下面是一个悬臂梁的瞬态分析例子,请结合这个简单例子指导一下。建立一个二维悬臂梁的模型,悬臂端给个初始位移,然后释放掉,0.4秒之后,再在悬臂端加一个竖向力,力的大小为t=0.4秒时悬臂端的速度,求终止时刻为0.8秒的这段时间内悬臂端的竖向位移响应。命令流如下,问题是在荷载步2和3之间进入了时间历程后处理,提取速度,再施加第3荷载步,进行求解,可这个时候第3个荷载步的计算不是在荷载步2的基础上进行的,如何使荷载步2和3求解连续起来,使荷载步2的计算结果为荷载步3计算的初始条件?还有什么更好提取速度的方法?急盼答复,感激不尽!!!
finish
/clear
/prep7
et,1,beam3
mp,ex,1,2.1e11
mp,prxy,1,0.3
mp,dens,1,7800
r,1,1e-4,1e-8/12,0.01
k,1
k,2,1
l,1,2
lesize,1,,,10
lmesh,1
d,1,all
finish
/solu
antype,trans
outres,basic,1
timint,off
solcontrol,off
deltim,4e-3,4e-3,4e-3,off
time,0.04
nsubst,10
d,2,uy,0.2
kbc,1
lswrite,1
timint,on
solcontrol,off
deltim,4e-3,4e-3,4e-3,off
!autots,on
time,0.4
ddele,2,all
nsubst,90
lswrite,2
lssolve,1,2
finish
/post26
nsol,2,7,u,y
plvar,2 !位移
derive,3,2,1,,vy2
plvar,3 !速度
derive,4,3,1,,ay2
plvar,4 !加速度
*DEL,ABCDE
*DIM,ABCDE,,100,3
VGET,ABCDE(1,1),2
vget,ABCDE(1,2),3
vget,ABCDE(1,3),4
*CFOPEN,1,txt
*VWRITE,ABCDE(1,1),ABCDE(1,2),ABCDE(1,3)
(1x,F9.4,1x,F9.4,1x,F9.4)
*CFCLOS
/solu
allsel,all
upcoord,1,off
finish
/solu
antype,trans
outres,basic,1
timint,on
deltim,4e-3,4e-3,4e-3,off
!!autots,on
time,0.8
f,7,fy,-ABCDE(100,2) !所加力大小与速度有关
kbc,0
nsubst,100
lswrite,3
lssolve,3
finish
/post26
nsol,5,7,u,y,yu2
plvar,5
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