梯形水坝的有限元分析.docx

一题目下图为一梯形水坝，载荷沿厚度均匀分布，其尺寸如图所示，厚度弹性模量E=79G 泊松 M=0.3,容重 p=3200.3m6m二划分单元及载荷处理三输入数据1. 基本参数⑴单元数NE=40;⑵节点数NJ=30;⑶支撑数NZ=14;⑷节点载荷数NPJ=7;⑸半宽带DD=14：⑹节点位移数NJ2=NJ*2=60;2. 其他参数⑴问题类型码LXM,LXM=⑵ 弹性常数E,弹性模量EO=79,泊松比p,MU=0.3;⑶ 容重 p,LOU=3200;⑷板厚tJE=l;⑸节点坐标数组AJZ.0 、101.5203304.5406L31.5061.51.571.5381.54.591.55.2510301131.512331334.51434.825154.50164.51.5174.53184.54.519602061.521632264.275237.50247.51.5257.53267.53.7527902891.529*3 >30AJZ(NJ,2)=AJZ(NJ,2)中的元素按节点整体码顺序输入,数组行号为节点整体码，每个节点的坐标值存一行，第一列存x的值.如X广AJZ(2,1) )\ =AJZ(4,2).⑹节点码数组JMJM(NE,3)=4 62 22 72 33 83 44 94 56 116 77 127 88 138 99 149 101111121213131612171318141616171718182020212122141520172118221924212522262224232824 2525 2925 26、26 2778899101012121313141415151717181819191921212222232325252626272729293030234L3678910111213141516171819202122232425262728293031323334353637383930/0JM (NE,3)中的元素按单元输入。

单元号为行号，每个单元的整体码存一行，局部码为 列号，局部码对应的整体码存放在相应的列7)支撑数组NZC.01234NZC(NZ)=56789NZC(NZ)中的元素按支撑对应的位移数从小到大排列⑻节点载荷数组PJ<000、0207637.510339937.5110229687.521PJ(NPJ+1,2+1)=0141487.531075337.541031237.547104900PJ(NPJ,2)中的元素排列原则为：一个节点载荷存一行，同一行中第一列是节点载荷值，第 二列是载荷对应的位移数程序及结果#iiiclude#iiiclude#define NE 40#define NJ 30#define NZ 10#define NPJ 7#define NJ2 60frdefine DD 14mt LXM=1;double E0=79000000000;double MU=0.3;double LOU=3200;double TE=1.0;//节点坐标数组double AJZ[NJ+l][3]={{0,0.0},{0,0,0},{0,L5,0},{0,3,0},{0,4.5,0},{0,6.0},\ {0,0,L5},{0,1.5,1.5},{0,3,1.5},{0,4.5,1.5},{0,5.25,1.5},{003},{0,1.5,3},\ {0,3,3},{0,4.5,3},{0,4.85,3},{0,045},{0,1.5,4.5},{0,3,4.5},{045,4.5},\ {0,0,6},{0,1.5,6},{0,3,6},{0,4.275,6},{0,0,7.5},{0,1.5,7.5},{0,3,7.5},\ {0,3.75,7.5},{0,0,9},{0,1.5,9},{0,3,9}};〃单元节点码数组intJM[NE+l][4]={{0,0,0,0},{0,l,7,6},{0,l,2,7},{0,2,8,7},{0,2,3,8},{0,3,9,8},\ {0,3,4,9},{0,4,10,9},{0,4,5,10},{0,6,12,11},{0,6,7,12},{0,7,13,12},{0,7,8,13},\ (0,8,14,13),{0,8,9,14),(0,9,15,14),(0,9,10,15),(0,11,17,16),(0,11,12,17),\ {0,12,18,17},{0,12,13,18},{0,13,19,18},{0,13,14,19},{0.14,15,19},{0,16,21,20},\ {0,16,17,21},{0,17,22,21},{0,17,18,22},{0』8,23,22},{0,18.19,23},{0,20,25,24},\ {0,20,21,25},{0,21,26,25},{0,21,22,26},{0,22,27,26},{0,22,23,27},{0,24,29,28},\ {0,24,25,29},{0,25,30,29},{0,25,26,30},{0,26,27,30}};mt NZC[NZ+l]={0』,2,3,4,5.6,7,8,9」0};double PJ[NPJ+l][2+l]={{0,0.0},{0,207637.5』},{0.339937.5,11},{0,229687.5,21}, {0,141487.5,31},{0,75337.5,41},{0,31237.5,47},{0,4900,55}};double AE,KZ[NJ2+1][DD+1],P[NJ2+1],S[3+1][6+1],KE[6+1][6+1];int IE,JE.ME;void DUGD(int,int);void main(){int NJ 1 jnaJ J,z,JO,ii jj,h,dh,E,l,zl,dl;double PE,c,SIGl,SIG2,SIG3.PYL.RYL,MAYL,MIYL.CETA;double WY[6+1],YL[3+1];//?-??????DUGD?-????u?9KZ^??6:fbr(i=O ;i<=NJ2; i++)for(j=0 j <=DD:j ++) KZ[i]|j]=0.0;}for(E=l;E<=NE;E++)DUGD(E,3);fbr(i=l;i<=3;i++) (fbr(ii= 1 ;ii<=2;ii++)dh=2*(JM[E][i]-l)+ii;fbr(j=lj<=3J++)for0j=ljjv=2jj++){zl=2*(JM[E][j]-l)+jj;dl=zl-dh+l；if(dl>0)KZ[dh][dl]=KZ[dh][dl]+KE[h][l];}1))}fbr(i=l ;i<=NJ2;i++)P[i]=O.O;if0)for(i=l ;iv=NP J;i++)(j=(mt)PJ[i][2];P[j]=PJ[i][l]；)}if(LOU>0)for(E=l;E<=NE;E++)(DUGD(E.l);PE=-LOU*(AE)*TE/3;?[2*正]=P[2* IE]+PE;P[2*JE]=P[2*JE]+PE;P[2*ME]=P[2*ME]+PE;)}fbr(i=l ;i<=NZ;i++)z=NZC[i];KZ[z][l]=1.0;for(j=2j<=DDj++)KZ[z]|j]=O.O;iRz!=l)if(z>DD) JO=DD;else JO=z;for(j=2j<=J0j++)KZ[z-j+l](j]=O.O;)P[z]=O.O;}NJ1=NJ2.1;for(k=l;k<=NJ 1 ;k++){if(NJ2>k+DD-l)IM=k+DD-l;else IM=NJ2;IN=k+l;fbr(i=IN;i<=DvI ;i++){l=i-k+l;c=KZ[k][l]/KZ[k][l];jn=DD-l+l;for(J=1J<=Jn；j++)(ni』+j・k;KZ[i]|j]=KZ[i]|j]-c*KZ[k][m];P[i]=P 国.c*P[k];))P [NJ2]=P[NJ2]/KZ[NJ2][ 1 ];for(i=NJ 1 ;i>=l ;i 一){if(DD>NJ2-i+l) J0=NJ2.i+l;else JO=DD;for(j=2j<=J0j++){h=j+i-l;P[i]=P[i]-KZ[i][j]*P[h];)P[i]=P[i]/KZ[i][l];)pnntR”\n");VW);pnntffJD Ufor(i= 1 ;i<=NJ ;i++)prmtf(n%2d%-9.6fifor(E=l;E<=NE;E++)(DUGD(E.2);fbr(i=l;iv=3;i++)(for(j=lj<=2j++)(h=2*(i-l)+j;dh=2*(JM[E][i]-l)+j;WY[h]=P[dh];}}fdr(i=l;i<=3;i++)(YL[i]=O;for(j=lj<=6j-H-)YL[i]=YL[i]+S[i][j]*WY|j];1SIG1=YL[1];SIG2=YL ⑵；SIG3=YL[3];PYL=(SIGl+SIG2)/2;RYL=sqrt(pow((SIGl-SIG2)/2.0,2)+pow(SIG3,2));MAYL=PYL+RYL;MIYL=PYL-RYL;if(SIG2==MIYL)CETA=O;elseCETA=90・57.29578。