KBC settings in ANSYS

 

Now we have an example of harmonic response:
/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 ,,,,,
FLSt, 2, 1, 1, Orde, 1
Fitem, 2, 7
F, Seven, FX, 100
Harfrq, 330,340
Nsubst, 100
KBC, 0
Solve
Finish
/Post26
Nsol, 2,7, U, X, ux_2
Store, new
Plvar, 2

When
Harfrq, 330,340
Nsubst, 100
Corresponding to 332.0 0.34591e-2
When
Harfrq, 331,333
Nsubst, 100
Corresponding to 332.0 0.864774e-2
Why are the results of the same frequency and nsubst different from those of harfrq ??
I hope you can help me. Thank you.

The above problems are caused by the definition of the load amplitude. KBC, 0 (ramped) indicates that the load amplitude increases with the frequency (substep). The load F = 100sin (wt), 100 is gradually added. If KBC, 1 (stepped), 100 remains unchanged at each frequency. Therefore, this problem should not occur when KBC gets 1. From: http://www.baisi.net/thread-734075-1-1.html