我做的简支梁自振频率问题很大

lee2009521

fini
/clear
/filename,simple-supported beam
/title,simple-supported beam
/prep7
et,1,beam3
r,1,0.08,0.107,0.2
mp,ex,1,3e10
mp,nuxy,1,0.2
mp,dens,1,2500
n,1,0,0,0
n,21,20,0,0
fill,1,21,19
type,1
mat,1
real,1
*do,i,1,20
e,i,i+1
*enddo
fini
/sol
antype,modal
modopt,lanb,6
mxpand,6
d,1,ux
d,1,uy
d,21,uy
solve
/post1
set,list
/eshape,1
set,first
pldisp,0
prnsol,u,y
set,next
pldisp,0
prnsol,u,y
set,next
pldisp,0
prnsol,u,y
set,next
pldisp,0
prnsol,u,y
set,next
pldisp,0
prnsol,u,y
set,next
pldisp,0
prnsol,u,y

振型不对，自振频率与数值解相差也很大，麻烦高手指点。

16443

r,1,0.08,0.107,0.2
mp,ex,1,3e10
mp,nuxy,1,0.2
mp,dens,1,2500
n,1,0,0,0
n,21,20,0,0

你的单位匹配么 ？
20米长的，截面才是r,1,0.08,0.107,0.2。

lee2009521

单位是r,1,0.08平米,0.107米的四次方,0.4米。高写错了不好意思，全部采用国际单位，算出来的一阶振型15.49但是根据欧拉梁的解析解w=n^2*pi^2sqrt(EI/(M(L^4)))，这里的m单位长读质量：m=2500*0.08,得到答案应该是9.81.

[ 本帖最后由 lee2009521 于 2009-9-24 21:52 编辑 ]

jxxansys

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