Local Stiffness Matrix Negative

3 vues (au cours des 30 derniers jours)
Ramses Young
Ramses Young le 10 Avr 2022
Hey Everyone,
I am trying to construct a small 5 element, 2 noded finite element model of a tapered beam using matlab.
I currently have an acceptable answer, but I was wondering if there was any way that the clever people of the forum may have a more elegant solution.
Currently I am creating a global stiffness matrix using a connectivity matrix, describing the nodal connections. Since it is a beam, the nodes just line up with eachother simply.
In my nested loop I currently have the code at the bottom as a way of looping and checking if it is the corner of the sub-matrix 2x2. Do you all know if there is a better way of doing this so i dont have to use a if statement? It seems excessive.
B=[1 2;2 3;3 4;4 5;5 6]; %connectivity matrix
E=10e6 %youngs modulus, psi
L=24 %length of beam, inches
kg=zeros(6,6)
A=[11449/15625; 7921/15625; 5041/15625; 2809/15625; 49/625]; %area of cross section at each node
for e=1:5 %number of elements
for i=1:length(B(1,:)) %number of nodes per element
for j=1:length(B(1,:)) %number of nodes per element
if i==2 & j==2
k=A(e)*E/(L/5);
elseif i==2 & j==1
k=-A(e)*E/(L/5);
elseif i==1 & j==2
k=-A(e)*E/(L/5) ;
else
k=A(e)*E/(L/5);
end
kg(B(e,i),B(e,j))=kg(B(e,i),B(e,j))+k;
end
end
end

Réponses (1)

Alan Stevens
Alan Stevens le 11 Avr 2022
How about replacing
if i==2 & j==2
k=A(e)*E/(L/5);
elseif i==2 & j==1
k=-A(e)*E/(L/5);
elseif i==1 & j==2
k=-A(e)*E/(L/5) ;
else
k=A(e)*E/(L/5);
end
by
k=A(e)*E/(L/5)*((i==j) - (i~=j));

Catégories

En savoir plus sur Programming dans Help Center et File Exchange

Produits


Version

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by