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注册: 2011-02-25
专业: 金属材料的制备科学与跨学

[求助] 平面八节点等参单元的前处理程序（fortran语言）

下面是平面八节点等参单元的前处理程序，没有什么注释，看的我有点懵，尤其对于网格划分这块不太明白。请哪位明白的老师或是同学帮我看看，谢谢！
题目：矩形薄板中含有一半径为a小圆孔，沿对边承受均布压力作用。给出了荷载集度，弹性模量，泊松比，圆孔直径，办款，板长，厚度的数据。
nf1：圆周环向单元划分的份数，nf2：沿圆周径向的层数，nf3：长边除去宽度后划分的份数
!输出节点编号，坐标，约束，节点荷载的程序
program zuobiao
character*64 fname1,fname2
dimension p(20000)
dimension x(1000),y(1000),jz(1000,2),mea(200,9),zhong(100,100,2)
write(*,'(a22)')'input data file:'
read(*,'(a)')fname1
write(*,'(a22)')'output data file:'
read(*,'(a)')fname2
open(1,file=fname1)
open(2,file=fname2,status='new')
read(1,11)a,b,c,tt,q
read(1,12)eo,vo
read(1,13)nf1,nf2,nf3
11 format(5f10.2)
12 format(2f10.2)
13 format(3i3)
close(1)
nm=1
fangle=3.141592654/(4*nf1)
fr=(b-a)/(4*nf2)
do 20 i=1,2*nf1+1
angle=(i-1)*fangle
r=a
nbei=1
do 20 j=1,2*nf2+1
   zhong(j,i,1)=r*cos(angle)
   zhong(j,i,2)=r*sin(angle)
   if(j.gt.nf2*2*nbei/3)nbei=nbei+1
   r=r+nbei*fr
20 continue
zhong(2*nf2+1,nf1+1,1)=b
zhong(2*nf2+1,nf1+1,2)=b
zhong(2*nf2,nf1+1,1)=(b+zhong(2*nf2-1,nf1+1,1))/2.
zhong(2*nf2,nf1+1,2)=(b+zhong(2*nf2-1,nf1+1,2))/2.
do 30 i=1,nf1-1
   zhong(2*nf2+1,i+1,1)=b
   zhong(2*nf2+1,i+1,2)=b*tan(i*fangle)
   zhong(2*nf2,i+1,1)=(b+zhong(2*nf2-1,i+1,1))/2.
   zhong(2*nf2,i+1,2)=(zhong(2*nf2+1,i+1,2)+zhong(2*nf2-1,i+1,2))/2.
zhong(2*nf2+1,2*nf1+1-i,2)=b
zhong(2*nf2+1,2*nf1+1-i,1)=zhong(2*nf2+1,i+1,2)
zhong(2*nf2,2*nf1+1-i,2)=(b+zhong(2*nf2-1,2*nf1+1-i,2))/2.
zhong(2*nf2,2*nf1+1-i,1)=(zhong(2*nf2+1,2*nf1+1-i,1)+zhong(2*nf2-1,2*nf1+1-i,1))/2.
30 continue
nn=1
do 40 i=1,2*nf2+1
   if(i/2.eq.i/2.)then
      k=2
   else
      k=1
   endif
   do 40 j=1,2*nf1+1,k
      x(nn)=zhong(i,j,1)
      y(nn)=zhong(i,j,2)
      nn=nn+1
40 continue
ne=1
do 50 i=1,nf2
  do 50 j=1,nf1
   k1=(3*nf1+2)*(i-1)+j*2-1
   k2=k1+nf1*2-j+2
   k3=k1+3*nf1+2
   mea(ne,1)=k1
   mea(ne,2)=k3
   mea(ne,3)=k3+2
   mea(ne,4)=k1+2
   mea(ne,5)=k2
   mea(ne,6)=k3+1
   mea(ne,7)=k2+1
   mea(ne,8)=k1+1
   ne=ne+1
50 continue
nc=1
do 60 i=0,nf2
   jz(nc,1)=1+i*(3*nf1+2)
   jz(nc,2)=2
   jz(nc+1,1)=2*nf1+1+i*(3*nf1+2)
   jz(nc+1,2)=1
   nc=nc+2
60 continue
do 70 i=0,nf2-1
   jz(nc,1)=2*nf1+2+i*(3*nf1+2)
   jz(nc,2)=2
   jz(nc+1,1)=3*nf1+2+i*(3*nf1+2)
   jz(nc+1,2)=1
   nc=nc+2
70 continue
do 80 i=1,nf1+1
   zhong(1,i,1)=b
   zhong(1,i,2)=zhong(2*nf2+1,i,2)
80 continue
  flong=(c-b)/(2*nf3)
  do 90 i=2,nf3*2+1
   do 90 j=1,nf1+1
   zhong(i,j,1)=b+(i-1)*flong
   zhong(i,j,2)=zhong(1,j,2)
90 continue
nn1=nn
do 100 i=2,2*nf3+1
   if(i/2.eq.i/2.)then
      k=2
   else
      k=1
   endif
   do 100 j=1,nf1+1,k
      x(nn)=zhong(i,j,1)
      y(nn)=zhong(i,j,2)
      nn=nn+1
100 continue
  do 110 i=1,nf3
do 110 j=1,nf1/2
   if(i.eq.1)then
   k4=nn1-2*nf1-1+(j-1)*2
   mea(ne,1)=k4
   mea(ne,4)=k4+2
   mea(ne,8)=k4+1
   mea(ne,5)=nn1+(j-1)
   mea(ne,7)=nn1+j
   mea(ne,2)=nn1+nf1/2+1+(j-1)*2
   mea(ne,6)=nn1+nf1/2+2+(j-1)*2
   mea(ne,3)=nn1+nf1/2+3+(j-1)*2
   ne=ne+1
   else
   k1=(3*nf1/2+2)*(i-1)+j*2-3-nf1+nn1
   k2=k1+nf1-j+2
   k3=k1+3*nf1/2+2
   mea(ne,1)=k1
   mea(ne,2)=k3
   mea(ne,3)=k3+2
   mea(ne,4)=k1+2
   mea(ne,5)=k2
   mea(ne,6)=k3+1
   mea(ne,7)=k2+1
   mea(ne,8)=k1+1
   ne=ne+1
   endif
110 continue
do 120 i=1,nf3
   jz(nc,1)=nn1+(i-1)*(3*nf1/2+2)
   jz(nc,2)=2
   jz(nc+1,1)=nn1+nf1/2+1+(i-1)*(3*nf1/2+2)
   jz(nc+1,2)=2
   nc=nc+2
120 continue
nc=nc-1
ne=ne-1
nn=nn-1
nn1=nn
do 500 i=1,20000
p(i)=0
500 continue
do 130 i=1,nf1
p((nn1-1)*2+1)=p((nn1-1)*2+1)+(y(nn1)-y(nn1-1))*q*0.5
p((nn1-2)*2+1)=p((nn1-2)*2+1)+(y(nn1)-y(nn1-1))*q*0.5
   nn1=nn1-1
130 continue
do 140 i=1,ne
   mea(i,9)=1
140 continue
write(2,460)nn,ne,nm,nc
write(2,465)(i,x(i),y(i),p(i*2-1),p(i*2),i=1,nn)
write(2,470)((jz(i,j),j=1,2),i=1,nc)
write(2,475)(i,(mea(i,j),j=1,9),i=1,ne)
write(2,480)eo,vo,tt
460 format(4i5)
465 format(i5,4f15.6)
470 format(5x,i5,5x,i5)
475 format(5x,i5,5x,9i5)
480 format(3f15.6)
end

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