please see this code and tel me where is the problem beacause i don't find it

hold on
%Nouveau programme le 21/02/2014
%%%%%%%%%%%%%Données physique du probleme%%%%%%%%%%%%%%%%%%%%%%%%%%
M=30; % masse de la poutre KG/ML??
EI=0.20*10^7; % Rigidité flexionnelle de la poutre N/m^2
ksi=0.4;
k=0.04*10^8; % rigidité du sol N/m^2
v=27.77; %vitesse m/s de chargement
n1=5;
c=0.05;
%nombre des modes
l=2*(3*pi)/2*(4*EI/k)^0.25; %m;longueur elastique de la poutre en m
%%%%%%%%%%%%%%%% Les conditions initiales %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
v0=0.001; %m/s;vitesse initiale
T0=0.0001; %m déplacement initiale
t1=l/v; %temps de parcours de la longueuer elastique
ww1=0;
Q=1; %chargement en Newton
n=23; % Nombre de points d'analyse
f=-21:1:21; %creation du vecteu englobant toute la longueur élastique
beta=1;
for m=1:5
wn(m)=((((m^4)*(pi^4)*EI)/(M*(l^4)))+(k/M))^0.5; % Pulsation propre du systeme
wd(m)=(wn(m)^2-beta^2)^0.5; % pseudo pulsation
omega(m)=m*pi*v/l;
for i=1:n
f(i)=i;
xx(i)=l/(n-1);
xx1(i)=xx(i)*(i-1);
phi(i,m)=sin(m*pi*(xx1(i))/l);
for j=1:n
tt(j)=l/v/(n-1);
tt1(j)=tt(j)*(j-1);
x1(j)=v*tt1(j)
A1=2*Q/(M*l);
A2(m)=1/(wd(m)^4+2*((beta^2-wd(m)^2)*omega(m)^2)+omega(m)^4); %(((w_d(m))^4)+2*((((ksi*w_d)^2)+((w_n(m)^2)))*(mu(m)^2))+mu(m)^4);
A3(m)=2*beta*omega(m);
A4(m)=(omega(m)/wd(m))*(((beta)^2)-(wd(m)^2)+omega(m)^2);
A5(m)=-A3(m);
A6(m)=wd(m)^2-omega(m)^2;
T11(j,m)=exp(-beta*(tt1(j)));
T1(j,m)=A1*A2(m)*(T11(j,m)*(A3(m)*cos(wd(m)*tt1(j))+A4(m)*sin(wd(m)*tt1(j)))+A5(m)*cos(omega(m)*tt1(j))+A6(m)*sin(omega(m)*tt1(j)));
ww1(i,j,m)=phi(i,m)*T1(j,m);
end
end
end
for i=1:n;
for j=1:n;
ww1(i,j)=0;
for m=1:n1;
ww1(i,j)=ww1(i,j)+ww1(i,j,m);
%ww1(i,j) = sum(ww1(i,j,:));
end
end
end
for i=1:n
www(i)=ww1(i,j)
j=5
end
%plot(www,'g')
for j=1:n
www1(j)=ww1(i,j)
i=12
end
plot(www1,'r')
%plot(ww1(10,:))

2 Comments

What difference is there between what you expect and what shows up when you run the program?
thank you to be Interest; Yes the curve is breaken and i have to plot it fonction of x the beam lenght not fonction the number of element 'i' and

Sign in to comment.

 Accepted Answer

The program is functioning well , here is the result :

2 Comments

yes but i dn't have the result i want the cuve is breaken
you mean the result interpretation? ok the code is technically correct, the problem then is in the constants, verify the constants /or change : c, ksi and the velocity v .

Sign in to comment.

More Answers (0)

Categories

Find more on Numerical Integration and Differential Equations in Help Center and File Exchange

Tags

No tags entered yet.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!