*enddo
!定义单元
*do,i,1,5
*do,j,1,8
k1=(i-1)*9+j
k2=k1+1
k3=k2+9
k4=k3-1
e,k1,k2,k3,k4
*enddo
*enddo
*do,i,6,6
*do,j,1,8
k1=(i-1)*9+j
k2=k1+1
k3=k2+9
k4=k3-1
e,k4,k3,k2,k1
*enddo
*enddo
!定义边界条件,四边简支,采用了等效边界条件
*do,i,1,9
d,i,uz,0
d,i+54,uz,0
*enddo
*do,j,1,7
k=(j-1)*9+1
d,k,uz,0
d,k+8,uz,0
*enddo
d,32,ux,0
d,32,uy,0
!定义荷载
*do,i,1,8
SFE,i,3,PRES,1,7461.25,7461.25,0,0
SFE,i,3,PRES,2,0,0,0,0
SFE,i+40,3,PRES,1,7461.25,7461.25,0,0
SFE,i+40,3,PRES,2,0,0,0,0
*enddo
!输出每个子步的结果
OUTRES,ALL,ALL,
antype,0 !分析类型
!设定大变形，应力刚化
NLGEOM,1
SSTIF,ON
NSUBST,20
!下面用弧长法求解下降段
arclen,ON
arctrm,u,120,32,uz
finish
/solu
SOLVE
/POST26
NSOL,2,32,U,Z,
Xvar,2
PLVAR,1, , , , , , , , , , |
置顶卡
