University of Florida/Eml4507/Team 7 Report 6
Problem 1
editOn our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
Find
editSolve the general eigenvalue problem by first converting it to the standard eigenvalue problem for the spring-mass-damper system given on p.53-13 of 'Fead.s13.sec.53b'.
Given
edit
Solution
editConverting the general eigenvalue problem to the standard eigenvalue problem and solving it
General eigenvalue form:
Standard eigenvalue form:
Where,
Calculating the determinant
Therefore, the Eigenvalues are:
The next step is finding the Eigenvectors
Letting and calculating
Letting and calculating
These results compare favorably with those from problem 5.6 using CALFEM.
Mass Orthogonality:
The eigenvectors are therefore proven to be mass-orthogonal.
Problem 2
editOn our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
Find
editThe Lowest 3 eigenpairs by transforming the generalized eigenvalue problem into a standard eigenvalue problem and plot the 3 lowest mode shapes in a gif image.
Given
edit
Solution
edit
clear
clc
%GIVENS
L=0.3;
E=100e9;
A=1e-4;
rho=5000;
ep=[E A];
%ELEMENT DOF
Edof=zeros(25,5);
for i=1:6
j=2*(i-1);
Edof(i,:)=[i,1+j,2+j,3+j,4+j];
end
for i=7:12
j=2*(i+1);
Edof(i,:)=[i,j-1,j,j+1,j+2];
end
for i=13:19
j=2*(i-13);
Edof(i,:)=[i,1+j,2+j,15+j,16+j];
end
for i=20:25
j=2*(i-20);
Edof(i,:)=[i,1+j,2+j,17+j,18+j];
end
%COORDINATES
ex=zeros(14,2);
ey=zeros(14,2);
for i=1:25
if i<7
ex(i,:)=[L*(i-1),L*i];
ey(i,:)=[0,0];
elseif i>6&&i<13
ex(i,:)=[L*(i-7),L*(i-6)];
ey(i,:)=[L,L];
elseif i>12&&i<20
ex(i,:)=[L*(i-13),L*(i-13)];
ey(i,:)=[0,L];
else
ex(i,:)=[L*(i-20),L*(i-19)];
ey(i,:)=[0,L];
end
end
%STIFFNESS AND MASS MATRIX
K=zeros(28);
M=zeros(28);
for i=1:25
E=ep(1);A=ep(2);
xt=ex(i,2)-ex(i,1);
yt=ey(i,2)-ey(i,1);
L=sqrt(xt^2+yt^2);
l=xt/L; m=yt/L;
ke=E*A/L*[l^2 l*m -l^2 -l*m;
l*m m^2 -l*m -m^2;
-l^2 -l*m l^2 l*m;
-l*m -m^2 l*m m^2;];
m=L*A*rho;
me=[m/2 0 0 0;
0 m/2 0 0;
0 0 m/2 0;
0 0 0 m/2];
edoft=Edof(i,2:5);
K(edoft,edoft)=K(edoft,edoft)+ke;
M(edoft,edoft)=M(edoft,edoft)+me;
end
%FIND THE EIGEN VALUES AND VECTORS
ndof=size(K,1);
freedof=[1:ndof]';
fixdof=[1 15 16]';
freedof(fixdof(:))=[];
Kred=K(freedof,freedof);
Mred=M(freedof,freedof);
Ks=Mred^(-1/2)*Kred*Mred^(-1/2);
[X1,D1]=eig(Ks);
X2=Mred^(-1/2)*X1;
for j=1:size(X2)
d=sqrt(X2(:,j)'*X2(:,j));
X3(:,j)=X2(:,j)/d;
end
[D1,i]=sort(diag(D1));
X4=X3(:,i);
eigenval=D1;
eigenvec=zeros(size(K,1),size(X4,1));
eigenvec(freedof,:)=X4;
%PLOT THE GIF
for kk=1:3
eigenvalx=eigenval(kk);
eigenvecx=eigenvec(:,kk);
w=sqrt(eigenvalx);
T=2*pi/w
dt=T/100;
t=0:dt:T;
figure(kk)
if kk==1;
filename = 'mode1 r5p2.gif';
elseif kk==2;
filename = 'mode2 r5p2.gif';
elseif kk==3;
filename = 'mode3 r5p2.gif';
end
for jj = 1:100
tt=t(jj);
d=eigenvecx*sin(w*tt);
d=d';
rr=Edof(:,2:5);
ed=zeros(25,4);
for i=1:25;
ed(i,1:4)=d(rr(i,:));
end
xd=(ex+ed(:,[1 3]))';
yd=(ey+ed(:,[2 4]))';
s1=['-' , 'k'];
axis manual
plot(xd,yd,s1)
set(gca, 'ylim', [-1.5 1.5], 'xlim', [0 3]);
drawnow
frame = getframe(1);
im = frame2im(frame);
[imind,cm] = rgb2ind(im,256);
if jj == 1;
imwrite(imind,cm,filename,'gif', 'Loopcount',inf);
else
imwrite(imind,cm,filename,'gif','WriteMode','append');
end
end
end
The resulting egienpairs are:
The mode shapes for the previous eigenpairs are"
Problem 3
editOn our honor, the solution to this problem was generated by referring to the solution to Problem 3.1 at these locations: Eml4507 Team7 Report3,
Team Negative Damping (3): Report 3, Eml4507.s13.team2/Report3, and EML4507.s13.team4ever.R3. Their solutions were not copied, but consulted.
Find
editUsing FE matlab code, with non-zero prescribed displacement degrees-of-freedom, using the same "connectivity array" given in the same problem statement, provide all of the unknown global displacement degrees-of-freedom, force components(reactions), and member forces. Plot the deformed shape superposed on the undeformed shape. Verify the results with CALFEM.
Compare the reactions in the case of non-zero prescribed disp dofs with the reactions in the case of zero prescribed disp dofs.
Given
edit
In the non-zero prescribed case:
In the zero prescribed case:
Solution
editMATLAB code:
%Displacements
d3= 0.02; %in m
d4= -0.01;
d5= -0.03;
d6= 0.05;
d7 = 0;
d8 = 0;
ph1= 0;
ph2 = 0;
dsmall = [d3; d4; d5; d6;]; %ph =placeholder, displacement unknown
theta=45;
%Property Matrix= [element number; EA/L; angle;];
P = [1:3; 10^5*206 10^5*206 10^5*206; -pi/6 pi/2 -5*pi/6;];
%angles refer to 1->3, 1->2, 1->4
%Global Coordinate System
GNC = [ 0, 0;
0, 1;
.866, -.5;
-.866, -.5;];
lm = [.866, -.5;
0 , 1;
-.866, -.5;];
%populate global force matrix
dof = 8; %4 nodes, 2 directions reach
F = zeros(dof,1);
F(1) = 20000*cos(pi/4);
F(2) = 20000*sin(pi/4);
%populate stiffness matrices
lsq1 = (lm(1,1))^2;
msq1 = (lm(1,2))^2;
ltimesm1 = lm(1,1)*lm(1,2);
Ke1 = [ lsq1, ltimesm1, 0, 0, -lsq1, -ltimesm1, 0, 0;
ltimesm1, msq1, 0, 0 -ltimesm1, -msq1, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;
-lsq1, -ltimesm1, 0, 0, lsq1, ltimesm1, 0, 0;
-ltimesm1, -msq1, 0, 0, ltimesm1, msq1, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;];
lsq2 = (lm(2,1))^2;
msq2 = (lm(2,2))^2;
ltimesm2 = lm(2,1)*lm(2,2);
Ke2 = [ lsq2, ltimesm2, -lsq2, -ltimesm2, 0, 0, 0, 0;
ltimesm2, msq2, -ltimesm2, -msq2, 0, 0, 0, 0;
-lsq2, -ltimesm2, lsq2, ltimesm2, 0, 0, 0, 0;
-ltimesm2, -msq2, ltimesm2, msq2, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;];
lsq3 = (lm(3,1))^2;
msq3 = (lm(3,2))^2;
ltimesm3 = lm(3,1)*lm(3,2);
Ke3 = [ lsq3, ltimesm3, 0, 0, 0, 0, -lsq3, -ltimesm3;
ltimesm3, msq3, 0, 0, 0, 0, -ltimesm3, -msq3;
0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0;
-lsq3, -ltimesm3, 0, 0, 0, 0, lsq3, ltimesm3;
-ltimesm3, -msq3, 0, 0, 0, 0, ltimesm3, msq3;];
Kglobal = P(2,1)*(Ke1+Ke2+Ke3)
%find unknown displacements, after reducing
Fsmall = [F(1) F(2)]';
Ksmall = zeros([2,6]);
for i=1:2
for j=1:6
Ksmall (i,j) = Kglobal((i+1),j);
end
end
unknownds = Ksmall\Fsmall
d1 = unknownds(1);
d2 = unknownds(2);
Ksmall2 = zeros([4,4]);
for i=1:4
for j=1:4
Ksmall2 (i,j) = Kglobal((i+1),j);
end
end
unknownFs = Ksmall2* dsmall
P1 = 10^5*206*.866*(d5-d1)-.5*(d6-d2)
P2 = 10^10*206*0*(d3-d1)-1*(d4-d2)
P3 = 10^10*206*-.866*(d7-d1)-.5*(d8-d2)
Resulting in:
Kglobal =
Columns 1 through 7
3.0898e+07 0 0 0 -1.5449e+07 8.9198e+06 -1.5449e+07
0 3.09e+07 0 -2.06e+07 8.9198e+06 -5.15e+06 -8.9198e+06
0 0 0 0 0 0 0
0 -2.06e+07 0 2.06e+07 0 0 0
-1.5449e+07 8.9198e+06 0 0 1.5449e+07 -8.9198e+06 0
8.9198e+06 -5.15e+06 0 0 -8.9198e+06 5.15e+06 0
-1.5449e+07 -8.9198e+06 0 0 0 0 1.5449e+07
-8.9198e+06 -5.15e+06 0 0 0 0 8.9198e+06
Column 8
-8.9198e+06
-5.15e+06
0
0
0
0
8.9198e+06
5.15e+06
unknownds =
0.00045767 0.00045767
unknownFs =
-1.339e+06
0
1.236e+06
-3.9818e+05
P1 =
-5.3519e+05
P2 =
1045.8
P3 =
22.884
In the zero case:
In the zero prescribed case:
Simply strike the rows and columns, resulting in:
unknownds =
0.00045767 0.00045767
All the outer node reactions are zero, since their displacements are zero.
P1 =
-0.34500e+05
P2 =
-0.94400e+05
P3 =
1.29000e+05
Problem 4
editOn our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
Find
editSolve the initial truss structure using truss FEs
Given
edit
Solution
edit
function FEAr6p4
nodelocation=[-1.5 0 8;
1.5 0 8;
-1.5 1.5 4;
1.5 1.5 4;
1.5 -1.5 4;
-1.5 -1.5 4;
-4 4 0;
4 4 0;
4 -4 0;
-4 -4 0];
%INPUT: ex = [-1.5 1.5]
% ey = [0 0] element node coordinates
% ez = [z1 z2]
%
% ep = [E A] E: Young's modulus
% A: cross section area
%
ex=[nodelocation(1,1) nodelocation(2,1);
nodelocation(4,1) nodelocation(1,1);
nodelocation(3,1) nodelocation(2,1);
nodelocation(5,1) nodelocation(1,1);
nodelocation(6,1) nodelocation(2,1);
nodelocation(4,1) nodelocation(2,1);
nodelocation(5,1) nodelocation(2,1);
nodelocation(3,1) nodelocation(1,1);
nodelocation(6,1) nodelocation(1,1);
nodelocation(6,1) nodelocation(3,1);
nodelocation(5,1) nodelocation(4,1);
nodelocation(4,1) nodelocation(3,1);
nodelocation(6,1) nodelocation(5,1);
nodelocation(10,1) nodelocation(3,1);
nodelocation(7,1) nodelocation(6,1);
nodelocation(9,1) nodelocation(4,1);
nodelocation(8,1) nodelocation(5,1);
nodelocation(7,1) nodelocation(4,1);
nodelocation(8,1) nodelocation(3,1);
nodelocation(10,1) nodelocation(5,1);
nodelocation(9,1) nodelocation(6,1);
nodelocation(10,1) nodelocation(6,1);
nodelocation(7,1) nodelocation(3,1);
nodelocation(8,1) nodelocation(4,1);
nodelocation(9,1) nodelocation(5,1);];
ey=[nodelocation(1,2) nodelocation(2,2);
nodelocation(4,2) nodelocation(1,2);
nodelocation(3,2) nodelocation(2,2);
nodelocation(5,2) nodelocation(1,2);
nodelocation(6,2) nodelocation(2,2);
nodelocation(4,2) nodelocation(2,2);
nodelocation(5,2) nodelocation(2,2);
nodelocation(3,2) nodelocation(1,2);
nodelocation(6,2) nodelocation(1,2);
nodelocation(6,2) nodelocation(3,2);
nodelocation(5,2) nodelocation(4,2);
nodelocation(4,2) nodelocation(3,2);
nodelocation(6,2) nodelocation(5,2);
nodelocation(10,2) nodelocation(3,2);
nodelocation(7,2) nodelocation(6,2);
nodelocation(9,2) nodelocation(4,2);
nodelocation(8,2) nodelocation(5,2);
nodelocation(7,2) nodelocation(4,2);
nodelocation(8,2) nodelocation(3,2);
nodelocation(10,2) nodelocation(5,2);
nodelocation(9,2) nodelocation(6,2);
nodelocation(10,2) nodelocation(6,2);
nodelocation(7,2) nodelocation(3,2);
nodelocation(8,2) nodelocation(4,2);
nodelocation(9,2) nodelocation(5,2);];
ez=[nodelocation(1,3) nodelocation(2,3);
nodelocation(4,3) nodelocation(1,3);
nodelocation(3,3) nodelocation(2,3);
nodelocation(5,3) nodelocation(1,3);
nodelocation(6,3) nodelocation(2,3);
nodelocation(4,3) nodelocation(2,3);
nodelocation(5,3) nodelocation(2,3);
nodelocation(3,3) nodelocation(1,3);
nodelocation(6,3) nodelocation(1,3);
nodelocation(6,3) nodelocation(3,3);
nodelocation(5,3) nodelocation(4,3);
nodelocation(4,3) nodelocation(3,3);
nodelocation(6,3) nodelocation(5,3);
nodelocation(10,3) nodelocation(3,3);
nodelocation(7,3) nodelocation(6,3);
nodelocation(9,3) nodelocation(4,3);
nodelocation(8,3) nodelocation(5,3);
nodelocation(7,3) nodelocation(4,3);
nodelocation(8,3) nodelocation(3,3);
nodelocation(10,3) nodelocation(5,3);
nodelocation(9,3) nodelocation(6,3);
nodelocation(10,3) nodelocation(6,3);
nodelocation(7,3) nodelocation(3,3);
nodelocation(8,3) nodelocation(4,3);
nodelocation(9,3) nodelocation(5,3);];
E=3*10^7; A=0.0218;
B=zeros(3, 25);
Kefinal=zeros(25,1);
for i=1:25
B(1,i)=ex(i,2)-ex(i,1) ;
B(2,i)= ey(i,2)-ey(i,1);
B(3,i)=ez(i,2)-ez(i,1) ;
for i=1:25
l=B(:,i);
L=sqrt(l'*l);
Kle=E*A/L*[1 -1;
-1 1];
n=l'/L
G=[ n zeros(size(n));
zeros(size(n)) n ];
Ke=G'*Kle*G
end
end
% b=B*B'
% L=sqrt(b)
% Kle=E*A/L*[1 -1;
% -1 1];
%
% n=B'/L;
% G=[ n zeros(size(n));
% zeros(size(n)) n ]
%
% Ke=G'*Kle*G;
OUTPUT:
Columns 1 through 4
1.0745 -2.3640 1.7192 -1.0745
-2.3640 5.2007 -3.7823 2.3640
1.7192 -3.7823 2.7508 -1.7192
-1.0745 2.3640 -1.7192 1.0745
2.3640 -5.2007 3.7823 -2.3640
-1.7192 3.7823 -2.7508 1.7192
Columns 5 through 6
2.3640 -1.7192
-5.2007 3.7823
3.7823 -2.7508
-2.3640 1.7192
5.2007 -3.7823
-3.7823 2.7508
Ke =
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
Ke =
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
Ke =
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
Ke =
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
Ke =
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
Ke =
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
Ke =
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
Ke =
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
Ke =
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
Ke =
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
NaN NaN NaN NaN NaN NaN
Ke =
Columns 1 through 3
218000 0 0
0 0 0
0 0 0
-218000 0 0
0 0 0
0 0 0
Columns 4 through 6
-218000 0 0
0 0 0
0 0 0
218000 0 0
0 0 0
0 0 0
Ke =
1.0e+004 *
Columns 1 through 4
4.1378 2.0689 -5.5171 -4.1378
2.0689 1.0345 -2.7585 -2.0689
-5.5171 -2.7585 7.3561 5.5171
-4.1378 -2.0689 5.5171 4.1378
-2.0689 -1.0345 2.7585 2.0689
5.5171 2.7585 -7.3561 -5.5171
Columns 5 through 6
-2.0689 5.5171
-1.0345 2.7585
2.7585 -7.3561
2.0689 -5.5171
1.0345 -2.7585
-2.7585 7.3561
Ke =
1.0e+004 *
Columns 1 through 4
4.1378 -2.0689 5.5171 -4.1378
-2.0689 1.0345 -2.7585 2.0689
5.5171 -2.7585 7.3561 -5.5171
-4.1378 2.0689 -5.5171 4.1378
2.0689 -1.0345 2.7585 -2.0689
-5.5171 2.7585 -7.3561 5.5171
Columns 5 through 6
2.0689 -5.5171
-1.0345 2.7585
2.7585 -7.3561
-2.0689 5.5171
1.0345 -2.7585
-2.7585 7.3561
Ke =
1.0e+004 *
Columns 1 through 4
4.1378 -2.0689 -5.5171 -4.1378
-2.0689 1.0345 2.7585 2.0689
-5.5171 2.7585 7.3561 5.5171
-4.1378 2.0689 5.5171 4.1378
2.0689 -1.0345 -2.7585 -2.0689
5.5171 -2.7585 -7.3561 -5.5171
Columns 5 through 6
2.0689 5.5171
-1.0345 -2.7585
-2.7585 -7.3561
-2.0689 -5.5171
1.0345 2.7585
2.7585 7.3561
Ke =
1.0e+004 *
Columns 1 through 4
4.1378 2.0689 5.5171 -4.1378
2.0689 1.0345 2.7585 -2.0689
5.5171 2.7585 7.3561 -5.5171
-4.1378 -2.0689 -5.5171 4.1378
-2.0689 -1.0345 -2.7585 2.0689
-5.5171 -2.7585 -7.3561 5.5171
Columns 5 through 6
-2.0689 -5.5171
-1.0345 -2.7585
-2.7585 -7.3561
2.0689 5.5171
1.0345 2.7585
2.7585 7.3561
Ke =
1.0e+005 *
Columns 1 through 4
0 0 0 0
0 0.1887 -0.5033 0
0 -0.5033 1.3422 0
0 0 0 0
0 -0.1887 0.5033 0
0 0.5033 -1.3422 0
Columns 5 through 6
0 0
-0.1887 0.5033
0.5033 -1.3422
0 0
0.1887 -0.5033
-0.5033 1.3422
Ke =
1.0e+005 *
Columns 1 through 4
0 0 0 0
0 0.1887 0.5033 0
0 0.5033 1.3422 0
0 0 0 0
0 -0.1887 -0.5033 0
0 -0.5033 -1.3422 0
Columns 5 through 6
0 0
-0.1887 -0.5033
-0.5033 -1.3422
0 0
0.1887 0.5033
0.5033 1.3422
Ke =
1.0e+005 *
Columns 1 through 4
0 0 0 0
0 0.1887 -0.5033 0
0 -0.5033 1.3422 0
0 0 0 0
0 -0.1887 0.5033 0
0 0.5033 -1.3422 0
Columns 5 through 6
0 0
-0.1887 0.5033
0.5033 -1.3422
0 0
0.1887 -0.5033
-0.5033 1.3422
Ke =
1.0e+005 *
Columns 1 through 4
0 0 0 0
0 0.1887 0.5033 0
0 0.5033 1.3422 0
0 0 0 0
0 -0.1887 -0.5033 0
0 -0.5033 -1.3422 0
Columns 5 through 6
0 0
-0.1887 -0.5033
-0.5033 -1.3422
0 0
0.1887 0.5033
0.5033 1.3422
Ke =
Columns 1 through 3
0 0 0
0 218000 0
0 0 0
0 0 0
0 -218000 0
0 0 0
Columns 4 through 6
0 0 0
0 -218000 0
0 0 0
0 0 0
0 218000 0
0 0 0
Ke =
Columns 1 through 3
0 0 0
0 218000 0
0 0 0
0 0 0
0 -218000 0
0 0 0
Columns 4 through 6
0 0 0
0 -218000 0
0 0 0
0 0 0
0 218000 0
0 0 0
Ke =
Columns 1 through 3
218000 0 0
0 0 0
0 0 0
-218000 0 0
0 0 0
0 0 0
Columns 4 through 6
-218000 0 0
0 0 0
0 0 0
218000 0 0
0 0 0
0 0 0
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1.0e+004 *
Columns 1 through 4
5.2007 -2.3640 3.7823 -5.2007
-2.3640 1.0745 -1.7192 2.3640
3.7823 -1.7192 2.7508 -3.7823
-5.2007 2.3640 -3.7823 5.2007
2.3640 -1.0745 1.7192 -2.3640
-3.7823 1.7192 -2.7508 3.7823
Columns 5 through 6
2.3640 -3.7823
-1.0745 1.7192
1.7192 -2.7508
-2.3640 3.7823
1.0745 -1.7192
-1.7192 2.7508
Ke =
1.0e+004 *
Columns 1 through 4
5.2007 2.3640 -3.7823 -5.2007
2.3640 1.0745 -1.7192 -2.3640
-3.7823 -1.7192 2.7508 3.7823
-5.2007 -2.3640 3.7823 5.2007
-2.3640 -1.0745 1.7192 2.3640
3.7823 1.7192 -2.7508 -3.7823
Columns 5 through 6
-2.3640 3.7823
-1.0745 1.7192
1.7192 -2.7508
2.3640 -3.7823
1.0745 -1.7192
-1.7192 2.7508
Ke =
1.0e+004 *
Columns 1 through 4
5.2007 2.3640 3.7823 -5.2007
2.3640 1.0745 1.7192 -2.3640
3.7823 1.7192 2.7508 -3.7823
-5.2007 -2.3640 -3.7823 5.2007
-2.3640 -1.0745 -1.7192 2.3640
-3.7823 -1.7192 -2.7508 3.7823
Columns 5 through 6
-2.3640 -3.7823
-1.0745 -1.7192
-1.7192 -2.7508
2.3640 3.7823
1.0745 1.7192
1.7192 2.7508
Ke =
1.0e+004 *
Columns 1 through 4
5.2007 -2.3640 -3.7823 -5.2007
-2.3640 1.0745 1.7192 2.3640
-3.7823 1.7192 2.7508 3.7823
-5.2007 2.3640 3.7823 5.2007
2.3640 -1.0745 -1.7192 -2.3640
3.7823 -1.7192 -2.7508 -3.7823
Columns 5 through 6
2.3640 3.7823
-1.0745 -1.7192
-1.7192 -2.7508
-2.3640 -3.7823
1.0745 1.7192
1.7192 2.7508
Ke =
1.0e+004 *
Columns 1 through 4
2.6865 2.6865 4.2984 -2.6865
2.6865 2.6865 4.2984 -2.6865
4.2984 4.2984 6.8775 -4.2984
-2.6865 -2.6865 -4.2984 2.6865
-2.6865 -2.6865 -4.2984 2.6865
-4.2984 -4.2984 -6.8775 4.2984
Columns 5 through 6
-2.6865 -4.2984
-2.6865 -4.2984
-4.2984 -6.8775
2.6865 4.2984
2.6865 4.2984
4.2984 6.8775
Ke =
1.0e+004 *
Columns 1 through 4
2.6865 -2.6865 4.2984 -2.6865
-2.6865 2.6865 -4.2984 2.6865
4.2984 -4.2984 6.8775 -4.2984
-2.6865 2.6865 -4.2984 2.6865
2.6865 -2.6865 4.2984 -2.6865
-4.2984 4.2984 -6.8775 4.2984
Columns 5 through 6
2.6865 -4.2984
-2.6865 4.2984
4.2984 -6.8775
-2.6865 4.2984
2.6865 -4.2984
-4.2984 6.8775
Ke =
1.0e+004 *
Columns 1 through 4
2.6865 2.6865 -4.2984 -2.6865
2.6865 2.6865 -4.2984 -2.6865
-4.2984 -4.2984 6.8775 4.2984
-2.6865 -2.6865 4.2984 2.6865
-2.6865 -2.6865 4.2984 2.6865
4.2984 4.2984 -6.8775 -4.2984
Columns 5 through 6
-2.6865 4.2984
-2.6865 4.2984
4.2984 -6.8775
2.6865 -4.2984
2.6865 -4.2984
-4.2984 6.8775
Ke =
1.0e+004 *
Columns 1 through 4
2.6865 -2.6865 -4.2984 -2.6865
-2.6865 2.6865 4.2984 2.6865
-4.2984 4.2984 6.8775 4.2984
-2.6865 2.6865 4.2984 2.6865
2.6865 -2.6865 -4.2984 -2.6865
4.2984 -4.2984 -6.8775 -4.2984
Columns 5 through 6
2.6865 4.2984
-2.6865 -4.2984
-4.2984 -6.8775
-2.6865 -4.2984
2.6865 4.2984
4.2984 6.8775
Problem 5
editOn our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
Find
editVibration analysis of 3-D truss system in R6.4
Given
edit
Solution
edit
clear
clc
E=3*10^7; A=0.0218;
rho= 7.3*10^(-4);
Edof=[1 1 2 3 4 5 6;
2 1 2 3 10 11 12 ;
3 4 5 6 7 8 9;
4 1 2 3 13 14 15;
5 4 5 6 16 17 18;
6 4 5 6 10 11 12;
7 4 5 6 13 14 15;
8 1 2 3 7 8 9;
9 1 2 3 16 17 18;
10 7 8 9 16 17 18;
11 10 11 12 13 14 15;
12 7 8 9 10 11 12;
13 13 14 15 16 17 18;
14 7 8 9 28 29 30;
15 16 17 18 19 20 21;
16 10 11 12 25 26 27;
17 13 14 15 22 23 24;
18 10 11 12 19 20 21;
19 7 8 9 22 23 24;
20 13 14 15 28 29 30;
21 16 17 18 25 26 27;
22 16 17 18 28 29 30;
23 7 8 9 19 20 21;
24 10 11 12 22 23 24;
25 13 14 15 25 26 27];
coord=[-1.5 0 8;
1.5 0 8;
-1.5 1.5 4;
1.5 1.5 4;
1.5 -1.5 4;
-1.5 -1.5 4;
-4 4 0;
4 4 0;
4 -4 0;
-4 -4 0];
ex=zeros(25,2);
ey=zeros(25,2);
ez=zeros(25,2);
for i=1:25
ex(i,1)=coord((Edof(i,2)+2)/3,1);
ex(i,2)=coord((Edof(i,5)+2)/3,1);
ey(i,1)=coord((Edof(i,3)+1)/3,2);
ey(i,2)=coord((Edof(i,6)+1)/3,2);
ez(i,1)=coord(Edof(i,4)/3,3);
ez(i,2)=coord(Edof(i,7)/3,3);
end
K=zeros(75,75);
M=zeros(75);
for i=1:25
dx=ex(i,2)-ex(i,1) ;
dy= ey(i,2)-ey(i,1);
dz=ez(i,2)-ez(i,1) ;
L=sqrt(dx^2+dy^2+dz^2);
Kle=E*A/L*[1 -1;
-1 1];
l=dx/L;
m=dy/L;
n=dz/L;
G=[l m n 0 0 0;
0 0 0 l m n];
mass=L*A*rho;
me=[mass/2 0 0 0 0 0;
0 mass/2 0 0 0 0;
0 0 mass/2 0 0 0;
0 0 0 mass/2 0 0;
0 0 0 0 mass/2 0;
0 0 0 0 0 mass/2];
Ke=G'*Kle*G;
S=Edof(i,2:7);
K(S,S)=K(S,S)+Ke;
M(S,S)=M(S,S)+me;
end
nd=size(K,1);
fdof=[1:nd]';
bc=[25 26 22 24 19 21 28 30 ]';
fdof(bc(:))=[];
[X,D]=eig(K(fdof,fdof),M(fdof,fdof));
nfdof=size(X,1);
eigenval=diag(D);
eigenvec=zeros(nd,nfdof);
eigenvec(fdof,:)=X;
%top
for kk=1:3
eigenvalx=eigenval(kk);
eigenvecx=eigenvec(:,kk);
w=sqrt(eigenvalx);
T=2*pi/w;
dt=T/100;
t=0:dt:T;
figure(kk)
if kk==1;
filename = 'mode1_top.gif';
elseif kk==2;
filename = 'mode2_top.gif';
elseif kk==3;
filename = 'mode3_top.gif';
end
for jj = 1:100
tt=t(jj);
d=eigenvecx*sin(w*tt);
d=d';
rr=Edof(:,2:7);
ed=zeros(25,6);
for i=1:25;
ed(i,1:6)=d(rr(i,:));
end
xd=(ex+ed(:,[1 4]))';
yd=(ey+ed(:,[2 5]))';
zd=(ez+ed(:,[3 6]))';
s1=['-' , 'k'];
axis manual
plot(xd,zd,s1)
set(gca, 'ylim', [-12 12], 'xlim', [-12 12]);
drawnow
frame = getframe(1);
im = frame2im(frame);
[imind,cm] = rgb2ind(im,256);
if jj == 1;
imwrite(imind,cm,filename,'gif', 'Loopcount',inf);
else
imwrite(imind,cm,filename,'gif','WriteMode','append');
end
end
end
%side
for kk=1:3
eigenvalx=eigenval(kk);
eigenvecx=eigenvec(:,kk)
w=sqrt(eigenvalx)
T=2*pi/w;
dt=T/100
t=0:dt:T;
figure(kk)
if kk==1;
filename = 'mode1_side.gif'
elseif kk==2;
filename = 'mode2_side.gif'
elseif kk==3;
filename = 'mode3_side.gif'
end
for jj = 1:100
tt=t(jj);
d=eigenvecx*sin(w*tt);
d=d';
rr=Edof(:,2:7);
ed=zeros(25,6);
for i=1:25;
ed(i,1:6)=d(rr(i,:));
end
xd=(ex+ed(:,[1 4]))';
yd=(ey+ed(:,[2 5]))';
zd=(ez+ed(:,[3 6]))';
s1=['-' , 'k'];
axis manual
plot(xd,yd,s1)
set(gca, 'ylim', [-12 12], 'xlim', [-12 12]);
drawnow
frame = getframe(1);
im = frame2im(frame);
[imind,cm] = rgb2ind(im,256);
if jj == 1;
imwrite(imind,cm,filename,'gif', 'Loopcount',inf);
else
imwrite(imind,cm,filename,'gif','WriteMode','append');
end
end
end
Top and Side view for Mode 1:
Top and Side view for Mode 2:
Top and Side view for Mode 3:
References
editContributing Team Members
edit| Problem Assignments | ||
|---|---|---|
| Problem # | Solved & Typed by | Reviewed by |
| 1 | Zeyn Kermani | All |
| 2 | Matthew Gidel | All |
| 3 | Joshua Plicque | All |
| 4 | Brandon Wright,Spencer Herran | All |
| 5 | Kelvin Li,Matthew Gidel | All |
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.