! 由於預應力在特徵值計算的過程中會同樣放大,因此需要特別處理
! 在本例中,當豎向力 force<=4.0e4 時,由於預應力的作用,柱子的
! 淨軸力爲拉力,因此無法得到屈曲荷載
! 我的解決方法:迭代,調整 force 大小,使得需要的特徵值屈曲 freq=1.
! 這樣就可以得到屈曲荷載且排除預應力放大幹擾
/com buckling analysis
fini
/CLEAR
/UNITS,SI
! 外荷載,可以取爲 1,4.0e4,4.1e4,113.e4 並比較其區別
!FORCE=4.1e4
FORCE=113.e4 265
/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
作者:XuJC