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c 直交異方性材料の直方体構造物を、直方六面体（ソリッド）要素の c 線形／弾性の剛性行列で解くプログラム。剛性行列は、 c Robert J. Melesh, Strucutural Analysis of Solids, c Jornal of the Structural Division, ASCE, c Vol. 89, No. ST4, 1963, pp. 205-223 c にあるものを用いた。但し、印刷ミスと思われる以下の箇所について、 c p.209: K11(6,5)の+2d'(5,5)を+2d(5,5)に c p.209: K11(6,8)の-4d'(5,5)を-4d(5,5)に c p.209: K22(3,1)の+4d(4,4)を+d(4,4)に c p.210: K33(5,4)の-d'(6,6)を+d'(6,6)に c p.211: K31左(8,1)の2を-2に c p.211: k31左(8,4)の-1を-2に c それぞれ修正した（対応箇所についてはプログラム中にも注意を書いた） c c 02/5/24版 c プログラムを作った人：後藤文彦 c 改造などは自由ですが、もとのプログラムが、 c http://gthmhk.virtualave.net/programoj/tyoku/tyoku.f c であることをプログラム中に明記して戴けると嬉しいです。 c バグの報告は、 c http://gthmhk.virtualave.net/denbin.html c にお願いします。 c program tyoku implicit real*8 (a-h, o-z) dimension bc(2700*3),f(2700*3),e66(6,6), & d(2700*3),ngame(100),nfair(100), & kkaisi(2700),kryou(2700),bu(2700),bv(2700),bw(2700), & lkaisi(2700),lryou(2700),fx(2700),fy(2700),fz(2700), & sm(3,3,8,8),s(2700*3,2700*3),n8(8), & st(2700*3,2700*3),df(2700*3) c c 入力可能な自由度の最大値：nmax nmax=2700*3 c 私の環境（cygwin の g77）だと配列の最大サイズは a(67108863) c つまり 2**26=67108864 未満。 c tyoku.f では自由度×自由度の行列を解くので、 c (2700*3)*(2700*3)=65610000 を最大サイズにしておく。 c 具体的な要素分割例としては、縦×横×軸長が、 c 6*6*75=2700, 8*8*42=2688<2700, 13*13*13=2199<2700 など。 c c 梁の断面の横の長さ:yoko, 縦の長さ:tate read(*,*) yoko read(*,*) tate c 梁の軸長：ziku read(*,*) ziku c a=yoko*tate xi=yoko*tate**3/12.d0 yi=tate*yoko**3/12.d0 c 断面横方向の要素分割数：nx read(*,*) nx c 一要素の横の長さの半分:yoko1 yoko1=yoko/float(nx)/2.d0 c 断面縦方向の要素分割数：ny read(*,*) ny c 一要素の縦の長さの半分：tate1 tate1=tate/float(ny)/2.d0 c 軸方向の要素分割数：nz read(*,*) nz c 一要素の軸長の半分：ziku1 ziku1=ziku/float(nz)/2.d0 c c 要素数：nyou nyou=nx*ny*nz c 節点数：nset nset=(nx+1)*(ny+1)*(nz+1) c 自由度数：nziyuu nziyuu=nset*3 c c ****** 境界条件 ******* do 100 i=1, nziyuu bc(i)=1.d0 100 continue read(*,*) kkasyo do 110 i=1,kkasyo read(*,*) kkaisi(i) read(*,*) kryou(i) read(*,*) bu(i) read(*,*) bv(i) read(*,*) bw(i) do 110 j=kkaisi(i)-1, kryou(i)-1 bc(3*j+1)=bu(i) bc(3*j+2)=bv(i) bc(3*j+3)=bw(i) 110 continue c ****** 載荷条件 ******* do 120 i=1, nziyuu f(i)=0.d0 120 continue read(*,*) lkasyo do 130 i=1,lkasyo read(*,*) lkaisi(i) read(*,*) lryou(i) read(*,*) fx(i) read(*,*) fy(i) read(*,*) fz(i) do 130 j=lkaisi(i)-1, lryou(i)-1 f(3*j+1)=fx(i) f(3*j+2)=fy(i) f(3*j+3)=fz(i) 130 continue c 応力−ひずみ行列：e66 c 最初は、e66の3*3のところに、ひずみ−応力行列を読み込む call zeromx(6,e66) do 140 i=1,3 do 140 j=1,3 read(*,*) e66(i,j) 140 continue c で、ヤング率やポアソン比を書き写してから exx=1.d0/e66(1,1) eyy=1.d0/e66(2,2) ezz=1.d0/e66(3,3) poixy=-e66(1,2)*exx poixz=-e66(1,3)*exx poiyz=-e66(2,3)*eyy c 逆行列を取って、応力−ひずみ行列に変換する call gyaku(e66) c せん断弾性係数を読み込む read(*,*) e66(4,4) read(*,*) e66(5,5) read(*,*) e66(6,6) c 一応、せん断弾性係数も書き写しておく gxy=e66(4,4) gxz=e66(5,5) gyz=e66(6,6) c 荷重増分：dp read(*,*) dp c c 出力条件： c 比較のために出力したい梁−柱理論値：nhikak c nhikak=1: w 方向の伸びを出力 c read(*,*) nhikak c 変位を画面出力したい節点の数：ngamen read(*,*) ngamen c 変位を画面出力したい節点の節点番号：ngame(i) do 200 i=1,ngamen read(*,*) ngame(i) 200 continue c 変位をファイル出力したい節点の数：nfairu read(*,*) nfairu c 変位をファイル出力したい節点の節点番号：nfair(i) do 210 i=1,nfairu read(*,*) nfair(i) 210 continue c c 増分ステップ数：nstep read(*,*) nstep c c ***********データ入力終わり*********** c c c Melesh(1963)の剛性行列をsmに書き込む call melesh(yoko1,tate1,ziku1,e66,sm) c call zeromx(nziyuu,s) c c 全要素のsmを、それぞれの要素の8節点の節点番号ごとに C 割り振ってsに書き込む do 1000 k=1,nz do 1000 j=1,ny do 1000 i=1,nx n8(8)=(nx+1)*(ny+1)*(k-1)+(nx+1)*(j-1)+i n8(4)=n8(8)+1 n8(7)=n8(8)+(nx+1) n8(3)=n8(7)+1 n8(5)=n8(8)+(nx+1)*(ny+1) n8(1)=n8(4)+(nx+1)*(ny+1) n8(6)=n8(7)+(nx+1)*(ny+1) n8(2)=n8(3)+(nx+1)*(ny+1) c c do 1000 l=1,3 do 1000 m=1,3 do 1000 ii=1,8 do 1000 jj=1,8 i24=(n8(ii)-1)*3 j24=(n8(jj)-1)*3 s(i24+l,j24+m)=s(i24+l,j24+m)+sm(l,m,ii,jj) 1000 continue c c c 境界条件を入れる do 1100 i=1,nziyuu do 1100 j=1,nziyuu s(i,j)=bc(i)*s(i,j) s(j,i)=bc(i)*s(j,i) 1100 continue do 1200 i=1,nziyuu if( bc(i).lt.1.d-3) then s(i,i)=1.d0 f(i)=0.d0 end if 1200 continue c p=0.d0 open(8,file='uvw.out') do 2000 l=1,nstep p=p+dp do 2010 i=1,nziyuu df(i)=f(i)*p do 2010 j=1,nziyuu st(i,j)=s(i,j) 2010 continue c c c 剛性方程式を解く call solve(st,d,df,nziyuu,nmax) c c 出力 c do 3000 j=1,ngamen nga=ngame(j)*3-3 write(*,*) '--------------------------------------------------' write(*,*) '節点番号：',ngame(j) goto (511,512,513), nhikak 511 write(*,125) ' f=',p,' PL/EA=',p*ziku/ezz/a goto 599 512 continue tawa1=p*ziku**3/3.d0/ezz/xi tk=10.d0*(1.d0+poiyz)/(12.d0+11.d0*poiyz) tawa2=tawa1+p*ziku/gyz/a/tk write(*,126) ' f=',p,' 梁理論=',tawa1,' せん断考慮=',tawa2 goto 599 513 write(*,125) ' f=',p,' PL^3/3EI=',p*ziku**3/3.d0/ezz/yi goto 599 write(*,115) ' f=',p 599 continue write(*,135) ' u,v,w=',d(nga+1),d(nga+2),d(nga+3) 3000 continue do 3010 j=1,nfairu nga=ngame(j)*3-3 c write(8,*) '--------------------------------------------------' c write(8,*) '節点番号：',nfair(j) write(8,145) p,d(nga+1),d(nga+2),d(nga+3) 3010 continue c 2000 continue close(8) 115 format(a,1pd13.5) 125 format(a,1pd13.5,a,1pd13.5) 126 format(a,1pd13.5,a,1pd13.5,a,1pd13.5) 135 format(a,1p3d13.5) 145 format(1p4d13.5) end c c c ********************************************************** c c ひずみ−応力行列の逆数を取って、 c 応力−ひずみ行列を求める。 c subroutine gyaku(a) implicit real*8(a-h, o-z) dimension a(6,6), b(3,3) det=a(1,1)*a(2,2)*a(3,3)+a(2,1)*a(3,2)*a(1,3) & +a(3,1)*a(1,2)*a(2,3)-a(2,1)*a(1,2)*a(3,3) & -a(3,1)*a(2,2)*a(1,3)-a(1,1)*a(3,2)*a(2,3) b(1,1)=(-1.d0)**(1+1)*(a(2,2)*a(3,3)-a(3,2)*a(2,3)) b(1,2)=(-1.d0)**(1+2)*(a(2,1)*a(3,3)-a(3,1)*a(2,3)) b(1,3)=(-1.d0)**(1+3)*(a(2,1)*a(3,2)-a(3,1)*a(2,2)) b(2,1)=(-1.d0)**(2+1)*(a(1,2)*a(3,3)-a(3,2)*a(1,3)) b(2,2)=(-1.d0)**(2+2)*(a(1,1)*a(3,3)-a(3,1)*a(1,3)) b(2,3)=(-1.d0)**(2+3)*(a(1,1)*a(3,2)-a(3,1)*a(1,2)) b(3,1)=(-1.d0)**(3+1)*(a(1,2)*a(2,3)-a(2,2)*a(1,3)) b(3,2)=(-1.d0)**(3+2)*(a(1,1)*a(2,3)-a(2,1)*a(1,3)) b(3,3)=(-1.d0)**(3+3)*(a(1,1)*a(2,2)-a(2,1)*a(1,2)) do 10 i=1,3 do 10 j=1,3 a(i,j)=b(j,i)/det 10 continue return end c c c 代数方程式を解くサブルーチン c 元は東京大学かどこかの大計のライブラリーか何かを書き換えたもの SUBROUTINE SOLVE (ST,DD,DF,N01,nmax) C C GAUSSIAN ELIMINATION METHOD C IMPLICIT REAL*8 (A-H,O-Z) DIMENSION ST(nmax,N01),DD(N01),DF(N01) c c C FORWARD ELIMINATION DO 100 I=1,N01-1 INEXT=I+1 ST(I,I)=1.D0/ST(I,I) DF(I)=DF(I)*ST(I,I) DO 110 J=INEXT,N01 ST(I,J)=ST(I,J)*ST(I,I) DF(J)=DF(J)-ST(J,I)*DF(I) DO 110 K=INEXT,N01 ST(K,J)=ST(K,J)-ST(K,I)*ST(I,J) 110 CONTINUE 100 CONTINUE C BACKWARD SUBSTITUTION DD(N01)=DF(N01)/ST(N01,N01) DP=DD(N01) DO 120 I=1,N01-1 K=N01-I DD(K)=DF(K) KNEXT=K+1 DO 120 J=KNEXT,N01 DD(K)=DD(K)-ST(K,J)*DD(J) 120 CONTINUE C RETURN END c c c 剛性行列 subroutine melesh(a,b,c,d,sm) implicit real*8(a-h, o-z) dimension d(6,6),sm(3,3,8,8), & s21a(8,8),s21b(8,8),s31a(8,8),s31b(8,8),s32a(8,8),s32b(8,8) c sm に Melesh(1963) の剛性行列そのままを与えて c メインプログラムで節点順に並び替えた剛性行列を s に入れる。 d11=d(1,1)*b**2*c**2 d22=d(2,2)*a**2*c**2 d33=d(3,3)*a**2*b**2 d44=d(4,4)*b**2*c**2 d55=d(5,5)*a**2*b**2 d66=d(6,6)*a**2*b**2 d44d=d(4,4)*a**2*c**2 d55d=d(5,5)*b**2*c**2 d66d=d(6,6)*a**2*c**2 d44dd=c*d(4,4) d55dd=b*d(5,5) d66dd=a*d(6,6) d12=c*d(1,2) d13=b*d(1,3) d23=a*d(2,3) c sm(1,1,1,1)= 4.d0*d11 +4.d0*d44d+4.d0*d55 sm(1,1,2,1)= 2.d0*d11 -4.d0*d44d+2.d0*d55 sm(1,1,3,1)= d11 -2.d0*d44d-2.d0*d55 sm(1,1,4,1)= 2.d0*d11 +2.d0*d44d-4.d0*d55 sm(1,1,5,1)=-4.d0*d11 +2.d0*d44d+2.d0*d55 sm(1,1,6,1)=-2.d0*d11 -2.d0*d44d +d55 sm(1,1,7,1)= -d11 -d44d -d55 sm(1,1,8,1)=-2.d0*d11 +d44d-2.d0*d55 c Melesh(63)ではsm(1,1,1,1)のd44dがd44になっている。 c sm(1,1,2,2)= 4.d0*d11 +4.d0*d44d+4.d0*d55 sm(1,1,3,2)= 2.d0*d11 +2.d0*d44d-4.d0*d55 sm(1,1,4,2)= d11 -2.d0*d44d-2.d0*d55 sm(1,1,5,2)=-2.d0*d11 -2.d0*d44d +d55 sm(1,1,6,2)=-4.d0*d11 +2.d0*d44d+2.d0*d55 sm(1,1,7,2)=-2.d0*d11 +d44d-2.d0*d55 sm(1,1,8,2)= -d11 -d44d -d55 c sm(1,1,3,3)= 4.d0*d11 +4.d0*d44d+4.d0*d55 sm(1,1,4,3)= 2.d0*d11 -4.d0*d44d+2.d0*d55 sm(1,1,5,3)= -d11 -d44d -d55 sm(1,1,6,3)=-2.d0*d11 +d44d-2.d0*d55 sm(1,1,7,3)=-4.d0*d11 +2.d0*d44d+2.d0*d55 sm(1,1,8,3)=-2.d0*d11 -2.d0*d44d +d55 c sm(1,1,4,4)= 4.d0*d11 +4.d0*d44d+4.d0*d55 sm(1,1,5,4)=-2.d0*d11 +d44d-2.d0*d55 sm(1,1,6,4)= -d11 -d44d -d55 sm(1,1,7,4)=-2.d0*d11 -2.d0*d44d +d55 sm(1,1,8,4)=-4.d0*d11 +2.d0*d44d+2.d0*d55 c sm(1,1,5,5)= 4.d0*d11 +4.d0*d44d+4.d0*d55 sm(1,1,6,5)= 2.d0*d11 -4.d0*d44d+2.d0*d55 sm(1,1,7,5)= d11 -2.d0*d44d-2.d0*d55 sm(1,1,8,5)= 2.d0*d11 +2.d0*d44d-4.d0*d55 c Melesh(1963)ではsm(1,1,6,5)とsm(1,1,8,5)のd55がd55dになっている c sm(1,1,6,6)= 4.d0*d11 +4.d0*d44d+4.d0*d55 sm(1,1,7,6)= 2.d0*d11 +2.d0*d44d-4.d0*d55 sm(1,1,8,6)= d11 -2.d0*d44d-2.d0*d55 c sm(1,1,7,7)= 4.d0*d11 +4.d0*d44d+4.d0*d55 sm(1,1,8,7)= 2.d0*d11 -4.d0*d44d+2.d0*d55 c sm(1,1,8,8)= 4.d0*d11 +4.d0*d44d+4.d0*d55 c c sm(2,2,1,1)= 4.d0*d22 +4.d0*d44 +4.d0*d66 sm(2,2,2,1)=-4.d0*d22 +2.d0*d44 +2.d0*d66 sm(2,2,3,1)=-2.d0*d22 +d44 -2.d0*d66 sm(2,2,4,1)= 2.d0*d22 +2.d0*d44 -4.d0*d66 sm(2,2,5,1)= 2.d0*d22 -4.d0*d44 +2.d0*d66 sm(2,2,6,1)=-2.d0*d22 -2.d0*d44 +d66 sm(2,2,7,1)= -d22 -d44 -d66 sm(2,2,8,1)= d22 -2.d0*d44 -2.d0*d66 c Melesh(63)では、sm(2,2,3,1)のd44が4.d0*d44になっている c sm(2,2,3,1)=-2.d0*d22 +4.d0*d44 -2.d0*d66 c sm(2,2,2,2)= 4.d0*d22 +4.d0*d44 +4.d0*d66 sm(2,2,3,2)= 2.d0*d22 +2.d0*d44 -4.d0*d66 sm(2,2,4,2)=-2.d0*d22 +d44 -2.d0*d66 sm(2,2,5,2)=-2.d0*d22 -2.d0*d44 +d66 sm(2,2,6,2)= 2.d0*d22 -4.d0*d44 +2.d0*d66 sm(2,2,7,2)= d22 -2.d0*d44 -2.d0*d66 sm(2,2,8,2)= -d22 -d44 -d66 c sm(2,2,3,3)= 4.d0*d22 +4.d0*d44 +4.d0*d66 sm(2,2,4,3)=-4.d0*d22 +2.d0*d44 +2.d0*d66 sm(2,2,5,3)= -d22 -d44 -d66 sm(2,2,6,3)= d22 -2.d0*d44 -2.d0*d66 sm(2,2,7,3)= 2.d0*d22 -4.d0*d44 +2.d0*d66 sm(2,2,8,3)=-2.d0*d22 -2.d0*d44 +d66 c sm(2,2,4,4)= 4.d0*d22 +4.d0*d44 +4.d0*d66 sm(2,2,5,4)= d22 -2.d0*d44 -2.d0*d66 sm(2,2,6,4)= -d22 -d44 -d66 sm(2,2,7,4)=-2.d0*d22 -2.d0*d44 +d66 sm(2,2,8,4)= 2.d0*d22 -4.d0*d44 +2.d0*d66 c sm(2,2,5,5)= 4.d0*d22 +4.d0*d44 +4.d0*d66 sm(2,2,6,5)=-4.d0*d22 +2.d0*d44 +2.d0*d66 sm(2,2,7,5)=-2.d0*d22 +d44 -2.d0*d66 sm(2,2,8,5)= 2.d0*d22 +2.d0*d44 -4.d0*d66 c sm(2,2,6,6)= 4.d0*d22 +4.d0*d44 +4.d0*d66 sm(2,2,7,6)= 2.d0*d22 +2.d0*d44 -4.d0*d66 sm(2,2,8,6)=-2.d0*d22 +d44 -2.d0*d66 c sm(2,2,7,7)= 4.d0*d22 +4.d0*d44 +4.d0*d66 sm(2,2,8,7)=-4.d0*d22 +2.d0*d44 +2.d0*d66 c sm(2,2,8,8)= 4.d0*d22 +4.d0*d44 +4.d0*d66 c c sm(3,3,1,1)= 4.d0*d33 +4.d0*d55d +4.d0*d66d sm(3,3,2,1)= 2.d0*d33 +2.d0*d55d -4.d0*d66d sm(3,3,3,1)=-2.d0*d33 +d55d -2.d0*d66d sm(3,3,4,1)=-4.d0*d33 +2.d0*d55d +2.d0*d66d sm(3,3,5,1)= 2.d0*d33 -4.d0*d55d +2.d0*d66d sm(3,3,6,1)= d33 -2.d0*d55d -2.d0*d66d sm(3,3,7,1)= -d33 -d55d -d66d sm(3,3,8,1)=-2.d0*d33 -2.d0*d55d +d66d c sm(3,3,2,2)= 4.d0*d33 +4.d0*d55d +4.d0*d66d sm(3,3,3,2)=-4.d0*d33 +2.d0*d55d +2.d0*d66d sm(3,3,4,2)=-2.d0*d33 +d55d -2.d0*d66d sm(3,3,5,2)= d33 -2.d0*d55d -2.d0*d66d sm(3,3,6,2)= 2.d0*d33 -4.d0*d55d +2.d0*d66d sm(3,3,7,2)=-2.d0*d33 -2.d0*d55d +d66d sm(3,3,8,2)= -d33 -d55d -d66d c sm(3,3,3,3)= 4.d0*d33 +4.d0*d55d +4.d0*d66d sm(3,3,4,3)= 2.d0*d33 +2.d0*d55d -4.d0*d66d sm(3,3,5,3)= -d33 -d55d -d66d sm(3,3,6,3)=-2.d0*d33 -2.d0*d55d +d66d sm(3,3,7,3)= 2.d0*d33 -4.d0*d55d +2.d0*d66d sm(3,3,8,3)= d33 -2.d0*d55d -2.d0*d66d c sm(3,3,4,4)= 4.d0*d33 +4.d0*d55d +4.d0*d66d c sm(3,3,5,4)=-2.d0*d33 -2.d0*d55d -d66d sm(3,3,5,4)=-2.d0*d33 -2.d0*d55d +d66d sm(3,3,6,4)= -d33 -d55d -d66d sm(3,3,7,4)= d33 -2.d0*d55d -2.d0*d66d sm(3,3,8,4)= 2.d0*d33 -4.d0*d55d +2.d0*d66d c Melesh(63)ではsm(3,3,5,4)の+d66dは-d66dになっているが、 c 答えが出ないので+d66dに変えた。 c sm(3,3,5,5)= 4.d0*d33 +4.d0*d55d +4.d0*d66d sm(3,3,6,5)= 2.d0*d33 +2.d0*d55d -4.d0*d66d sm(3,3,7,5)=-2.d0*d33 +d55d -2.d0*d66d sm(3,3,8,5)=-4.d0*d33 +2.d0*d55d +2.d0*d66d c sm(3,3,6,6)= 4.d0*d33 +4.d0*d55d +4.d0*d66d sm(3,3,7,6)=-4.d0*d33 +2.d0*d55d +2.d0*d66d sm(3,3,8,6)=-2.d0*d33 +d55d -2.d0*d66d c sm(3,3,7,7)= 4.d0*d33 +4.d0*d55d +4.d0*d66d sm(3,3,8,7)= 2.d0*d33 +2.d0*d55d -4.d0*d66d c sm(3,3,8,8)= 4.d0*d33 +4.d0*d55d +4.d0*d66d c c do 10 n=1,3 do 10 i=1,8 do 15 j=1,i sm(n,n,i,j)=sm(n,n,i,j)/18.d0/a/b/c 15 continue do 10 j=1,i-1 sm(n,n,j,i)=sm(n,n,i,j) 10 continue c c s21a(1,1)=-2.d0 s21a(2,1)= 2.d0 s21a(3,1)= 1.d0 s21a(4,1)=-1.d0 s21a(5,1)=-2.d0 s21a(6,1)= 2.d0 s21a(7,1)= 1.d0 s21a(8,1)=-1.d0 c s21a(1,3)=-1.d0 s21a(2,3)= 1.d0 s21a(3,3)= 2.d0 s21a(4,3)=-2.d0 s21a(5,3)=-1.d0 s21a(6,3)= 1.d0 s21a(7,3)= 2.d0 s21a(8,3)=-2.d0 c s21b(1,1)=-2.d0 s21b(2,1)=-2.d0 s21b(3,1)=-1.d0 s21b(4,1)=-1.d0 s21b(5,1)= 2.d0 s21b(6,1)= 2.d0 s21b(7,1)= 1.d0 s21b(8,1)= 1.d0 c s21b(1,3)= 1.d0 s21b(2,3)= 1.d0 s21b(3,3)= 2.d0 s21b(4,3)= 2.d0 s21b(5,3)=-1.d0 s21b(6,3)=-1.d0 s21b(7,3)=-2.d0 s21b(8,3)=-2.d0 c s31a(1,1)= 2.d0 s31a(2,1)= 1.d0 s31a(3,1)=-1.d0 s31a(4,1)=-2.d0 s31a(5,1)= 2.d0 s31a(6,1)= 1.d0 s31a(7,1)=-1.d0 s31a(8,1)=-2.d0 c Melesh(63)ではs31a(8,1)は2になっているが、i=1,8まで c 足しても0にならないので -2 の間違いだと思う。 c s31a(1,2)= 1.d0 s31a(2,2)= 2.d0 s31a(3,2)=-2.d0 s31a(4,2)=-1.d0 s31a(5,2)= 1.d0 s31a(6,2)= 2.d0 s31a(7,2)=-2.d0 s31a(8,2)=-1.d0 c s31b(1,1)= 2.d0 s31b(2,1)= 1.d0 s31b(3,1)= 1.d0 s31b(4,1)= 2.d0 s31b(5,1)=-2.d0 s31b(6,1)=-1.d0 s31b(7,1)=-1.d0 s31b(8,1)=-2.d0 c s31b(1,2)= 1.d0 s31b(2,2)= 2.d0 s31b(3,2)= 2.d0 s31b(4,2)= 1.d0 s31b(5,2)=-1.d0 s31b(6,2)=-2.d0 s31b(7,2)=-2.d0 s31b(8,2)=-1.d0 c s32a(1,1)=-2.d0 s32a(2,1)=-2.d0 s32a(3,1)= 2.d0 s32a(4,1)= 2.d0 s32a(5,1)=-1.d0 s32a(6,1)=-1.d0 s32a(7,1)= 1.d0 s32a(8,1)= 1.d0 c s32a(1,5)=-1.d0 s32a(2,5)=-1.d0 s32a(3,5)= 1.d0 s32a(4,5)= 1.d0 s32a(5,5)=-2.d0 s32a(6,5)=-2.d0 s32a(7,5)= 2.d0 s32a(8,5)= 2.d0 c s32b(1,1)=-2.d0 s32b(2,1)= 2.d0 s32b(3,1)= 2.d0 s32b(4,1)=-2.d0 s32b(5,1)=-1.d0 s32b(6,1)= 1.d0 s32b(7,1)= 1.d0 s32b(8,1)=-1.d0 c s32b(1,5)=-1.d0 s32b(2,5)= 1.d0 s32b(3,5)= 1.d0 s32b(4,5)=-1.d0 s32b(5,5)=-2.d0 s32b(6,5)= 2.d0 s32b(7,5)= 2.d0 s32b(8,5)=-2.d0 c do 20 i=1,8 s21a(i,2)= s21a(i,1) s21a(i,5)=-s21a(i,1) s21a(i,6)=-s21a(i,1) s21a(i,4)= s21a(i,3) s21a(i,7)=-s21a(i,3) s21a(i,8)=-s21a(i,3) c s21b(i,2)=-s21b(i,1) s21b(i,5)= s21b(i,1) s21b(i,6)=-s21b(i,1) s21b(i,4)=-s21b(i,3) s21b(i,7)= s21b(i,3) s21b(i,8)=-s21b(i,3) c s31a(i,4)= s31a(i,1) s31a(i,5)=-s31a(i,1) s31a(i,8)=-s31a(i,1) s31a(i,3)= s31a(i,2) s31a(i,6)=-s31a(i,2) s31a(i,7)=-s31a(i,2) c s31b(i,4)=-s31b(i,1) s31b(i,5)= s31b(i,1) s31b(i,8)=-s31b(i,1) s31b(i,3)=-s31b(i,2) s31b(i,6)= s31b(i,2) s31b(i,7)=-s31b(i,2) c s32a(i,2)=-s32a(i,1) s32a(i,3)=-s32a(i,1) s32a(i,4)= s32a(i,1) s32a(i,6)=-s32a(i,5) s32a(i,7)=-s32a(i,5) s32a(i,8)= s32a(i,5) c s32b(i,2)= s32b(i,1) s32b(i,3)=-s32b(i,1) s32b(i,4)=-s32b(i,1) s32b(i,6)= s32b(i,5) s32b(i,7)=-s32b(i,5) s32b(i,8)=-s32b(i,5) 20 continue c Melesh(63)ではs31a(8,4)=-1.d0となっているが、これだと c (8,j),j=1,8まで総和したときに0にならない。 c do 30 i=1,8 do 30 j=1,8 sm(2,1,i,j)=( d12*s21a(i,j) + d44dd*s21b(i,j) )/12.d0 sm(3,1,i,j)=( d13*s31a(i,j) + d55dd*s31b(i,j) )/12.d0 sm(3,2,i,j)=( d23*s32a(i,j) + d66dd*s32b(i,j) )/12.d0 30 continue do 40 i=1,8 do 40 j=1,8 sm(1,2,i,j)=sm(2,1,j,i) sm(1,3,i,j)=sm(3,1,j,i) sm(2,3,i,j)=sm(3,2,j,i) 40 continue c return end c subroutine zeromx(n,a) implicit real*8(a-h,o-z) dimension a(n,n) do 10 i=1,n do 10 j=1,n a(i,j)=0.d0 10 continue return end c