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FINITE ELEMENT METHOD (FEM) coding using C PROGRAMMING

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Akash Gupta

The document outlines a project on finite element method (FEM) structural analysis of bicycle frames using C programming, focusing on calculating node displacements and rotations for men's and women's bicycles. It details the input parameters, FEM steps, program structure, and outputs, which include nodal displacements, reactions, and a comparison of results with ANSYS. The analysis confirms that results obtained from the C program align with those produced by ANSYS.

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ADVANCED MECHANICS OF MATERIALS COURSE PROJECT ON FINITE ELEMENT METHOD STRUCTURAL ANALYSIS OF BICYCLE (FRAME ELEMENT) USING “C PROGRAMMING” SUBMITTED BY: AKASH KUMAR GUPTA MDS17M004
PROBLEM STATEMENT:- DETERMINE THE DISPLACEMENT AND ROTATIONS AT NODES FOR THE FRAME SHOWN IN BELOW FIGURES: CASE 1:- CYCLE (MEN) Discretization along with the connectivity of elements E=30e6 psi I1=0.01in4 I2=I3=I4=I5=0.02in4 I6=I7= 0.1 in4 A1=0.1in2 A2=A3=A4=A5=0.15in2 A6=A7=0.3in2
FEM STEP FOLLOWED IN C CODE:  DISCRETIZATION.  SELECTION OF INTERPOLATION MODEL.  EVALUATION OF ELEMENT STIFFNESS MATRIX AND FORCE VECTOR.  ASSEMBLY OF ELEMENT STIFFNESS MATRIX.  APPLY BOUNDARY CONDITION.  NODAL DISPLACEMENT.  EVALUATION OF SECONDARY VARIABLES. Formulation of L matrix
Formulation of k matrix Local stiffness matrix of each element:- INPUT of the program :- please enter number of elements 7 please enter number of nodes 6 enter the value of E30000000 enter AREA and INERTIA of 1th element respectively 0.1 0.01 enter AREA and INERTIA of 2th element respectively 0.15 0.02 enter AREA and INERTIA of 3th element respectively 0.15 0.02 enter AREA and INERTIA of 4th element respectively 0.15
0.02 enter AREA and INERTIA of 5th element respectively 0.15 0.02 enter AREA and INERTIA of 6th element respectively 0.3 0.1 enter AREA and INERTIA of 7th element respectively 0.3 0.1 enter X and Y co-ordinates of 1th node 0 0 enter X and Y co-ordinates of 2th node 10 15 enter X and Y co-ordinates of 3th node 20 0 enter X and Y co-ordinates of 4th node 30 15 enter X and Y co-ordinates of 5th node 32 11.25 enter X and Y co-ordinates of 6th node 38 0 enter ni and nj co-ordinates of 1th element 1 2
enter ni and nj co-ordinates of 2th element 2 3 enter ni and nj co-ordinates of 3th element 1 3 enter ni and nj co-ordinates of 4th element 2 4 enter ni and nj co-ordinates of 5th element 3 5 enter ni and nj co-ordinates of 6th element 4 5 enter ni and nj co-ordinates of 7th element 5 6  Then after input enter the nodes where you wants to put constraints please enter back wheel node first 1 6  Then after input please enter the forces on back seat and its notes respectively -125 2 please enter the forces on front handle and its notes respectively -25 4 OUTPUT of the program :- <<<<<<<UX of each node>>>>>> node 1 => 0.000000000
node 2 => 0.001919842 node 3 => 0.000291206 node 4 => 0.001703861 node 5 => 0.003210961 node 6 => 0.014042133 <<<<<<<UY of each node>>>>>> node 1 => 0.000000000 node 2 => -0.002132138 node 3 => -0.003045308 node 4 => -0.006676688 node 5 => -0.005851189 node 6 => 0.000000000 <<<<<<<rotZ of each node>>>>>> node 1 => -0.000092717 node 2 => -0.000229784 node 3 => -0.000297664 node 4 => 0.000323999 node 5 => 0.000518149 node 6 => 0.001189210 <<<<<<<Reaction force>>>>>> node 1 => 97.368492126 node 6 => 52.632865906 COMPARISON OF RESULTS WITH ANSYS : NODAL DISPALCEMENT:
REACTION AT SUPPORT:
 The same code gives the displacement & reaction of cycle(women) . CASE 2- DETERMINE THE DISPLACEMENT AND ROTATIONS AT NODES FOR THE FRAME SHOWN IN BELOW FIGURES: CASE 2- CYCLE (WOMEN)
Discretization along with the connectivity of elements E=10e6 psi I1=0.01in4 I2=I3=I4=I5=0.02in4 I6=I7=I8= 0.1 in4 A1=0.1in2 A2=A3=A4=A5=0.15in2 A6=A7=A8=0.3in2 INPUT of the program :- please enter number of elements 8 please enter number of nodes 7 enter the value of E10000000 enter AREA and INERTIA of 1th element respectively 0.1 0.01 enter AREA and INERTIA of 2th element respectively 0.15 0.02
enter AREA and INERTIA of 3th element respectively 0.15 0.02 enter AREA and INERTIA of 4th element respectively 0.15 0.02 enter AREA and INERTIA of 5th element respectively 0.15 0.02 enter AREA and INERTIA of 6th element respectively 0.3 0.1 enter AREA and INERTIA of 7th element respectively 0.3 0.1 enter AREA and INERTIA of 8th element respectively 0.3 0.1 enter X and Y co-ordinates of 1th node 0 0 enter X and Y co-ordinates of 2th node 10 15 enter X and Y co-ordinates of 3th node 18 3.75 enter X and Y co-ordinates of 4th node 20 0 enter X and Y co-ordinates of 5th node 30
15 enter X and Y co-ordinates of 6th node 32 11.25 enter X and Y co-ordinates of 7th node 38 0 enter ni and nj co-ordinates of 1th element 1 2 enter ni and nj co-ordinates of 2th element 2 3 enter ni and nj co-ordinates of 3th element 1 4 enter ni and nj co-ordinates of 4th element 3 5 enter ni and nj co-ordinates of 5th element 4 6 enter ni and nj co-ordinates of 6th element 5 6 enter ni and nj co-ordinates of 7th element 6 7 enter ni and nj co-ordinates of 8th element 3 4  Then after input
enter the nodes where you wants to put constraints please enter back wheel node first 1 7  Then after input please enter the forces on back seat and its notes respectively -125 2 please enter the forces on fronthandle and its notes respectively -25 5 OUTPUT of the program :- <<<<<<<UX of each node>>>>>> node 1 => 0.000000000 node 2 => 0.049084820 node 3 => -0.006489811 node 4 => 0.000885768 node 5 => -0.047025576 node 6 => -0.036962006 node 7 => 0.013382180 <<<<<<<UY of eachnode>>>>>> node 1 => 0.000000000 node 2 => -0.035041552 node 3 => -0.073752053 node 4 => -0.069679081 node 5 => -0.032445088 node 6 => -0.027073935 node 7 => 0.000000000 <<<<<<<rotZ of eachnode>>>>>> node 1 => -0.004590009 node 2 => -0.006706568 node 3 => 0.001747432 node 4 => 0.001976273 node 5 => 0.002551198 node 6 => 0.003141104 node 7 => 0.005154393 <<<<<<<Reaction force node>>>>>> node 1 => 97.372192383 node 7 => 52.635559082
COMPARISON OF RESULTS WITH ANSYS : NODAL DISPALCEMENT: REACTION AT SUPPORT:
Conclusion:- FEM Analysis of cycle structure (frame element) have conducted using C programming code and its output results are same as result generated by ANSYS. Program:- #include<stdio.h> #include<conio.h> #include <math.h> int main() { int ele,nod; printf("please enter number of elements "); /*total number of element of a structure */ scanf("%d",&ele); printf("n"); printf("please enter number of nodes "); scanf("%d",&nod);
float X[100],Y[100],le[100],l[100],m[100],L[100][20][20],KL[100][20][20],KG[100][20][20],res[100][20][20],LT[100] [20][20],KGL[100][100],keff[100][100],matrix[50][50],inv[100][100]; int ni[100],nj[100],N[100][100]; float A[20] ; float I[20] ; int E; float F[50]; float Q[50]; printf("enter the value of E"); scanf("%d",&E); for(int i=0;i<ele;i++) {printf("enter AREA and INERTIA of %dth element respectively ",i+1); scanf("%f%f",&A[i],&I[i]); printf("n"); } for(int i=0;i<nod;i++) {printf("enter X and Y co-ordinates of %dth node ",i+1); scanf("%f%f",&X[i],&Y[i]); printf("n"); } for(int i=0;i<ele;i++) {printf("enter ni and nj co-ordinates of %dth element ",i+1); /*enter the connectivity of each element i.e. connetivity table which store in ni & nj seperately */ scanf("%d%d",&ni[i],&nj[i]); // ni and nj are 1D array printf("n"); } for(int i=0;i<ele;i++) /* the store value of ni and nj provide us the dof of each element*/ // here a 2D array name "N[][]" will store the dof of each elemnt at the respective row // means dof of 1st element will store in 1st row of N[][] matrix and so on .... {int k=2,p=2;
for(int j=0;j<6;j++) {if(k>=0) {N[i][j]=3*ni[i]-k; k--; } else if(p>=0) {N[i][j]=3*nj[i]-p; p--; } } } for(int i=0;i<ele;i++) {for(int j=0;j<6;j++) {printf("%dt",N[i][j]); }printf("n"); } for(int i=0;i<ele;i++) {le[i]=sqrt((pow((X[nj[i]-1]-X[ni[i]-1]),2)+(pow((Y[nj[i]-1]-Y[ni[i]-1]),2)))); // calculation of length , l and m value of each element l[i]=(X[nj[i]-1]-X[ni[i]-1])/le[i]; m[i]=(Y[nj[i]-1]-Y[ni[i]-1])/le[i]; } printf("le matrix"); for(int i=0;i<ele;i++) { printf("%f",le[i]); printf("t"); } printf("n"); printf("l matrix"); for(int i=0;i<ele;i++) { printf("%f",l[i]); printf("t"); }
printf("n"); printf("m matrix"); for(int i=0;i<ele;i++) { printf("%f",m[i]); printf("t"); } printf("n"); for(int k=0;k<ele;k++) // a 3 dimension array name "L[][][]" will store the value of L matrix of each element {for(int i=0;i<6;i++) //means L[0][][] stores the value of L matrix of first element and so on ..... {for(int j=0;j<6;j++) { if(i==0&&j==0) {L[k][i][j]=l[k]; } else if(i==0&&j==1) {L[k][i][j]=m[k]; } else if(i==1&&j==0) {L[k][i][j]=-1*m[k]; } else if(i==1&&j==1) {L[k][i][j]=l[k]; } else if(i==2&&j==2) {L[k][i][j]=1; } else if(i==3&&j==3) {L[k][i][j]=l[k]; } else if(i==4&&j==3) {L[k][i][j]=-1*m[k]; } else if(i==3&&j==4) {L[k][i][j]=m[k];
} else if(i==4&&j==4) {L[k][i][j]=l[k]; } else if(i==5&&j==5) {L[k][i][j]=1; } else {L[k][i][j]=0; } } } } for(int k=0;k<ele;k++) {for(int i=0;i<6;i++) {for(int j=0;j<6;j++) {printf("%f",L[k][i][j]); }printf("n"); }printf("n"); } for(int k=0;k<ele;k++) { float P[6][6]= { {((E*A[k])/le[k]),0,0,(-1*(E*A[k])/le[k]),0,0}, {0,(12*E*I[k]/(pow(le[k],3))),(6*E*I[k]/(pow(le[k],2))),0,(- 12*E*I[k]/(pow(le[k],3))),(6*E*I[k]/(pow(le[k],2)))}, {0,(6*E*I[k]/(pow(le[k],2))),(4*E*I[k]/(pow(le[k],1))),0,(- 6*E*I[k]/(pow(le[k],2))),(2*E*I[k]/(pow(le[k],1)))}, {((-1*E*A[k])/le[k]),0,0,((E*A[k])/le[k]),0,0}, {0,(-12*E*I[k]/(pow(le[k],3))),(-6*E*I[k]/(pow(le[k],2))),0,(12*E*I[k]/(pow(le[k],3))),(- 6*E*I[k]/(pow(le[k],2)))}, {0,(6*E*I[k]/(pow(le[k],2))),(2*E*I[k]/(pow(le[k],1))),0,(- 6*E*I[k]/(pow(le[k],2))),(4*E*I[k]/(pow(le[k],1)))} }; for(int i=0;i<6;i++) {for(int j=0;j<6;j++) {KL[k][i][j]=P[i][j];
} } } for(int z=0;z<ele;z++) { for (int i = 0; i < 6; i++) { for (int j = 0; j < 6; j++) { res[z][i][j] = 0; for (int k = 0; k< 6; k++) { res[z][i][j]= res[z][i][j] + KL[z][i][k]*L[z][k][j]; } } } } for(int k=0;k<ele;k++) {for(int i=0;i<6;i++) {for(int j=0;j<6;j++) {LT[k][i][j]=L[k][j][i]; } } } for(int z=0;z<ele;z++) { for (int i = 0; i < 6; i++) { for (int j = 0; j< 6; j++) { KG[z][i][j] = 0; for (int k = 0; k < 6; k++) { KG[z][i][j] = KG[z][i][j]+LT[z][i][k]*res[z][k][j]; } } } }
printf("n"); printf(" <<<<<<<<local stiffness matrix of each element>>>>>>>"); printf("n"); for(int k=0;k<ele;k++) {for(int i=0;i<6;i++) {for(int j=0;j<6;j++) {printf("%f",KG[k][i][j]); printf("t"); } printf("n"); } printf("n"); } printf("n"); for(int i=0;i<18;i++) {for(int j=0;j<18;j++) {KGL[i][j]=0; } } printf("<<<<<<<< GLOBAL stiffness matrix>>>>>>> "); printf("n"); for(int k=0;k<ele;k++) {for(int i=0;i<6;i++) {for(int j=0;j<6;j++) {KGL[N[k][i]-1][N[k][j]-1]=KGL[N[k][i]-1][N[k][j]-1]+KG[k][i][j]; } } } printf("n"); for(int i=0;i<18;i++) {for(int j=0;j<18;j++) {printf("%.2f",KGL[i][j]);
printf("t"); }printf("n"); } printf("n"); int p1,p2; printf("enter the nodes where you wants to put constraints please enter back wheel node first"); scanf("%d%d",&p1,&p2); int w1,w2,w3,w4,w5; w1=3*p1-3; w2=3*p1-2; w3=3*p2-2; w4=3*nod; w5=w4-3; int v=0; for(int i=0;i<w4;i++) { int p=0; for(int j=0;j<w4;j++) { if(i!=w1&&i!=w2&&j!=w1&&j!=w2&&i!=w3&&j!=w3) {keff[v][p]=KGL[i][j]; p++; } if(p==w5) { v++; } } } printf("n"); for(int i=0;i<w5;i++) {for(int j=0;j<w5;j++) {printf("%.2f",keff[i][j]);
printf("t"); }printf("n"); } for(int i=0;i<w5;i++) {for(int j=0;j<w5;j++) { matrix[i][j]=keff[i][j]; } } printf("n"); int n=w5,i,j,k; float a,ratio; for(i = 0; i < n; i++){ for(j = n; j < 2*n; j++){ if(i==(j-n)) matrix[i][j] = 1.0; else matrix[i][j] = 0.0; } } for(i = 0; i < n; i++){ for(j = 0; j < n; j++){ if(i!=j){ ratio = matrix[j][i]/matrix[i][i]; for(k = 0; k < 2*n; k++){ matrix[j][k] -= ratio * matrix[i][k]; } } } } for(i = 0; i < n; i++){
a = matrix[i][i]; for(j = 0; j < 2*n; j++){ matrix[i][j] /= a; } } for(int i=0;i<w5;i++) {for(int j=0;j<w5;j++) {inv[i][j]=matrix[i][j+n]; } } printf("n"); printf("inverse of reduce stiffness matrix is: n"); for(int i = 0; i < w5; i++){ for(int j = 0; j < w5; j++){ printf("%.9f", inv[i][j]); printf("t"); } printf("n"); } printf("n"); float load1,load2; int l1,l2; printf("please enter the forces on back seat and its notes respectively "); scanf("%f%d",&load1,&l1); printf("n"); printf("please enter the forces on front handle and its notes respectively "); scanf("%f%d",&load2,&l2); for(int i=0;i<w4;i++) {F[i]=0; } F[(3*l1)-2]=load1; F[(3*l2)-2]=load2;
int k1=0; float F1[50]; for(int i=0;i<w4;i++) { if(i!=w1&&i!=w2&&i!=w3) {F1[k1]=F[i]; k1++; } } printf("n"); printf("force matrix"); printf("n"); for(int i=0;i<w5;i++) { printf("%f",F1[i]); printf("n"); } for(int i=0;i<w5;i++) { Q[i]=0; for(int j=0;j<w5;j++) {Q[i]=Q[i]+inv[i][j]*F1[j]; } } printf("n"); printf(" <<<<<<<<<<displacement matrix>>>>>>>"); printf("n"); for(int i=0;i<w5;i++) { printf("%.9f",Q[i]); printf("n"); }
int kc=0; float QG[50]; for(int i=0;i<w4;i++) { if(i==w1||i==w2||i==w3) {QG[i]=0; } else {QG[i]=Q[kc]; kc++; } } printf("n"); for(int i=0;i<w4;i++) { printf("%.9f",QG[i]); printf("n"); } float Re[50]; for (int i = 0; i < w4; i++) { Re[i]=0; for (int j = 0; j < w4; j++) Re[i]= Re[i] + KGL[i][j]*QG[j]; }
printf("n"); printf("<<<<<<<UX of each node>>>>>>"); printf("n"); int koi=0; for(int i=0;i<nod;i++) { printf("node %d => %.9f",i+1,QG[koi]); koi=koi+3; printf("n"); } printf("n"); printf("<<<<<<<UY of each node>>>>>>"); printf("n"); int Toi=1; for(int i=0;i<nod;i++) { printf("node %d => %.9f",i+1,QG[Toi]); Toi=Toi+3; printf("n"); } printf("n"); printf("<<<<<<<rotZ of each node>>>>>>"); printf("n"); int aoi=2; for(int i=0;i<nod;i++) { printf("node %d => %.9f",i+1,QG[aoi]); aoi=aoi+3; printf("n"); }
printf("n"); printf("<<<<<<<Reaction force node>>>>>>"); printf("n"); printf("node %d => %.9f",p1,Re[w2]); printf("n"); printf("node %d => %.9f",p2,Re[w3]); return 0; }

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FINITE ELEMENT METHOD (FEM) coding using C PROGRAMMING

  • 1. ADVANCED MECHANICS OF MATERIALS COURSE PROJECT ON FINITE ELEMENT METHOD STRUCTURAL ANALYSIS OF BICYCLE (FRAME ELEMENT) USING “C PROGRAMMING” SUBMITTED BY: AKASH KUMAR GUPTA MDS17M004
  • 2. PROBLEM STATEMENT:- DETERMINE THE DISPLACEMENT AND ROTATIONS AT NODES FOR THE FRAME SHOWN IN BELOW FIGURES: CASE 1:- CYCLE (MEN) Discretization along with the connectivity of elements E=30e6 psi I1=0.01in4 I2=I3=I4=I5=0.02in4 I6=I7= 0.1 in4 A1=0.1in2 A2=A3=A4=A5=0.15in2 A6=A7=0.3in2
  • 3. FEM STEP FOLLOWED IN C CODE:  DISCRETIZATION.  SELECTION OF INTERPOLATION MODEL.  EVALUATION OF ELEMENT STIFFNESS MATRIX AND FORCE VECTOR.  ASSEMBLY OF ELEMENT STIFFNESS MATRIX.  APPLY BOUNDARY CONDITION.  NODAL DISPLACEMENT.  EVALUATION OF SECONDARY VARIABLES. Formulation of L matrix
  • 4. Formulation of k matrix Local stiffness matrix of each element:- INPUT of the program :- please enter number of elements 7 please enter number of nodes 6 enter the value of E30000000 enter AREA and INERTIA of 1th element respectively 0.1 0.01 enter AREA and INERTIA of 2th element respectively 0.15 0.02 enter AREA and INERTIA of 3th element respectively 0.15 0.02 enter AREA and INERTIA of 4th element respectively 0.15
  • 5. 0.02 enter AREA and INERTIA of 5th element respectively 0.15 0.02 enter AREA and INERTIA of 6th element respectively 0.3 0.1 enter AREA and INERTIA of 7th element respectively 0.3 0.1 enter X and Y co-ordinates of 1th node 0 0 enter X and Y co-ordinates of 2th node 10 15 enter X and Y co-ordinates of 3th node 20 0 enter X and Y co-ordinates of 4th node 30 15 enter X and Y co-ordinates of 5th node 32 11.25 enter X and Y co-ordinates of 6th node 38 0 enter ni and nj co-ordinates of 1th element 1 2
  • 6. enter ni and nj co-ordinates of 2th element 2 3 enter ni and nj co-ordinates of 3th element 1 3 enter ni and nj co-ordinates of 4th element 2 4 enter ni and nj co-ordinates of 5th element 3 5 enter ni and nj co-ordinates of 6th element 4 5 enter ni and nj co-ordinates of 7th element 5 6  Then after input enter the nodes where you wants to put constraints please enter back wheel node first 1 6  Then after input please enter the forces on back seat and its notes respectively -125 2 please enter the forces on front handle and its notes respectively -25 4 OUTPUT of the program :- <<<<<<<UX of each node>>>>>> node 1 => 0.000000000
  • 7. node 2 => 0.001919842 node 3 => 0.000291206 node 4 => 0.001703861 node 5 => 0.003210961 node 6 => 0.014042133 <<<<<<<UY of each node>>>>>> node 1 => 0.000000000 node 2 => -0.002132138 node 3 => -0.003045308 node 4 => -0.006676688 node 5 => -0.005851189 node 6 => 0.000000000 <<<<<<<rotZ of each node>>>>>> node 1 => -0.000092717 node 2 => -0.000229784 node 3 => -0.000297664 node 4 => 0.000323999 node 5 => 0.000518149 node 6 => 0.001189210 <<<<<<<Reaction force>>>>>> node 1 => 97.368492126 node 6 => 52.632865906 COMPARISON OF RESULTS WITH ANSYS : NODAL DISPALCEMENT:
  • 9.  The same code gives the displacement & reaction of cycle(women) . CASE 2- DETERMINE THE DISPLACEMENT AND ROTATIONS AT NODES FOR THE FRAME SHOWN IN BELOW FIGURES: CASE 2- CYCLE (WOMEN)
  • 10. Discretization along with the connectivity of elements E=10e6 psi I1=0.01in4 I2=I3=I4=I5=0.02in4 I6=I7=I8= 0.1 in4 A1=0.1in2 A2=A3=A4=A5=0.15in2 A6=A7=A8=0.3in2 INPUT of the program :- please enter number of elements 8 please enter number of nodes 7 enter the value of E10000000 enter AREA and INERTIA of 1th element respectively 0.1 0.01 enter AREA and INERTIA of 2th element respectively 0.15 0.02
  • 11. enter AREA and INERTIA of 3th element respectively 0.15 0.02 enter AREA and INERTIA of 4th element respectively 0.15 0.02 enter AREA and INERTIA of 5th element respectively 0.15 0.02 enter AREA and INERTIA of 6th element respectively 0.3 0.1 enter AREA and INERTIA of 7th element respectively 0.3 0.1 enter AREA and INERTIA of 8th element respectively 0.3 0.1 enter X and Y co-ordinates of 1th node 0 0 enter X and Y co-ordinates of 2th node 10 15 enter X and Y co-ordinates of 3th node 18 3.75 enter X and Y co-ordinates of 4th node 20 0 enter X and Y co-ordinates of 5th node 30
  • 12. 15 enter X and Y co-ordinates of 6th node 32 11.25 enter X and Y co-ordinates of 7th node 38 0 enter ni and nj co-ordinates of 1th element 1 2 enter ni and nj co-ordinates of 2th element 2 3 enter ni and nj co-ordinates of 3th element 1 4 enter ni and nj co-ordinates of 4th element 3 5 enter ni and nj co-ordinates of 5th element 4 6 enter ni and nj co-ordinates of 6th element 5 6 enter ni and nj co-ordinates of 7th element 6 7 enter ni and nj co-ordinates of 8th element 3 4  Then after input
  • 13. enter the nodes where you wants to put constraints please enter back wheel node first 1 7  Then after input please enter the forces on back seat and its notes respectively -125 2 please enter the forces on fronthandle and its notes respectively -25 5 OUTPUT of the program :- <<<<<<<UX of each node>>>>>> node 1 => 0.000000000 node 2 => 0.049084820 node 3 => -0.006489811 node 4 => 0.000885768 node 5 => -0.047025576 node 6 => -0.036962006 node 7 => 0.013382180 <<<<<<<UY of eachnode>>>>>> node 1 => 0.000000000 node 2 => -0.035041552 node 3 => -0.073752053 node 4 => -0.069679081 node 5 => -0.032445088 node 6 => -0.027073935 node 7 => 0.000000000 <<<<<<<rotZ of eachnode>>>>>> node 1 => -0.004590009 node 2 => -0.006706568 node 3 => 0.001747432 node 4 => 0.001976273 node 5 => 0.002551198 node 6 => 0.003141104 node 7 => 0.005154393 <<<<<<<Reaction force node>>>>>> node 1 => 97.372192383 node 7 => 52.635559082
  • 14. COMPARISON OF RESULTS WITH ANSYS : NODAL DISPALCEMENT: REACTION AT SUPPORT:
  • 15. Conclusion:- FEM Analysis of cycle structure (frame element) have conducted using C programming code and its output results are same as result generated by ANSYS. Program:- #include<stdio.h> #include<conio.h> #include <math.h> int main() { int ele,nod; printf("please enter number of elements "); /*total number of element of a structure */ scanf("%d",&ele); printf("n"); printf("please enter number of nodes "); scanf("%d",&nod);
  • 16. float X[100],Y[100],le[100],l[100],m[100],L[100][20][20],KL[100][20][20],KG[100][20][20],res[100][20][20],LT[100] [20][20],KGL[100][100],keff[100][100],matrix[50][50],inv[100][100]; int ni[100],nj[100],N[100][100]; float A[20] ; float I[20] ; int E; float F[50]; float Q[50]; printf("enter the value of E"); scanf("%d",&E); for(int i=0;i<ele;i++) {printf("enter AREA and INERTIA of %dth element respectively ",i+1); scanf("%f%f",&A[i],&I[i]); printf("n"); } for(int i=0;i<nod;i++) {printf("enter X and Y co-ordinates of %dth node ",i+1); scanf("%f%f",&X[i],&Y[i]); printf("n"); } for(int i=0;i<ele;i++) {printf("enter ni and nj co-ordinates of %dth element ",i+1); /*enter the connectivity of each element i.e. connetivity table which store in ni & nj seperately */ scanf("%d%d",&ni[i],&nj[i]); // ni and nj are 1D array printf("n"); } for(int i=0;i<ele;i++) /* the store value of ni and nj provide us the dof of each element*/ // here a 2D array name "N[][]" will store the dof of each elemnt at the respective row // means dof of 1st element will store in 1st row of N[][] matrix and so on .... {int k=2,p=2;
  • 17. for(int j=0;j<6;j++) {if(k>=0) {N[i][j]=3*ni[i]-k; k--; } else if(p>=0) {N[i][j]=3*nj[i]-p; p--; } } } for(int i=0;i<ele;i++) {for(int j=0;j<6;j++) {printf("%dt",N[i][j]); }printf("n"); } for(int i=0;i<ele;i++) {le[i]=sqrt((pow((X[nj[i]-1]-X[ni[i]-1]),2)+(pow((Y[nj[i]-1]-Y[ni[i]-1]),2)))); // calculation of length , l and m value of each element l[i]=(X[nj[i]-1]-X[ni[i]-1])/le[i]; m[i]=(Y[nj[i]-1]-Y[ni[i]-1])/le[i]; } printf("le matrix"); for(int i=0;i<ele;i++) { printf("%f",le[i]); printf("t"); } printf("n"); printf("l matrix"); for(int i=0;i<ele;i++) { printf("%f",l[i]); printf("t"); }
  • 18. printf("n"); printf("m matrix"); for(int i=0;i<ele;i++) { printf("%f",m[i]); printf("t"); } printf("n"); for(int k=0;k<ele;k++) // a 3 dimension array name "L[][][]" will store the value of L matrix of each element {for(int i=0;i<6;i++) //means L[0][][] stores the value of L matrix of first element and so on ..... {for(int j=0;j<6;j++) { if(i==0&&j==0) {L[k][i][j]=l[k]; } else if(i==0&&j==1) {L[k][i][j]=m[k]; } else if(i==1&&j==0) {L[k][i][j]=-1*m[k]; } else if(i==1&&j==1) {L[k][i][j]=l[k]; } else if(i==2&&j==2) {L[k][i][j]=1; } else if(i==3&&j==3) {L[k][i][j]=l[k]; } else if(i==4&&j==3) {L[k][i][j]=-1*m[k]; } else if(i==3&&j==4) {L[k][i][j]=m[k];
  • 19. } else if(i==4&&j==4) {L[k][i][j]=l[k]; } else if(i==5&&j==5) {L[k][i][j]=1; } else {L[k][i][j]=0; } } } } for(int k=0;k<ele;k++) {for(int i=0;i<6;i++) {for(int j=0;j<6;j++) {printf("%f",L[k][i][j]); }printf("n"); }printf("n"); } for(int k=0;k<ele;k++) { float P[6][6]= { {((E*A[k])/le[k]),0,0,(-1*(E*A[k])/le[k]),0,0}, {0,(12*E*I[k]/(pow(le[k],3))),(6*E*I[k]/(pow(le[k],2))),0,(- 12*E*I[k]/(pow(le[k],3))),(6*E*I[k]/(pow(le[k],2)))}, {0,(6*E*I[k]/(pow(le[k],2))),(4*E*I[k]/(pow(le[k],1))),0,(- 6*E*I[k]/(pow(le[k],2))),(2*E*I[k]/(pow(le[k],1)))}, {((-1*E*A[k])/le[k]),0,0,((E*A[k])/le[k]),0,0}, {0,(-12*E*I[k]/(pow(le[k],3))),(-6*E*I[k]/(pow(le[k],2))),0,(12*E*I[k]/(pow(le[k],3))),(- 6*E*I[k]/(pow(le[k],2)))}, {0,(6*E*I[k]/(pow(le[k],2))),(2*E*I[k]/(pow(le[k],1))),0,(- 6*E*I[k]/(pow(le[k],2))),(4*E*I[k]/(pow(le[k],1)))} }; for(int i=0;i<6;i++) {for(int j=0;j<6;j++) {KL[k][i][j]=P[i][j];
  • 20. } } } for(int z=0;z<ele;z++) { for (int i = 0; i < 6; i++) { for (int j = 0; j < 6; j++) { res[z][i][j] = 0; for (int k = 0; k< 6; k++) { res[z][i][j]= res[z][i][j] + KL[z][i][k]*L[z][k][j]; } } } } for(int k=0;k<ele;k++) {for(int i=0;i<6;i++) {for(int j=0;j<6;j++) {LT[k][i][j]=L[k][j][i]; } } } for(int z=0;z<ele;z++) { for (int i = 0; i < 6; i++) { for (int j = 0; j< 6; j++) { KG[z][i][j] = 0; for (int k = 0; k < 6; k++) { KG[z][i][j] = KG[z][i][j]+LT[z][i][k]*res[z][k][j]; } } } }
  • 21. printf("n"); printf(" <<<<<<<<local stiffness matrix of each element>>>>>>>"); printf("n"); for(int k=0;k<ele;k++) {for(int i=0;i<6;i++) {for(int j=0;j<6;j++) {printf("%f",KG[k][i][j]); printf("t"); } printf("n"); } printf("n"); } printf("n"); for(int i=0;i<18;i++) {for(int j=0;j<18;j++) {KGL[i][j]=0; } } printf("<<<<<<<< GLOBAL stiffness matrix>>>>>>> "); printf("n"); for(int k=0;k<ele;k++) {for(int i=0;i<6;i++) {for(int j=0;j<6;j++) {KGL[N[k][i]-1][N[k][j]-1]=KGL[N[k][i]-1][N[k][j]-1]+KG[k][i][j]; } } } printf("n"); for(int i=0;i<18;i++) {for(int j=0;j<18;j++) {printf("%.2f",KGL[i][j]);
  • 22. printf("t"); }printf("n"); } printf("n"); int p1,p2; printf("enter the nodes where you wants to put constraints please enter back wheel node first"); scanf("%d%d",&p1,&p2); int w1,w2,w3,w4,w5; w1=3*p1-3; w2=3*p1-2; w3=3*p2-2; w4=3*nod; w5=w4-3; int v=0; for(int i=0;i<w4;i++) { int p=0; for(int j=0;j<w4;j++) { if(i!=w1&&i!=w2&&j!=w1&&j!=w2&&i!=w3&&j!=w3) {keff[v][p]=KGL[i][j]; p++; } if(p==w5) { v++; } } } printf("n"); for(int i=0;i<w5;i++) {for(int j=0;j<w5;j++) {printf("%.2f",keff[i][j]);
  • 23. printf("t"); }printf("n"); } for(int i=0;i<w5;i++) {for(int j=0;j<w5;j++) { matrix[i][j]=keff[i][j]; } } printf("n"); int n=w5,i,j,k; float a,ratio; for(i = 0; i < n; i++){ for(j = n; j < 2*n; j++){ if(i==(j-n)) matrix[i][j] = 1.0; else matrix[i][j] = 0.0; } } for(i = 0; i < n; i++){ for(j = 0; j < n; j++){ if(i!=j){ ratio = matrix[j][i]/matrix[i][i]; for(k = 0; k < 2*n; k++){ matrix[j][k] -= ratio * matrix[i][k]; } } } } for(i = 0; i < n; i++){
  • 24. a = matrix[i][i]; for(j = 0; j < 2*n; j++){ matrix[i][j] /= a; } } for(int i=0;i<w5;i++) {for(int j=0;j<w5;j++) {inv[i][j]=matrix[i][j+n]; } } printf("n"); printf("inverse of reduce stiffness matrix is: n"); for(int i = 0; i < w5; i++){ for(int j = 0; j < w5; j++){ printf("%.9f", inv[i][j]); printf("t"); } printf("n"); } printf("n"); float load1,load2; int l1,l2; printf("please enter the forces on back seat and its notes respectively "); scanf("%f%d",&load1,&l1); printf("n"); printf("please enter the forces on front handle and its notes respectively "); scanf("%f%d",&load2,&l2); for(int i=0;i<w4;i++) {F[i]=0; } F[(3*l1)-2]=load1; F[(3*l2)-2]=load2;
  • 25. int k1=0; float F1[50]; for(int i=0;i<w4;i++) { if(i!=w1&&i!=w2&&i!=w3) {F1[k1]=F[i]; k1++; } } printf("n"); printf("force matrix"); printf("n"); for(int i=0;i<w5;i++) { printf("%f",F1[i]); printf("n"); } for(int i=0;i<w5;i++) { Q[i]=0; for(int j=0;j<w5;j++) {Q[i]=Q[i]+inv[i][j]*F1[j]; } } printf("n"); printf(" <<<<<<<<<<displacement matrix>>>>>>>"); printf("n"); for(int i=0;i<w5;i++) { printf("%.9f",Q[i]); printf("n"); }
  • 26. int kc=0; float QG[50]; for(int i=0;i<w4;i++) { if(i==w1||i==w2||i==w3) {QG[i]=0; } else {QG[i]=Q[kc]; kc++; } } printf("n"); for(int i=0;i<w4;i++) { printf("%.9f",QG[i]); printf("n"); } float Re[50]; for (int i = 0; i < w4; i++) { Re[i]=0; for (int j = 0; j < w4; j++) Re[i]= Re[i] + KGL[i][j]*QG[j]; }
  • 27. printf("n"); printf("<<<<<<<UX of each node>>>>>>"); printf("n"); int koi=0; for(int i=0;i<nod;i++) { printf("node %d => %.9f",i+1,QG[koi]); koi=koi+3; printf("n"); } printf("n"); printf("<<<<<<<UY of each node>>>>>>"); printf("n"); int Toi=1; for(int i=0;i<nod;i++) { printf("node %d => %.9f",i+1,QG[Toi]); Toi=Toi+3; printf("n"); } printf("n"); printf("<<<<<<<rotZ of each node>>>>>>"); printf("n"); int aoi=2; for(int i=0;i<nod;i++) { printf("node %d => %.9f",i+1,QG[aoi]); aoi=aoi+3; printf("n"); }
  • 28. printf("n"); printf("<<<<<<<Reaction force node>>>>>>"); printf("n"); printf("node %d => %.9f",p1,Re[w2]); printf("n"); printf("node %d => %.9f",p2,Re[w3]); return 0; }