现在做了一个谐响应的例子:
/PREP7
ET,1,BEAM3
R,1,0.0025,0.05**4/12,0.05
MP,EX,1,2.01E9
MP,DENS,1,15
Mp,NUXY,1,0.3
K,1
K,2,1
K,3,0.5,0.866
LSTR,1,2
LSTR,1,3
LSTR,2,3
LESIZE,ALL,,,5
LMESH,1
LMESH,2
LMESH,3
FINISH
/SOL
ANTYPE,3
HROPT,FULL
D,1, , , , , ,UX,UY, , , ,
FLST,2,1,1,ORDE,1
FITEM,2,2
!*
/GO
D,2, , , , , ,UY, , , , ,
FLST,2,1,1,ORDE,1
FITEM,2,7
F,7,FX,100
HARFRQ,330,340
NSUBST,100
KBC,0
solve
FINISH
/POST26
NSOL,2,7,U,X,UX_2
STORE,NEW
PLVAR,,2
当
HARFRQ,330,340
NSUBST,100
时,对应于332.0 0.34591E-2
当
HARFRQ,331,333
NSUBST,100
时,对应于332.0 0.864774E-2
为什么相同的频率、NSUBST,由于HARFRQ的不同而结果不同??
希望大家帮帮忙,谢谢了先
/PREP7
ET,1,BEAM3
R,1,0.0025,0.05**4/12,0.05
MP,EX,1,2.01E9
MP,DENS,1,15
Mp,NUXY,1,0.3
K,1
K,2,1
K,3,0.5,0.866
LSTR,1,2
LSTR,1,3
LSTR,2,3
LESIZE,ALL,,,5
LMESH,1
LMESH,2
LMESH,3
FINISH
/SOL
ANTYPE,3
HROPT,FULL
D,1, , , , , ,UX,UY, , , ,
FLST,2,1,1,ORDE,1
FITEM,2,2
!*
/GO
D,2, , , , , ,UY, , , , ,
FLST,2,1,1,ORDE,1
FITEM,2,7
F,7,FX,100
HARFRQ,330,340
NSUBST,100
KBC,0
solve
FINISH
/POST26
NSOL,2,7,U,X,UX_2
STORE,NEW
PLVAR,,2
当
HARFRQ,330,340
NSUBST,100
时,对应于332.0 0.34591E-2
当
HARFRQ,331,333
NSUBST,100
时,对应于332.0 0.864774E-2
为什么相同的频率、NSUBST,由于HARFRQ的不同而结果不同??
希望大家帮帮忙,谢谢了先
以上问题是载荷幅值定义方式导致的。KBC,0(Ramped)表示载荷幅值随频率(子步)逐渐增加,提到的载荷F=100sin(wt),100是逐步加上去的,如果KBC,1(stepped),100在各个频率是保持不变。故KBC取1时应该没有这个问题。