用shell63单元分析简支梁振动频率出错！！！

学界雏鹰

我用shell63单元分析一简支梁的横向振动，shell63的关键字设置为keyopt1=2，即指考虑弯曲刚度，在solve时
出错——‘There no active degrees of freedom’！！！但是若将关键字设置为keyopt1=0，即同时考虑弯曲和膜刚度，计算结果没有问题，不知是哪里出了问题？？？？？？请各位专家指教！！！我写的程序如下：

finish
/clear,nostart

keyw,pr_set,1
keyw,pr_struc,1
*set,num,199      !沿梁长度方向划分的单元数量

/prep7
et,1,shell63,2      !keyopt1=2表示只考虑弯曲刚度-bending stiffness
r,1,0.01,0.01,0.01,0.01
mp,ex,1,2.1e11
mp,prxy,1,0.3
mp,dens,1,7800
rectng,0,1,-0.005,0.005
wpstyle,,,,,,,,1
wpstyle,0.05,0.1,-1,1,0.003,0,0,,5
wprota,0,90,0
asbw,1
wpstyle,,,,,,,,0
wpcsys,-1,0
esize,1/num      !单元尺寸
mshape,0,2D
mshkey,1
amesh,all
allsel

d,all,uz,,,,,rotx,roty      !约束面外平动和面内转动自由度
allsel

nsel,s,loc,x,0
nsel,r,loc,y,-0.005
d,all,ux,,,,,uy      !左端简支约束

nsel,s,loc,x,1
nsel,r,loc,y,-0.005
d,all,uy,,,      !右端铰支

allsel
finish

/solu
antype,modal
modopt,lanb,30,0,99999999,,off
mxpand,30,0,99999999
solve
finish

tq092mq22

看看，这东西怎样！！