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发表于 2003-12-22 20:52:50
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[求助]哪位大侠能帮我解决下图的速度梯度的问题!
我经过一个多星期的摸索终于弄出来了,当然了也少不了流体中文网以前先驱们发过的挨到一点边的贴子啦,谢谢了!APDL传上,供大家讨论
!* 螺棱面上的速度约束
ASEL,S, , ,2
NSLA,S,1
*get,tolmax2,node,,count
*dim,nnum2,,tolmax2,3, !dim an array to restore node
*get,maxnum2,node,,num,max !get the max node number
nnum2(1,1)=maxnum2
nnum2(1,2)=(ny(nnum2(1,1))+rs)*w*cos(sa)
nnum2(1,3)=(ny(nnum2(1,1))+rs)*w*sin(sa)
D,nnum2(1,1),Vx,-nnum2(1,2)
D,nnum2(1,1),Vz,nnum2(1,3)
*do,j2,2,tolmax2
*get,nnum2(j2,1),node,nnum2(j2-1,1),nxtl,
nnum2(j2,2)=(ny(nnum2(j2,1))+rs)*w*cos(sa) !the velocity of x direction
nnum2(j2,3)=(ny(nnum2(j2,1))+rs)*w*sin(sa) !the velocity of z direction
D,nnum2(j2,1),Vx,-nnum2(j2,2)
D,nnum2(j2,1),Vz,nnum2(j2,3)
*enddo
ASEL,S, , ,4
NSLA,S,1
*get,tolmax4,node,,count
*dim,nnum4,,tolmax4,3, !dim an array to restore node
*get,maxnum4,node,,num,max!get the max node number
nnum4(1,1)=maxnum4
nnum4(1,2)=(ny(nnum4(1,1))+rs)*w*cos(sa)
nnum4(1,3)=(ny(nnum4(1,1))+rs)*w*sin(sa)
D,nnum4(1,1),Vx,-nnum4(1,2)
D,nnum4(1,1),Vz,nnum4(1,3)
*do,i4,2,tolmax4
*get,nnum4(i4,1),node,nnum4(i4-1,1),nxtl,
nnum4(i4,2)=(ny(nnum4(i4,1))+rs)*w*cos(sa) !the velocity of x direction
nnum4(i4,3)=(ny(nnum4(i4,1))+rs)*w*sin(sa) !the velocity of z direction
D,nnum4(i4,1),Vx,nnum4(i4,2)
D,nnum4(i4,1),Vz,nnum4(i4,3)
*enddo
allsel
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