求教beam188释放自由度问题

题目是这样的，在做第二题的时候，全部采用beam188单元，然后对节点进行自由度释放，可是做不出来。所以求教我这段代码哪里错了
! 六角星型穹顶的几何非线性分析
finish
/clear
/prep7
r1=500
r2=250
b=62.16
a=20
d=10
et,1,beam188
mp,ex,1,3030.0$mp,prxy,1,3030/1096/2-1
sectype,1,beam,csolid
secdata,d/2
csys,1
k,1,,,a+b
k,2,r2,,b
kgen,6,2,7,,,60
k,8,r1,-30
kgen,6,8,8,,,60
csys,0
*do,i,2,6,1
l,i,i+6
l,i,i+7
*enddo
l,7,8
l,7,13
*do,i,2,7,1
l,1,i
*enddo
*do,i,2,6,1
l,i,i+1
*enddo
l,7,2
lesize,all,,,20
lmesh,all
*do,i,1,5
lsel,s,,,2*i+1,2*i+2
lsel,s,,,i+13,i+13
lsel,s,,,i+18,i+19
esll
endrelease,,,ball
*enddo
lsel,s,,,1,2
lsel,s,,,13,13
lsel,s,,,19,24,5
esll
endrelease,,,ball
lsel,s,,,13,18
esll
endrelease,,,ball
allsel

*do,i,8,13,1
dk,i,all
*enddo
finish
/solu
antype,0
nlgeom,1
outres,all,all
nsubst,200
fk,1,fz,-600
arclen,1
solve
finish
/post26
node1=node(kx(1),ky(1),kz(1))
nsol,2,node1,u,z,d1uz
prod,3,2,,,,,,-1,1,1
prod,4,1,,,,,,600,1,1
xvar,3
plvar,4
node2=node(kx(2),ky(2),kz(2))
nsol,2,node2,u,z,dduz
prod,3,2,,,,,,-1,1,1
prod,4,1,,,,,,600,1,1
xvar,3
plvar,4