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1、有限元课程设计姓名：XXX专业：工程力学081学号：XXXXXXXXX指导老师：XXX2011年6月29日一、题目已知：悬臂梁的材料参数为E=190GPa =0.25 t=10 cm =0 尺寸参数如图所示。要求：1.单元数目不得少于20个。2.采用矩形单元计算求解。3.计算结果并给出变形图、应力分布图、单元划分图。二、力学分析1.题目可以看做是平面应力问题故有LXM=02.单元划分图3.简化成力学模型根据力学模型可以写出支承数组NZC、节点载荷数组PJ 三、程序代码#include #include #define NE 40#define NJ 55#define NZ 10#define
2、 NPJ 14#define NJ2 110#define DD 26int LXM=0;double E0=190000000000;double MU=0.25;double LOU=0.0;double TE=0.1;double AJZNJ+13=0,0,0,0,0,0,0,0.5,0,0,1,0,0,1.5,0,0,2,0,0,2.5,0,0,3,0,0,3.5,0,0,4,0,0,4.5,0,0,5,0,0,0,0.25,0,0.5,0.25,0,1,0.25,0,1.5,0.25,0,2,0.25,0,2.5,0.25,0,3,0.25,0,3.5,0.25,0,4,0.25,0
3、,4.5,0.25,0,5,0.25,0,0,0.5,0,0.5,0.5,0,1,0.5,0,1.5,0.5,0,2,0.5,0,2.5,0.5,0,3,0.5,0,3.5,0.5,0,4,0.5,0,4.5,0.5,0,5,0.5,0,0,0.75,0,0.5,0.75,0,1,0.75,0,1.5,0.75,0,2,0.75,0,2.5,0.75,0,3,0.75,0,3.5,0.75,0,4,0.75,0,4.5,0.75,0,5,0.75,0,0,1,0,0.5,1,0,1,1,0,1.5,1,0,2,1,0,2.5,1,0,3,1,0,3.5,1,0,4,1,0,4.5,1,0,5,
4、1;int JMNE+15=0,0,0,0,0,0,1,2,13,12,0,2,3,14,13,0,3,4,15,14,0,4,5,16,15,0,5,6,17,16,0,6,7,18,17,0,7,8,19,18,0,8,9,20,19,0,9,10,21,20,0,10,11,22,21,0,21,22,33,32,0,20,21,32,31,0,19,20,31,30,0,18,19,30,29,0,17,18,29,28,0,16,17,28,27,0,15,16,27,26,0,14,15,26,25,0,13,14,25,24,0,12,13,24,23,0,23,24,35,34
5、,0,24,25,36,35,0,25,26,37,36,0,26,27,38,37,0,27,28,39,38,0,28,29,40,39,0,29,30,41,40,0,30,31,42,41,0,31,32,43,42,0,32,33,44,43,0,43,44,55,54,0,42,43,54,53,0,41,42,53,52,0,40,41,52,51,0,39,40,51,50,0,38,39,50,49,0,37,38,49,48,0,36,37,48,47,0,35,36,47,46,0,34,35,46,45;int NZCNZ+1=0,1,2,23,24,45,46,67,
6、68,89,90;double PJNPJ+12+1=0,0,0,0,-104166.7,21,0,-125000,43,0,125000,87,0,250000,89,0,500000,91,0,500000,93,0,500000,95,0,500000,97,0,500000,99,0,500000,101,0,500000,103,0,500000,105,0,500000,107,0,354166.7,109;double AE,KZNJ2+1DD+1,PNJ2+1,S3+18+1,KE8+18+1,SZ3+132+1;int IE,JE,ME,PE;void DUGD(int,in
7、t);FILE *fp1,*fp2;void main()int NJ1,k,IN,IM,jn,m,i,j,z,JO,ii,jj,h,dh,E,l,zl,dl,n;double F,c,SIG1,SIG2,SIG3,PYL,RYL,MAYL,MIYL,CETA;double WY8+1,YL3+1;fp1=fopen(data1.doc,wb);fp2=fopen(data2.doc,wb);if(LXM!=0)E0=E0/(1.0-MU*MU);MU=MU/(1.0-MU); for(i=0;i=NJ2;i+)for(j=0;j=DD;j+)KZij=0.0;for(E=1;E=NE;E+)
8、DUGD(E,3);for(i=1;i=4;i+)for(ii=1;ii=2;ii+)h=2*(i-1)+ii;dh=2*(JMEi-1)+ii;for(j=1;j=4;j+)for(jj=1;jj0)KZdhdl=KZdhdl+KEhl;for(i=1;i0)for(i=1;i0)for(E=1;E=NE;E+)DUGD(E,1);F=-LOU*(AE)*TE/4;P2*IE=P2*IE+F;P2*JE=P2*JE+F;P2*ME=P2*ME+F;P2*PE=P2*PE+F;for(i=1;i=NZ;i+)z=NZCi;KZz1=1.0;for(j=2;jDD)JO=DD;elseJO=z;
9、for(j=2;j=JO;j+)KZz-j+1j=0.0;Pz=0.0;NJ1=NJ2-1;for(k=1;kk+DD-1)IM=k+DD-1;elseIM=NJ2;IN=k+1;for(i=IN;i=IM;i+)l=i-k+1;c=KZkl/KZk1;jn=DD-l+1;for(j=1;j=1;i-)if(DD=NJ2-i+1)JO=NJ2-i+1;elseJO=DD;for(j=2;j=JO;j+)h=j+i-1;Pi=Pi-KZij*Ph;Pi=Pi/KZi1;printf(n);printf(JD U Vn);fprintf(fp1,JD U Vn);for(i=1;i=NJ;i+)p
10、rintf(%d %-9.6f %-9.6fn,i,P2*i-1,P2*i);fprintf(fp1,%d %-11.10f %-11.10fn,i,P2*i-1,P2*i);for(E=1;E=NE;E+)DUGD(E,2);for(i=1;i=4;i+)for(j=1;j=2;j+)h=2*(i-1)+j;dh=2*(JMEi-1)+j;WYh=Pdh; for(n=1;n=4;n+) for(i=1;i=3;i+) YLi=0; for(j=1;j=8;j+) YLi=YLi+SZi8*(n-1)+j*WYj; SIG1=YL1; SIG2=YL2; SIG3=YL3; PYL=(SIG1+SIG2)/2; RYL=sqrt(pow(SIG1-SIG2)/2.0,2)+pow(SIG3,2); MAYL=PYL+RYL; MIYL=PYL-RYL; if(SIG2=MIYL) CETA=0; else CETA=90-57.29578*atan2(SIG3,(SIG2-MIYL); printf(n); printf(E=%dn,E); printf(sx=%-9.6f sy=%
