[問題] 矩陣的疊加已回收
for i=1:Member_length
E=Material_Properties(i,1);
A=Cross_Sectional_Properties(i,1);
I=Cross_Sectional_Properties(i,2);
L=L_data(i,1);
i
K=[E*A/L 0 0 -E*A/L 0 0;
0 12*E*I/L^3 6*E*I/L^2 0 -12*E*I/L^L^3 6*E*I/L^2;
0 6*E*I/L^2 4*E*I/L 0 -6*E*I/L^2 2*E*I/L;
-E*A/L 0 0 E*A/L 0 0;
0 -12*E*I/L^3 -6*E*I/L^2 0 12*E*I/L^L^3 -6*E*I/L^2;
0 6*E*I/L^2 2*E*I/L 0 -6*E*I/L^2 4*E*I/L]
%以上算的為每根桿件的勁度矩陣in Global cor.
S=zeros(i);
S(Member(i,1),Member(i,1))=K(1,1);
S(Member(i,1),Member(i,2))=K(1,2);
S(Member(i,1),Member(i,3))=K(1,3);
S(Member(i,1),Member(i,4))=K(1,4);
S(Member(i,1),Member(i,5))=K(1,5);
S(Member(i,1),Member(i,6))=K(1,6);
%以上為member stiffness matrix 1th row
S(Member(i,2),Member(i,1))=K(2,1);
S(Member(i,2),Member(i,2))=K(2,2);
S(Member(i,2),Member(i,3))=K(2,3);
S(Member(i,2),Member(i,4))=K(2,4);
S(Member(i,2),Member(i,5))=K(2,5);
S(Member(i,2),Member(i,6))=K(2,6);
%以上為member stiffness matrix 2th row
S(Member(i,3),Member(i,1))=K(3,1);
S(Member(i,3),Member(i,2))=K(3,2);
S(Member(i,3),Member(i,3))=K(3,3);
S(Member(i,3),Member(i,4))=K(3,4);
S(Member(i,3),Member(i,5))=K(3,5);
S(Member(i,3),Member(i,6))=K(3,6);
%以上為member stiffness matrix 3th row
S(Member(i,4),Member(i,1))=K(4,1);
S(Member(i,4),Member(i,2))=K(4,2);
S(Member(i,4),Member(i,3))=K(4,3);
S(Member(i,4),Member(i,4))=K(4,4);
S(Member(i,4),Member(i,5))=K(4,5);
S(Member(i,4),Member(i,6))=K(4,6);
%以上為member stiffness matrix 4th row
S(Member(i,5),Member(i,1))=K(5,1);
S(Member(i,5),Member(i,2))=K(5,2);
S(Member(i,5),Member(i,3))=K(5,3);
S(Member(i,5),Member(i,4))=K(5,4);
S(Member(i,5),Member(i,5))=K(5,5);
S(Member(i,5),Member(i,6))=K(5,6);
%以上為member stiffness matrix 5th row
S(Member(i,6),Member(i,1))=K(6,1);
S(Member(i,6),Member(i,2))=K(6,2);
S(Member(i,6),Member(i,3))=K(6,3);
S(Member(i,6),Member(i,4))=K(6,4);
S(Member(i,6),Member(i,5))=K(6,5);
S(Member(i,6),Member(i,6))=K(6,6);
%以上為member stiffness matrix 6th row
S
%對應到每根桿件各個code number
end
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小弟的問題是現在每個i值都會對應出一個S矩陣
不知道有沒有寫法將每個 i值對應的S矩陣疊成一個大的 SS矩陣
SS=S1+S2+S3+.......
麻煩各位高手了
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