I can Not include my condition

Hi everybody,
I want to put a condition in my code, This is the code:
for i = 1:ne
nodnum=zeros(1,nen);
for j = 1 : nen
check=Dof(:,1:nend)-ones(n,1)*Edof(i,(j-1)*nend+2:j*nend+1);
[indx,dum]=find(check==0);
nodnum(j)=indx(1);
end
Ex(i,:)=Coord(nodnum,1)';
if nsd>1
Ey(i,:)=Coord(nodnum,2)';
end
if nsd>2
Ez(i,:)=Coord(nodnum,3)';
end
Le(i)=sqrt((Ex(i,1)-Ex(i,2))^2+(Ey(i,1)-Ey(i,2))^2+(Ez(i,1)-Ez(i,2))^2);
and this is that I want to put as a condition,
(Le(i)/Ep(i,6))>200
please help me out

2 comentarios

Zoltán Csáti
Zoltán Csáti el 8 de Dic. de 2014
Why don't you just put that condition to the specific place with an if structure?
Hamid
Hamid el 8 de Dic. de 2014
Editada: Hamid el 8 de Dic. de 2014
Thank you Zoltán Csáti
I don't know how to do this because I want Ex,Ey,Ez obtain base on
(Le(i)/Ep(i,6))>200

Iniciar sesión para comentar.

 Respuesta aceptada

Thorsten
Thorsten el 8 de Dic. de 2014
if Le(i)/Ep(i,6) > 200
% do something
end

6 comentarios

Hamid
Hamid el 8 de Dic. de 2014
Editada: Hamid el 8 de Dic. de 2014
Thank you Thorsten
I want Ex,Ey,Ez obtain base on
(Le(i)/Ep(i,6))>200
Thorsten
Thorsten el 8 de Dic. de 2014
Editada: Thorsten el 8 de Dic. de 2014
Ok. But Le depends on Ex, Ey, Ez, so you have to compute Ex, Ey, Ez first. Then you can set the values to NaN (not a number):
if Le(i)/Ep(i,6) > 200
Ex(i,:)= nan(1, nen);
Ey(i,:)= nan(1, nen);
Ez(i,:)= nan(1, nen);
end
Hamid
Hamid el 8 de Dic. de 2014
Dear Thorsten
I get this error, What should I do?
??? Undefined function or method 'nans' for input arguments of type 'double'.
Error in ==> coordxtr at 54
Ex(i,:)= nans(1, nen);
Thank you very very much.
Thorsten
Thorsten el 8 de Dic. de 2014
Sorry, should be
nan
I have corrected it above.
Hamid
Hamid el 8 de Dic. de 2014
Dear Thorsten
Ex,Ey,Ez are coordinates for elements, They must be numbers otherwise it has no meaning.
Thanks.
Hamid
Hamid el 8 de Dic. de 2014
PLEASE HELP ME Thorsten

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Etiquetas

Preguntada:

el 8 de Dic. de 2014

Comentada:

el 8 de Dic. de 2014

Community Treasure Hunt

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

Start Hunting!

Translated by