双层柱面网壳,为何模态分析中自振频率为8hz?
正确的结果为2.8hz.
希望诸位大侠指点,小弟不胜感激!!!!!
/prep7
local,11,1
cscir,11,1
f=8.892
h=2
r=(f*f+380.25)/(2*f)
b=(atan((380.25-f*f)/(39*f)))*180/3.14159
n,1,r,b
n,14,r,180-b
fill,1,14
/view,1,1,1,1
/ang,1
/rep,fast
local,11,
ngen,22,14,1,14,1,,,3
local,11,1
n,309,r-h,(90+12*b)/13,1.5
ngen,13,1,309,309,,,(180-2*b)/13
local,11
ngen,21,13,309,321,1,,,3
et,1,link8
mp,ex,1,2.06e11
mp,dens,1,7890
mp,nuxy,1,0.3
MP,DAMP,1,0.02
r,1,703e-6
type,1
mat,1
real,1
*do,i,1,14,1
e,i,i+14
*enddo
egen,21,14,1,14,1
*do,i,1,13,1
e,i,i+1
*enddo
egen,22,14,295,307,1
r,2,543e-6
type,1
mat,1
real,2
*do,i,309,321,1
e,i,i+13
*enddo
egen,20,13,581,593,1
*do,i,309,320,1
e,i,i+1
*enddo
egen,21,13,841,852,1
r,3,433e-6
type,1
mat,1
real,3
nx=13
ny=21
nnup=308
*do,i,1,ny
*do,j,1,nx
k1=(i-1)*nx+j+nnup
k2=k1-nnup+i-1
e,k1,k2
e,k1,k2+1
e,k1,k2+nx+1
e,k1,k2+nx+2
*enddo
*enddo
allsel,all
acel,,9.82
local,11,1
nsel,s,loc,x,r
local,11
f,all,fy,-1000
local,11,1
nsel,s,loc,z,0
nsel,a,loc,z,63
nsel,a,loc,y,b
nsel,a,loc,y,180-b
local,11
f,all,fy,-500
/REP
local,11,1
nsel,s,loc,x,r-h
local,11
f,all,fy,-500
local,11,1
nsel,s,loc,z,1.5
nsel,a,loc,z,61.5
nsel,a,loc,y,(90+12*b)/13
nsel,a,loc,y,(2250-12*b)/13
local,11,
f,all,fy,-250
allsel,all
GPLOT !进行总绘图
/PBC,ALL, ,1 !显示边界条件及值
/rep
/solu
nsel,s,node,,2,13,1
nsel,a,node,,296,307,1
d,all,uy,0
nsel,all
d,1,all,,,14,13
d,295,all,,,308,13
nsel,s,node,,1,295,14
local,11,1
nsel,a,loc,y,180-b
d,all,ux,0
d,all,uy,0
d,all,uz,0
allsel
fini |