有限元編程實(shí)驗(yàn)報(bào)告

西北工業(yè)大學(xué)有限元上機(jī)實(shí)驗(yàn)報(bào)告實(shí)驗(yàn)內(nèi)容：懸臂梁有限元分析姓名：班級(jí)：學(xué)號(hào)：指導(dǎo)老師：完成時(shí)間：成績(jī)：實(shí)驗(yàn)?zāi)康?.用有限元法計(jì)算懸臂梁給定點(diǎn)的應(yīng)力和位移。2.通過編制程序進(jìn)行有限元分析更深入掌握有限元的思想和計(jì)算方法。實(shí)驗(yàn)內(nèi)容求解矩形懸臂梁?jiǎn)栴}，分析示例程序，并進(jìn)行調(diào)試檢查；計(jì)算圖示受拉平板點(diǎn)的位移和點(diǎn)的應(yīng)力，已知圖中4mm，1.0mm，a=1.0mm；材料常數(shù)E＝200GPa，0.30，均布載荷10N/mm2。將上述矩形懸臂梁?jiǎn)栴}密化并求解；求解三角形薄板問題；設(shè)有正方形薄板，在沿對(duì)角線頂點(diǎn)上作用有壓力，載荷沿厚度均布且為2kN/m，板厚h=1。試計(jì)算板的應(yīng)力值。將上述三角形問題密化并求解。程序設(shè)計(jì)示例程序的分析與調(diào)試；(1)示例程序是C程序，求解平面應(yīng)力問題，程序主要步驟是：eq\o\ac(○,1)單元離散化：本程序采用三角形單元離散化；eq\o\ac(○,2)挑選差值函數(shù)：本程序采用插值函數(shù)法；eq\o\ac(○,3)單元性質(zhì)分析；eq\o\ac(○,4)列總剛度方程：程序中包括形成總剛度矩陣、、加載荷、位移邊界條件處理（置大數(shù)法）；eq\o\ac(○,5)求解方程組：本程序采用高斯消主元法求解方程組；eq\o\ac(○,6)進(jìn)一步計(jì)算：示例程序包含求解單元應(yīng)力和各結(jié)點(diǎn)應(yīng)力；(2)示例程序重要參數(shù)解釋說明：k[20][20]：總剛度矩陣r[20]：載荷列陣，即各節(jié)點(diǎn)x、y方向所受外載荷ee：彈性模量e1：等效彈性模量vv,：泊松比h：厚度q：外載荷dip[20]：位移列陣即各節(jié)點(diǎn)x、y方向位移str[3][8]：?jiǎn)卧獞?yīng)力矩陣，即各單元σx、σy、τxy，后一個(gè)腳標(biāo)是單元編號(hào)strn[7][10]：節(jié)點(diǎn)應(yīng)力矩陣，后一個(gè)腳標(biāo)是結(jié)點(diǎn)編號(hào)iee:單元個(gè)數(shù)ie[8][3]：結(jié)點(diǎn)編號(hào)，前一個(gè)腳標(biāo)是單元編號(hào)，即每個(gè)單元的i、j、mnpoin：結(jié)點(diǎn)個(gè)數(shù)b[3]：bi、bj、bmc[3]：ci、cj、cmarea：?jiǎn)卧娣einl：受力結(jié)點(diǎn)個(gè)數(shù)ir[20]：受力結(jié)點(diǎn)編號(hào)ind：已知位移結(jié)點(diǎn)個(gè)數(shù)id[20]：已知位移結(jié)點(diǎn)編號(hào)iffix[2]：約束方向，即[10]表示x正方向，[-10]表示x負(fù)方向，[01]表示y正方向，[0-1]表示y負(fù)方向(3)輸入文件說明：eq\o\ac(○,1)INIT.DATA10：結(jié)點(diǎn)數(shù)8：?jiǎn)卧獢?shù)200000.0：彈性模量0.3：泊松比2：受力結(jié)點(diǎn)數(shù)5.0：外載荷大小1.0：厚度2：已知位移結(jié)點(diǎn)個(gè)數(shù)1：已知位移結(jié)點(diǎn)編號(hào)2：已知位移結(jié)點(diǎn)編號(hào)0：1結(jié)點(diǎn)x方向位移0：1結(jié)點(diǎn)y方向位移0：2結(jié)點(diǎn)x方向位移0：2結(jié)點(diǎn)y方向位移eq\o\ac(○,2)COORD.DAT00：1結(jié)點(diǎn)坐標(biāo)01：2結(jié)點(diǎn)坐標(biāo)10：3結(jié)點(diǎn)坐標(biāo)11：4結(jié)點(diǎn)坐標(biāo)20：5結(jié)點(diǎn)坐標(biāo)21：6結(jié)點(diǎn)坐標(biāo)30：7結(jié)點(diǎn)坐標(biāo)31：8結(jié)點(diǎn)坐標(biāo)40：9結(jié)點(diǎn)坐標(biāo)41：10結(jié)點(diǎn)坐標(biāo)eq\o\ac(○,3)IE.DAT132：1單元三個(gè)結(jié)點(diǎn)（逆時(shí)針）234：2單元三個(gè)結(jié)點(diǎn)（逆時(shí)針）354：3單元三個(gè)結(jié)點(diǎn)（逆時(shí)針）456：4單元三個(gè)結(jié)點(diǎn)（逆時(shí)針）576：5單元三個(gè)結(jié)點(diǎn)（逆時(shí)針）678：6單元三個(gè)結(jié)點(diǎn)（逆時(shí)針）798：7單元三個(gè)結(jié)點(diǎn)（逆時(shí)針）8910：8單元三個(gè)結(jié)點(diǎn)（逆時(shí)針）eq\o\ac(○,4)IR.DAT9：受力結(jié)點(diǎn)編號(hào)10：受力結(jié)點(diǎn)編號(hào)eq\o\ac(○,5)IFFIX.DAT10：x方向10：x方向矩形懸臂梁?jiǎn)栴}的密化密化程序#include"stdio.h"intm,n;voidwriteinit(FILE*fp);voidwritecoord(FILE*fp);voidwriteie(FILE*fp);voidwriteir(FILE*fp);voidwriteiffix(FILE*fp);intmain(){ printf("Pleaseenterthelinem&&therown:\n"); scanf("%d%d",&m,&n); FILE*fpinit,*fpcoord,*fpie,*fpir,*fpiffix; fpinit=fopen("INIT.DAT","w"); fpcoord=fopen("COORD.DAT","w"); fpie=fopen("IE.DAT","w"); fpir=fopen("IR.DAT","w"); fpiffix=fopen("IFFIX.DAT","w"); writeinit(fpinit); writecoord(fpcoord); writeie(fpie); writeir(fpir); writeiffix(fpiffix); printf("WriteSuccess!!!\n"); return0;}voidwriteinit(FILE*fp){ intnum_of_point=m*n; intnum_of_unit=(m-1)*(n-1)*2; floattanxinmoliang=200000.0; floatposongbi=0.3; intshoulijiedianshu=n; floatwaizaihe=5.0; floathoudu=1.0; intweiyijiediangeshu=n; fprintf(fp,"%d\n",num_of_point); fprintf(fp,"%d\n",num_of_unit); fprintf(fp,"%.1f\n",tanxinmoliang); fprintf(fp,"%.1f\n",posongbi); fprintf(fp,"%d\n",shoulijiedianshu); fprintf(fp,"%.1f\n",waizaihe); fprintf(fp,"%.1f\n",houdu); fprintf(fp,"%d\n",weiyijiediangeshu); inti,j; for(i=0;i<n;++i) { fprintf(fp,"%d\n",i+1); } j=0; for(i=0;i<n;++i) { fprintf(fp,"%d\n%d",j,j); if(i!=n-1) { fprintf(fp,"\n"); } } fclose(fp);}voidwritecoord(FILE*fp){ inti,j; for(i=0;i<m;++i) { for(j=0;j<n;++j) { fprintf(fp,"%f%f",float(i*4)/float(m-1),float(j)/float(n-1)); if(!(i==m-1&&j==n-1)) { fprintf(fp,"\n"); } } } fclose(fp);}voidwriteie(FILE*fp){ intpoint1,point2,point3; intline,row; for(line=0;line<m-1;++line) { for(row=0;row<n-1;++row) { point1=line*n+row+1; point2=point1+n; point3=point1+1; fprintf(fp,"%d%d%d\n",point1,point2,point3); point1=line*n+row+2; point2=point1+n-1; point3=point2+1; fprintf(fp,"%d%d%d",point1,point2,point3); if(!(line==m-2&&row==n-2)) { fprintf(fp,"\n"); } } } fclose(fp);}voidwriteir(FILE*fp){ inti; inttemp0; for(i=1;i<=n;++i) { temp0=(m-1)*n+i; fprintf(fp,"%d",temp0); if(i!=n) { fprintf(fp,"\n"); } } fclose(fp);}voidwriteiffix(FILE*fp){ inti; inttemp0=1,temp1=0; for(i=1;i<=n;++i) { fprintf(fp,"%d%d",temp0,temp1); if(i!=n) { fprintf(fp,"\n"); } } fclose(fp);}密化后以6*3個(gè)結(jié)點(diǎn)為例，得到的輸入文件為：原輸入文件密化后文件INIT.DATA108200000.00.325.01.02120000INIT.DATA1820200000.00.335.01.03123000000COORD.DAT00011011202130314041COORD.DAT0.0000000.0000000.0000000.5000000.0000001.0000000.8000000.0000000.8000000.5000000.8000001.0000001.6000000.0000001.6000000.5000001.6000001.0000002.4000000.0000002.4000000.5000002.4000001.0000003.2000000.0000003.2000000.5000003.2000001.0000004.0000000.0000004.0000000.5000004.0000001.000000IE.DAT1322343544565766787988910IE.DAT142245253356475578586689710881011811991112101311111314111412121415131614141617141715151718IR.DAT910IR.DAT161718IFFIX.DAT1010IFFIX.DAT101010源程序的修改示例程序修改后//初始化；doublek[20][20],r[20];floatx[10],y[10],ee,e1,vv,h,q,dip[20];floate2,str[3][8],strn[7][10];intiee,ie[8][3];intnpoin;floatb[3],c[3],area;intinl,ir[20],ind,id[20],iffix[2];inti,j,m,nn,n,add;floata,tt;intnmax,nmax1;floataa,bb,smax,smax1;intii;//初始化；doublek[36][36],r[36];floatx[18],y[18],ee,e1,vv,h,q,dip[36];floate2,str[3][20],strn[7][18];intiee,ie[20][3];intnpoin;floatb[3],c[3],area;intinl,ir[36],ind,id[36],iffix[3];inti,j,m,nn,n,add;floata,tt;intnmax,nmax1;floataa,bb,smax,smax1;intii;求解三角形薄板問題；由于兩個(gè)對(duì)角線均為該板的對(duì)稱軸（圖3-1（a）），所以只需要去薄板的四分之一作為計(jì)算對(duì)象（圖3-1（b））。根據(jù)板的受力情況，應(yīng)按平面應(yīng)力問題處理。、對(duì)示例程序修改示例程序修改后//初始化；doublek[20][20],r[20];floatx[10],y[10],ee,e1,vv,h,q,dip[20];floate2,str[3][8],strn[7][10];intiee,ie[8][3];intnpoin;floatb[3],c[3],area;intinl,ir[20],ind,id[20],iffix[2];inti,j,m,nn,n,add;floata,tt;intnmax,nmax1;floataa,bb,smax,smax1;intii;//初始化；doublek[12][12],r[12];floatx[6],y[6],ee,e1,vv,h,q,dip[12];floate2,str[3][4],strn[7][6];intiee,ie[4][3];intnpoin;floatb[3],c[3],area;intinl,ir[12],ind,id[12],iffix[2];inti,j,m,nn,n,add;floata,tt;intnmax,nmax1;floataa,bb,smax,smax1;intii;//位移邊界條件;printf("thenumberoftheKNOWNdisplacements:\n");fscanf(fp1,"%i",&ind);printf("nodeNUMBERoftheKNOWNdisplacements:\n");for(i=0;i<ind;i++){fscanf(fp1,"%i",&id[i]);}printf("pleaseinputtheKNOWNdisplacement:\n");for(i=0;i<ind;i++){fscanf(fp1,"%f",&dip[2*id[i]-2]);fscanf(fp1,"%f",&dip[2*id[i]-1]);//printf("dip[id[i]]:%f,%f",dip[2*id[i]-1],dip[2*id[i]]);}//位移邊界條件;printf("thenumberoftheKNOWNdisplacements:\n");fscanf(fp1,"%i",&ind);printf("nodeNUMBERoftheKNOWNdisplacements:\n");for(i=0;i<ind;i++){fscanf(fp1,"%i",&id[i]);}printf("pleaseinputtheKNOWNdisplacement:\n");for(i=0;i<(ind+1)/2;i++){fscanf(fp1,"%f",&dip[2*id[i]-2]);}for(i=(ind-1)/2;i<ind;i++){fscanf(fp1,"%f",&dip[2*id[i]-1]);//printf("dip[id[i]]:%f,%f",dip[2*id[i]-1],dip[2*id[i]]);}//置大數(shù)法；for(i=0;i<ind;i++){k[2*id[i]-2][2*id[i]-2]=1.0e+20;k[2*id[i]-1][2*id[i]-1]=1.0e+20;r[2*id[i]-2]=dip[2*id[i]-2]*1.0e+20;r[2*id[i]-1]=dip[2*id[i]-1]*1.0e+20;}//置大數(shù)法；for(i=0;i<(ind+1)/2;i++){k[2*id[i]-2][2*id[i]-2]=1.0e+20;r[2*id[i]-2]=dip[2*id[i]-2]*1.0e+20;}for(i=(ind-1)/2;i<ind;i++){k[2*id[i]-1][2*id[i]-1]=1.0e+20;r[2*id[i]-1]=dip[2*id[i]-1]*1.0e+20;}、輸入文件INIT.DATA6410111.0512456000000COORD.DAT020111001020IE.DAT123245253356IR.DAT1IFFIX.DAT0-1三角形問題的密化密化程序#include"stdio.h"intm;voidwriteinit(FILE*fp);voidwritecoord(FILE*fp);voidwriteie(FILE*fp);voidwriteir(FILE*fp);voidwriteiffix(FILE*fp);intmain(){ printf("Pleaseenterthem\n"); scanf("%d",&m); FILE*fpinit,*fpcoord,*fpie,*fpir,*fpiffix; fpinit=fopen("INIT.DAT","w"); fpcoord=fopen("COORD.DAT","w"); fpie=fopen("IE.DAT","w"); fpir=fopen("IR.DAT","w"); fpiffix=fopen("IFFIX.DAT","w"); writeinit(fpinit); writecoord(fpcoord); writeie(fpie); writeir(fpir); writeiffix(fpiffix); printf("WriteSuccess!!!\n"); return0;}voidwriteinit(FILE*fp){ intnum_of_point=m*(m+1)/2; intnum_of_unit=(m-1)*(m-1); floattanxinmoliang=1; floatposongbi=0; intshoulijiedianshu=1; floatwaizaihe=1.0; floathoudu=1.0; intweiyijiediangeshu=2*m-1; fprintf(fp,"%d\n",num_of_point); fprintf(fp,"%d\n",num_of_unit); fprintf(fp,"%.1f\n",tanxinmoliang); fprintf(fp,"%.1f\n",posongbi); fprintf(fp,"%d\n",shoulijiedianshu); fprintf(fp,"%.1f\n",waizaihe); fprintf(fp,"%.1f\n",houdu); fprintf(fp,"%d\n",weiyijiediangeshu); inti,j,a=1; for(i=0;i<m;i++) { a=a+i; fprintf(fp,"%d\n",a); } for(i=(m*m-m)/2+1;i<m*(m+1)/2;i++){fprintf(fp,"%d\n",i+1);} j=0; for(i=0;i<m*2;i++) { fprintf(fp,"%d\n",j); } fclose(fp);}voidwritecoord(FILE*fp){ inti,j; for(i=0;i<m;i++) { for(j=0;j<=i;j++) { fprintf(fp,"%f%f",(float)j*2/(m-1),(float)(m-1-i)*2/(m-1));if(!(i==2*m-1)) { fprintf(fp,"\n"); } } } }voidwriteie(FILE*fp){ inti; intpre_of_start=1; intstart_of_linei=1; for(i=1;i<m;++i) { pre_of_start+=i-1; start_of_linei+=i; intnum_of_linei=2*i-1; intk; for(k=0;k<i-1;++k) { intposition01,position02,position03; intposition11,position12,position13; position01=pre_of_start+k; position02=start_of_linei+k; position03=position02+1; position11=position01; position12=position03; position13=position11+1; fprintf(fp,"%d%d%d\n",position01,position02,position03); fprintf(fp,"%d%d%d\n",position11,position12,position13); } intlast01,last02,last03; last01=pre_of_start+i-1; last02=start_of_linei+i-1; last03=last02+1; fprintf(fp,"%d%d%d\n",last01,last02,last03); }fclose(fp);}voidwriteir(FILE*fp){fprintf(fp,"%d",1);fclose(fp);}voidwriteiffix(FILE*fp){fprintf(fp,"%d%d",0,-1);fclose(fp); }程序修改//初始化；doublek[20][20],r[20];floatx[10],y[10],ee,e1,vv,h,q,dip[20];floate2,str[3][9],strn[7][10];intiee,ie[9][3];intnpoin;floatb[3],c[3],area;intinl,ir[20],ind,id[20],iffix[2];inti,j,m,nn,n,add;floata,tt;intnmax,nmax1;floataa,bb,smax,smax1;intii;實(shí)驗(yàn)結(jié)果實(shí)驗(yàn)結(jié)果在數(shù)據(jù)文件DATA.DAT中，以下僅列出各結(jié)點(diǎn)位移和個(gè)單元應(yīng)力：懸臂梁?jiǎn)栴}例題密化后（6*3個(gè)結(jié)點(diǎn)）theDISPLACEMENTis:dip[0]=5.000000e-020dip[1]=1.601282e-020dip[2]=5.000000e-020dip[3]=-1.601282e-020dip[4]=4.418334e-005dip[5]=3.761885e-006dip[6]=5.116970e-005dip[7]=-1.074825e-005dip[8]=9.399339e-005dip[9]=-3.101618e-006dip[10]=1.012079e-004dip[11]=-1.808562e-005dip[12]=1.439872e-004dip[13]=-1.031212e-005dip[14]=1.512092e-004dip[15]=-2.531160e-005dip[16]=1.939870e-004dip[17]=-1.753397e-005dip[18]=2.012092e-004dip[19]=-3.253394e-005dip[0]=3.986541e-020dip[1]=1.954374e-020dip[2]=7.026924e-020dip[3]=4.106339e-022dip[4]=3.986536e-020dip[5]=-1.995436e-020dip[6]=5.394375e-005dip[7]=7.321941e-006dip[8]=5.647431e-005dip[9]=-2.905583e-006dip[10]=6.171772e-005dip[11]=-1.394909e-005dip[12]=1.129990e-004dip[13]=1.768494e-006dip[14]=1.168552e-004dip[15]=-9.602640e-006dip[16]=1.213256e-004dip[17]=-2.076239e-005dip[18]=1.731804e-004dip[19]=-4.905183e-006dip[20]=1.767229e-004dip[21]=-1.623111e-005dip[22]=1.812967e-004dip[23]=-2.714038e-005dip[24]=2.342838e-004dip[25]=-1.129789e-005dip[26]=2.356151e-004dip[27]=-2.277566e-005dip[28]=2.415437e-004dip[29]=-3.226052e-005dip[30]=3.000494e-004dip[31]=-1.720884e-005dip[32]=2.891694e-004dip[33]=-2.776719e-005dip[34]=3.097623e-004dip[35]=-4.352307e-005theSTRESSis:str[0][0]=9.7106str[1][0]=2.9132str[2][0]=0.2894str[0][1]=10.2894str[1][1]=0.1848str[2][1]=-0.2894str[0][2]=9.9906str[1][2]=0.0951str[2][2]=0.0095str[0][3]=10.0095str[1][3]=0.0060str[2][3]=-0.0095str[0][4]=9.9997str[1][4]=0.0031str[2][4]=0.0003str[0][5]=10.0003str[1][5]=0.0002str[2][5]=-0.0003str[0][6]=10.0000str[1][6]=0.0001str[2][6]=0.0000str[0][7]=10.0000str[1][7]=0.0000str[2][7]=-0.0000theSTRESSis:str[0][0]=14.8197str[1][0]=4.4459str[2][0]=0.7040str[0][1]=14.1662str[1][1]=0.1589str[2][1]=0.1099str[0][2]=15.5149str[1][2]=4.6545str[2][2]=-0.2794str[0][3]=15.4991str[1][3]=0.2323str[2][3]=-0.5346str[0][4]=14.8753str[1][4]=0.3716str[2][4]=-0.1447str[0][5]=15.0887str[1][5]=-0.0219str[2][5]=-0.0507str[0][6]=15.1319str[1][6]=0.1222str[2][6]=0.1627str[0][7]=14.9042str[1][7]=0.0073str[2][7]=0.0326str[0][8]=15.0339str[1][8]=-0.0383str[2][8]=-0.0484str[0][9]=14.9536str[1][9]=-0.0443str[2][9]=-0.0924str[0][10]=14.9755str[1][10]=0.0288str[2][10]=0.0504str[0][11]=15.0370str[1][11]=0.1474str[2][11]=0.0904str[0][12]=15.2931str[1][12]=0.0576str[2][12]=-0.0697str[0][13]=14.6656str[1][13]=-0.1914str[2][13]=-0.4245str[0][14]=14.7406str[1][14]=0.0585str[2][14]=0.0744str[0][15]=15.3006str[1][15]=0.7962str[2][15]=0.4198str[0][16]=16.5539str[1][16]=0.3751str[2][16]=-0.3635str[0][17]=13.3204str[1][17]=-0.2272str[2][17]=-2.1538str[0][18]=13.4620str[1][18]=0.2446str[2][18]=0.4321str[0][19]=16.6637str[1][19]=-1.3033str[2][19]=2.0852三角形問題原題密化后（以一邊4個(gè)結(jié)