|
由于预应力在特征值计算的过程中会同样放大，因此需要特别处理
! 在本例中，当竖向力force<=4.0e4时，由于预应力的作用，柱子的
! 净轴力为拉力，因此无法得到屈曲荷载
! 我的解决方法：迭代，调整force大小，使得需要的特征值屈曲freq=1.
! 这样就可以得到屈曲荷载且排除预应力放大干扰
/com buckling analysis
fini
/CLEAR
/UNITS,SI
! 外荷载，可以取为1,4.0e4,4.1e4,113.e4并比较其区别
!FORCE=4.1e4
FORCE=113.e4
/PREP7
!*
ET,1,BEAM4
ET,2,LINK10
R,1,0.1*0.2,0.2*0.1**3/12,0.1*0.2**3/12,0.2,0.1, ,
R,2,0.01*0.01,2e-3,
MPTEMP,1,0
MPDATA,EX,1,,200e9
MPDATA,PRXY,1,,0.27
MPDATA,DENS,1,,7800
k,1,
k,2,0,0,1
k,3,0,0,-10
l,1,3
l,1,2
lsel,,,,1,
latt,1,1,1
ALLSEL,ALL
lsel,,,,2,
latt,1,2,2
ALLSEL,ALL
lsel,,,,1,
LESIZE,all,0.3, , , , , , ,1
lsel,,,,2,
LESIZE,all, , ,1 , , , , ,1
ALLSEL,ALL
LMESH,ALL
FINISH
!/ESHAPE,1.0
!*
/SOLU
DK,3, , , ,0,UX,UY,UZ, ROTX,ROTY ,ROTZ ,
DK,2, , , ,0,UX,UY,UZ, , , ,
FK,1,FZ,-FORCE
ANTYPE,0
!NLGEOM,1
PSTRES,ON
SOLVE
FINISH
/SOLUTION
ANTYPE,1
BUCOPT,SUBSP,6,0,0
SUBOPT,0,0,0,0,0,ALL
SOLVE
FINISH
/POST1
PLDISP,0
/USER, 1
/VIEW, 1, 0.460197348251 , -0.540061973684 , 0.704664079717
/ANG, 1, -16.0547547534
/REPLO
SET,NEXT
/REPLOT
/AUTO, 1
!SET,LIST
/REP
|