資源簡介
給出了三種動態子結構方法的程序,對比了自由界面模態綜合法,固定界面模態綜合法,雙協調自由界面模態綜合法三種方法。-Shows three dynamic substructure method of procedure, compared to the free interface modal synthesis method, fixed-interface modal synthesis method, two-coordinate free interface modal synthesis method of three methods.

代碼片段和文件信息
%%%%%%%%%%%%%雙協調自由界面
clear;
b1=10;h1=10;b2=0.7;h2=0.7;
l1=50;A1=100;M1=4e5;EI1=1220;den1=M1/l1;I1=b1*(h1^3)/12;E1=EI1/I1;E1=7e10;
n1=50;%the?number?of?the?segments?in?length;
le1=l1/n1;
l2=5;A2=0.49;M2=3.1e4;EI2=218;den2=M2/l2;I2=b2*(h2^3)/12;E2=EI2/I2;E2=7e10;
n2=5;
le2?=?l2?/?n2;
%尋找單元的節點號
for?i?=?1:n1
????Element1(i:)?=?[i?1?+?i];
end
for?i?=?1:n2
????Element2(i:)?=?[i?1?+?i];
end
%單元節點的坐標號
for?i?=?1:n1
????for?j?=?1:2
????????U1(((j-1)?*?3?+?1):(j?*?3)i)?=?[(Element1(ij)?*?3?-2);(Element1(ij)?*?3?-1);(Element1(ij)?*?3)];
????end
end
for?i?=?1:n2
????for?j?=?1:2
????????U2(((j-1)?*?3?+?1):(j?*?3)i)?=?[(Element2(ij)?*?3?-2);(Element2(ij)?*?3?-1);(Element2(ij)?*?3)];
????end
end
Node1?=?3*(n1?+?1);%總的位移坐標數
Node2?=?3*(n2?+?1);
%邊界點號
BN1?=?[2?5];?BN2=[2?5];
syms?x?y?z
X1?=?1?-?(x/le1);X2=x/le1;
g1=X1;g2=1-3*X2*X2+2*(X2^3);g3=le1*(X2-2*(X2^2)+X2^3);g4=X2;g5=3*(X2^2)-2*(X2^3);g6=le1*(-X2^2+X2^3);
N1=[g1?0?0?g4?0?0];N2=[0?g2?g3?0?g5?g6];N=[N1;N2];
NN1?=?transpose(N)?*?N;
%單元質量陣
intx?=?int(NN1?x0?le1);
Me1?=?double(den1??*?intx);Me2?=?double(den2??*?intx);
B1=diff(N1x);B2=-y*diff(N2x2);B=[B1;B2];
BB1=B‘*E1*B;
intx=int(BB1x0le1);
inty=int(intxy-h1/2h1/2);
intz=int(intyz0b1);
Ke1?=?double(intz);
BB2=B‘*E2*B;
intx=int(BB2x0le2);
inty=int(intxy-h2/2h2/2);
intz=int(intyz0b2);
Ke2?=?double(intz);
%組集(同一根梁內的組集)
MM1?=?zeros(Node1Node1);MM2?=?zeros(Node2Node2);
KK1?=?zeros(Node1Node1);KK2?=?zeros(Node2Node2);
for?k?=?1:n1
????for?i?=?1:6
????????for?j?=?1:6
????????????MM1(U1(ik)U1(jk))?=?MM1(U1(ik)U1(jk))?+?Me1(ij);
????????????KK1(U1(ik)U1(jk))?=?KK1(U1(ik)U1(jk))?+?Ke1(ij);
????????end
????end
end
for?k?=?1:n2
????for?i?=?1:6
????????for?j?=?1:6
????????????MM2(U2(ik)U2(jk))?=?MM2(U2(ik)U2(jk))?+?Me2(ij);
????????????KK2(U2(ik)U2(jk))?=?KK2(U2(ik)U2(jk))?+?Ke2(ij);
????????end
????end
end
M1?=?MM1;M2=MM2;
K1?=?KK1;K2=KK2;
e1=eye(length(MM1));
e1=[e1;e1(4:5:);e1(13:14:)];
e1(13:14:)=[];e1(4:5:)=[];
MM1=e1*MM1*inv(e1);KK1=e1*KK1*inv(e1);
M1?=?MM1;
K1?=?KK1;%153行列
B=zeros(length(MM1)1);B(end-1)=1;B=e1*B;
C1=zeros(1length(MM1));C1(end-1)=1;C1=C1*inv(e1);
C2=zeros(1length(MM1));
e2=eye(length(MM2));
e2=[e2;e2(4:5:);e2(13:14:)];
e2(13:14:)=[];e2(4:5:)=[];
MM2=e2*MM2*inv(e2);KK2=e2*KK2*inv(e2);
M2=MM2;
K2=KK2;%18行列
%計算頻率和振型
disp(‘正在計算特征值……‘);
[V1D1]?=?eig(K1M1);%K1*V1=M1*V1*D1
[V2D2]?=?eig(K2M2);
w1=eig(K1M1).^(1/2);w2=eig(K2M2).^(1/2);%所有特征值組成的向量
u1=w1/6.28;u2=w2/6.28;
%B=V1(:1:n1+1)‘*B;
%C1=C1*V1(:1:n1+1);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%求剩余模態
nn1=length(M1);nn2=length(K2);%nn1=153???nn2=18
Bo1=zeros(nn1);
Bo1(:5:nn1)=[];
Bo1(nn1-31)=1;
Bo1(nn1-22)=1;
Bo1(nn1-13)=1;
Bo1(nn14)=1;
Fk1=V1(:1:n1+1);
Fd1=V1(:n1+2:nn1);
Kd1=Fd1‘*K1*Fd1;
pasi1=Fd1*inv(Kd1)*Fd1‘*Bo1;
Bo2=zeros(nn2);
Bo2(:5:nn2)=[];
Bo2(nn2-
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件???????4546??2011-07-26?14:05??ziyoujiemian.m
?????文件???????5421??2011-07-26?16:21??gaijinjiangjie.m
?????文件???????5631??2011-07-26?15:13??gudingjiemian.m
-----------?---------??----------?-----??----
????????????????15598????????????????????3
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