1、 kna5=knode*1 if (kna5/2*2 .lt. kna5) kna5=kna5+1 knb4=kelem*1 if (knb4/2*2 .lt. knb4) knb4=knb4+1 knb2=kvar1*1 if (knb2/2*2 .lt. knb2) knb2=knb2+1 knb3=kvar1*1 if (knb3/2*2 .lt. knb3) knb3=knb3+1 kna0=1 kna1=kna1+kna0 kna2=kna2+kna1 kna3=kna3+kna2 kna4=kna4+kna3 kna5=kna5+kna4 if (kna5-1.gt.2500000
2、00) thenexceed memory of array iamemory of ia = 250000000memory needed = ,kna5, in prgram start stop 55555 endif knb0=1 knb1=knb1+knb0 knb2=knb2+knb1 knb3=knb3+knb2 knb4=knb4+knb3 if (knb4-1.gt.125000000) thenexceed memory of array ibmemory of ib = 125000000,knb4, call start(knode,kdgof,kcoor,kvar,
3、*kelem,maxt,kvar1,ia(kna0),ia(kna1),ia(kna2), *ia(kna3),ia(kna4),ib(knb0),ib(knb1),ib(knb2), *ib(knb3), *filename) end subroutine start(knode,kdgof,kcoor,kvar, *kelem,maxt,kvar1,u0,u1,u2, *coor,inodvar,nodvar,numcol,lm,node, character*12 filename(20) DIMENSION NODVAR(KDGOF,KNODE),COOR(KCOOR,KNODE),R
4、(3), * U0(KDGOF,KNODE),U1(KDGOF,KNODE),U2(KDGOF,KNODE), * INODVAR(KNODE),node(kelem) DIMENSION NUMCOL(KVAR1),LM(KVAR1) CHARACTER*1 MATERIAL logical filflgC .C . KDGOF NUMBER OF D.O.FC . KNODE NUMBER OF NODESC . INODVAR ID DATAC . NODVAR DENOTE THE EQUATION NUMBER CORRESPONDING THE D.O.FC . U0 U1 U2
5、INITIAL VALUEC . COOR COORDINATESC . NODE ELEMENT NODAL CONNECTION6 FORMAT (1X, 15I4)7 FORMAT (1X,8F9.3)C.OPEN ID file OPEN (1,FILE=,FORM=UNFORMATTED,STATUS=OLD READ (1) NUMNOD,NODDOF,(NODVAR(I,J),I=1,NODDOF),J=1,NUMNOD) CLOSE (1) call chms(kdgof,knode,NODVAR)c WRITE(*,*) NUMNOD =,NUMNOD, NODDOF =,N
6、ODDOFc WRITE (*,*) ID =c WRITE (*,6) (NODVAR(I,J),I=1,NODDOF),J=1,NUMNOD)C. GET THE NATURAL NODAL ORDER DO 12 N=1,KNODE INODVAR(N)=N12 CONTINUEC. OPEN ORDER.NOD FILE AND READ THE NODAL ORDER IF THE FILE EXIST inquire(file=ORDER.NOD,exist=filflg) if (filflg) then READ (1) (INODVAR(I),I=1,NUMNOD) CLOS
7、E(1) WRITE(*,*) NODORDER = WRITE(*,6) (INODVAR(I),I=1,NUMNOD)C. GET NV BY ID NEQ=0 DO 20 JNOD=1,NUMNOD J=INODVAR(JNOD) DO 18 I=1,NODDOF IF (NODVAR(I,J).NE.1) GOTO 18 NEQ = NEQ + 1 NODVAR(I,J) = NEQ18 CONTINUE20 CONTINUE DO 30 JNOD=1,NUMNOD DO 28 I=1,NODDOF IF (NODVAR(I,J).GE.-1) GOTO 28 N = -NODVAR(
8、I,J)-1 NODVAR(I,J) = NODVAR(I,N)28 CONTINUE30 CONTINUEC. OPEN AND WRITE THE NV FILE OPEN(8,STATUS=unknown,FILE= ,FORM= WRITE(8) (NODVAR(I,J),I=1,NODDOF),J=1,NUMNOD) CLOSE(8)c WRITE(*,6) (NODVAR(I,J),I=1,NODDOF),J=1,NUMNOD)C. WRITE THE BOUNDAY CONDITION FILE BFD ACCORDING TO THE DISP0 FILEC.OPEN DISP
9、0 FILE OPEN(1,FILE= READ(1) NUMNOD,NODDOF,(U0(I,J),I=1,NODDOF),J=1,NUMNOD)C.OPEN BFD FILE WRITE(1) (U0(I,J),I=1,NODDOF),J=1,NUMNOD)C. GET THE INITIAL TIME FROM TIME0 FILEC.OPEN TIME0 FileFORMATTED READ(1,*) T0,TMAX,DT TIME = T0 IT = 0 TMAX,DT,TIME,IT =,TMAX,DT,TIME,ITC.OPEN TIME File WRITE(1) TMAX,D
10、T,TIME,ITC.OPEN COOR file READ (1) NUMNOD,NCOOR,(COOR(I,J),I=1,NCOOR),J=1,NUMNOD)COOR =c WRITE(*,7) (COOR(I,J),I=1,NCOOR),J=1,NUMNOD)C. GET THE INITIAL VALUE FROM THE DATA FILES BY PREPROCESSORdisp1 open(16,file=,status=old read(16) numnod,noddof,(U0(J,N),J=1,NODDOF),N=1,NUMNOD) close(16)disp2 read(
11、16) numnod,noddof,(U1(J,N),J=1,NODDOF),N=1,NUMNOD)disp3 read(16) numnod,noddof,(U2(J,N),J=1,NODDOF),N=1,NUMNOD) U0 = c WRITE(*,(6F13.3) (U0(J,N),J=1,NODDOF),N=1,NUMNOD)C WRITE(*,*) U1 = C WRITE(*,) (U1(J,N),J=1,NODDOF),N=1,NUMNOD)C. COMPUTE THE INITIAL VALUE BY BOUND.FOR zo = 0.0d0c DO 321 N=1,NUMNO
12、Dc DO 100 J=1,NCOORc100 R(J) = COOR(J,N)c DO 200 J=1,NODDOFc U0(J,N) = BOUND(R,zo,J)c U1(J,N) = BOUND1(R,zo,J)c U2(J,N) = BOUND2(R,zo,J)c200 CONTINUEc321 CONTINUEC.OPEN AND WRITE THE INITIAL VALUE FILE UNOD WRITE(1) (U0(I,J),J=1,NUMNOD),I=1,NODDOF), * (U1(I,J),J=1,NUMNOD),I=1,NODDOF), * (U2(I,J),J=1
13、,NUMNOD),I=1,NODDOF), * (U0(I,J),J=1,NUMNOD),I=1,NODDOF)c. open IO file open(21,file=formatted read(21, (1a) material read(21,*) numtyp close(21) DO I=1,NEQ NUMCOL(i)=1 ENDDOC.OPEN ELEM0 file OPEN (3,FILE= NUMEL=0 KELEM=0 KEMATE=0 DO 2000 ITYP=1,NUMTYPC.INPUT ENODE READ (3) NUM,NNODE, * (NODE(I-1)*N
14、NODE+J),J=1,NNODE),I=1,NUM)cc WRITE(*,*) NUM =,NUM, NNODE =,NNODENODE =cc WRITE(*,6) (NODE(I-1)*NNODE+J),J=1,NNODE),I=1,NUM) IF (KELEM.LT.NUM*NNODE) KELEM = NUM*NNODE NNE = NNODE IF (MATERIAL.EQ.Y .OR. MATERIAL.EQ.y) THEN READ (3) MMATE,NMATE IF (KEMATE.LT.MMATE*NMATE) KEMATE = MMATE*NMATE NNE = NNE
15、-1 ENDIFMMATE =,MMATE, NMATE =,NMATE DO 1000 NE=1,NUM L=0 DO 700 INOD=1,NNE NODI=NODE(NE-1)*NNODE+INOD) DO 600 IDGF=1,KDGOF INV=NODVAR(IDGF,NODI) IF (INV.LE.0) GOTO 600 L=L+1 LM(L)=INV600 CONTINUE700 CONTINUE NUMEL=NUMEL+1C WRITE (*,*) L,LM =,LC WRITE (*,(1X,15I5) (LM(I),I=1,L) if (l.gt.0) call ACLH
16、(NEQ,NUMCOL,l,lm)1000 continue2000 CONTINUEc CLOSE(1) CLOSE(3) call BCLH(NEQ,NUMCOL) MAXA=NUMCOL(NEQ)C.OPEN SYS File OPEN (2,FILE= WRITE(2) NUMEL,NEQ,NUMTYP,MAXA,KELEM,KEMATE CLOSE (2) OPEN(2,FILE= write(2) (NUMCOL(I),I=1,NEQ) CLOSE(2)c write(*,*) NEQ,NUMCOL=,NEQc write(*,6) (NUMCOL(i),i=1,NEQ) END
17、subroutine chms(kdgof,knode,id) dimension id(kdgof,knode),ms(1000),is(1000) do 1000 k=1,kdgof m = 0 do 800 n=1,knode if (id(k,n).le.-1) id(k,n)=-1 if (id(k,n).le.1) goto 800 j=id(k,n) j0=0 if (m.gt.0) then do i=1,m if (j.eq.ms(i) j0=is(i) enddo if (j0.eq.0) then m=m+1 ms(m)=j is(m)=n id(k,n)=1 else
18、id(k,n)=-j0-1800 continue return SUBROUTINE ACLH(NEQ,NUMCOL,ND,LM) DIMENSION LM(ND),NUMCOL(NEQ) LS=LM(1)+1 DO 100 I=1,ND110 IF(LM(I)-LS) 120,100,100120 LS=LM(I)100 CONTINUE DO 200 I=1,ND II=LM(I) ME=II-LS IF(ME.GT.NUMCOL(II) NUMCOL(II)=ME200 CONTINUE RETURN SUBROUTINE BCLH(NEQ,NUMCOL) DIMENSION NUMC
19、OL(NEQ)C NUMCOL(1) = 1 DO 490 I=2,NEQ490 NUMCOL(I) = NUMCOL(I) + NUMCOL(I-1) + 13. eshell3da.for,Galerkin法求解位移场的主程序 common /cc/ ic(62500000) MAXT=250000000/2/2 read(2) NUMEL,NEQ,NUMTYP,MAXA,KELEM,KEMATE IF (MAXA.GT.MAXT) THENMATRIX A EXCEED CORE MEMERY . ,MAXAREQUIRED CORE MEMERY . ,MAXT STOP 0000 K
20、VAR=KNODE*KDGOF KCOOR=3C KELEM=31250000KNODE,KDGOF,KVAR,KCOOR,KELEM = WRITE(*,(1X,6I7) KNODE,KDGOF,KVAR,KCOOR,KELEM kna1=kdgof*knode*1 if (kna1/2*2 .lt. kna1) kna1=kna1+1 knc1=kdgof*knode*2 knc2=kcoor*knode*2 knc7=kdgof*knode*2 knc3=neq*2 knb1=maxa*2 knb2=maxa*2 kna2=neq*1 if (kna2/2*2 .lt. kna2) kna2=kna2+1 knc6=kemate*2 kna3=kelem*1 if (kna3/2*2 .lt. kna3) kna3=kna3+1 knc8=100000*2 knc5=neq*2 knc4=kdgof*knode*2 if (kna3-1.gt.125000000) then,kna3, in prgram eshell3da if (knb2-1.gt.250000000) thenexc
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