisp,0
anmode,10,0.5e-1
set,next
pldisp,0
anmode,10,0.5e-1
finish

疑问? 如何用cyclic代替以上的cycgen与cycsol宏命令;节点坐标系怎么理解 ,nrotat命令

finish
/clear
/filname,modal2
/title,modal analysis of a gear

/prep7
et,1,shell63
et,2,solid45
mp,ex,1,2e8
mp,nuxy,1,0.3
mp,dens,1,7.8e-6

r1=30
r2=r1+10
r3=r1+30
r4=r1+50
r5=r1+70
h2=6
h1=h2+2
h3=h2+2
ang=15
ang1=3
ang2=6

k,1,r1
k,2,r4
k,3,r4,,h3
k,4,r3,,h3
k,5,r3,,h2
k,6,r2,,h2
k,7,r2,,h1
k,8,r1,,h1
a,1,2,3,4,5,6,7,8      
k,10
k,11,,,1
vrot,1,,,,,,10,11,ang  

csys,1     
k,50,r5,ang2
k,51,r5,ang-ang2
k,52,(r4+r3)/2,ang-ang1
k,53,(r4+r3)/2,ang1
nums,area,20  
a,50,51,52,53

lsel,s,loc,x,r1  
lsel,r,loc,y     
*get,line_num,line,,num,max  
vdrag,20,,,,,,line_num 
vadd,all    

asel,s,loc,x,r3,r5
asel,r,loc,z,h3
aadd,all 
asel,s,loc,z 
aadd,all
alls  

cyclic,24,ang,1

smrt,4   
asel,s,loc,y  
amesh,all

ngen,2,100,all,,,,ang     
agen,2,all,,,,ang 
alls
numm,node   
numm,kp
type,2
mshap,1,3d
vmesh,all
acle,all  

alls
save,,db
nsel,s,loc,x,r1
d,all,all
alls
finish

/solu
antype,2
modopt,lanb,5
mxpand,5
solve
finish

/post1
/cycexpand  
/replot
/view,,1,1,1
/ang,1
/rep
set,list,2   
set,first    
pldisp,0     
anmode,10,0.5e-1  
set,next     
anmode,10,0.5e-1
set,next
pldisp,0
anmode,10,0.5e-1
set,next
pldisp,0
anmode,10,0.5e-1
set,next
pldisp,0
anmode,10,0.5e-1
finish

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转载自blog.csdn.net/qq_37083038/article/details/100133450