我在MATLAB中几乎是新手。 我完全不知道如何更改我的代码以正确执行。 这是关于分析3d桁架的代码，其中增加了对特殊点的负载。
function D=DataT3D
%m number of elements
%n number of nodes
m=25;n=10;
%coordinates of nodes [(X Y Z) for each node]
Coord=[-37.5 0 200;37.5 0 200;-37.5 37.5 100;37.5 37.5 100;37.5 -37.5
100;-37.5 -37.5 100;
-100 100 0;100 100 0;100 -100 0;-100 -100 0];
%conection of the nodes [first in coordinates is the first node and ...]
Con=[1 2;1 4;2 3;1 5;2 6;2 4;2 5;1 3;1 6;3 6;4 5;3 4;5 6;3 10;6 7;4 9;5 8;4
7;3 8;5 10;6 9;
6 10;3 7;4 8;5 9]; Con(:,3:4)=0;
%Re degrees of freedom for each node (free=0 & fixed=1)
Re=ones(n,6);
Re(1:6,1:3)=zeros(6,3);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% concentrated loads on nodes
Load=zeros(n,6);
Load([1,2,3,6],1:3)=[1 -10 -10;0 -10 -10;0.5 0 0;0.6 0 0];
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% uniform loads in local coordinate system
w=zeros(m,3);
% E: material elastic modules G:shear elastic modules J:torsional constant
E=ones(1,m)*1e4;nu=0.3;G=E/(2*(1+nu));
% A:cross sectional area and Iy Iz: moment of inertia
A=ones(1,m)*0.5;Iz=ones(1,m);Iy=ones(1,m);J=ones(1,m);
%St: settlement of supports & displacements of free nodes
St=zeros(n,6); be=zeros(1,m);
% All of the variables are transposed and stored in a structure array in the
name of D
D=struct('m',m,'n',n,'Coord',Coord','Con',Con','Re',Re',...
'Load',Load','w',w','E',E','G',G','A',A','Iz',Iz','Iy',...
Iy','J',J','St',St','be',be');
end
此代码作为函数运行另一个代码。
function [Q,V,R]=MSA(D);
m=D.m;
n=D.n;
% the matrix to store K*T for each member 12*12*m
Ni=zeros(12,12,m);
% global stiffness matrix of the structure 6n*6n
S=zeros(6*n);
% element fixed end forces in global coordinate 6n*1
Pf=S(:,1);
% internal forces and moments in local coordinate system for each member
% 12*m
Q=zeros(12,m);
% element fixed end forces in local coordinate for each member 12*m
Qfi=Q;
% member code numbers* (mcn) in global stiffness matrix for each member
% 12*m
Ei=Q;
for i=1:m
% connectivity and release of the both member ends 4*1
H=D.Con(:,i);
% difference of beginning and end nodes coordinates 3*1
C=D.Coord(:,H(2))-D.Coord(:,H(1));
% member code numbers (mcn) in global stiffness matrix for a member
% 1*12
e=[6*H(1)-5:6*H(1),6*H(2)-5:6*H(2)];
c=D.be(i);
[a,b,L]=cart2sph(C(1),C(3),C(2));
ca=cos(a); sa=sin(a); cb=cos(b); sb=sin(b); cc=cos(c); sc=sin(c);
r=[1 0 0;0 cc sc;0 -sc cc]*[cb sb 0;-sb cb 0;0 0 1]*[ca 0 sa;0 1 0;-sa 0
ca];
% transformation matrix related to the
% coordinate transformation which in considering member orientation
% 12*12
T=kron(eye(4),r);
co=2*L*[6/L 3 2*L L];
x=D.A(i)*L^2; y=D.Iy(i)*co; z=D.Iz(i)*co;
g=D.G(i)*D.J(i)*L^2/D.E(i);
% local stiffness matrix for each member
K1=diag([x,z(1),y(1)]);
K2=[0 0 0;0 0 z(2);0 -y(2) 0];
K3=diag([g,y(3),z(3)]);
K4=diag([-g,y(4),z(4)]);
K=D.E(i)/L^3*[K1 K2 -K1 K2;K2' K3 -K2' K4;-K1 -K2 K1 -K2;K2' K4 -K2' K3];
% uniform loads in local coordinate system for each member 1*3
w=D.w(:,i)';
% local fixed-end force vector for a member, corresponding to external
% loads 12*1
Qf=-L^2/12*[6*w/L 0 -w(3) w(2) 6*w/L 0 w(3) -w(2)]';
% local fixed-end force vector for a member, corresponding to support
% displacements 12*1
Qfs=K*T*D.St(e)';
A=diag([0 -0.5 -0.5]);
B(2,3)=1.5/L;
B(3,2)=-1.5/L;
W=diag([1,0,0]);
Z=zeros(3);
M=eye(12);
p=4:6;
q=10:12;
% type of member release* 0 1 2 3
% M: A matrix for modifying stiffness matrix and local fixed-end force
vector of a released member ends such K=M*K , Qf=M*Qf and Qfs=M*Qfs
switch 2*H(3)+H(4)
case 0;B=2*B/3; % released at both ends
M(:,[p,q])=[-B -B;W Z;B B;Z W];
case 1; % released at the beginning
M(:,p)=[-B;W;B;A];
case 2; % released at the end
M(:,q)=[-B;A;B;W];
end
K=M*K;Ni(:,:,i)=K*T;
% global stiffness matrix of the structure 6n*6n
S(e,e)=S(e,e)+T'*Ni(:,:,i);
Qfi(:,i)=M*Qf;
% element fixed end forces in global coordinate 6n*1
Pf(e)=Pf(e)+T'*M*(Qf+Qfs);
% member code numbers* (mcn) in global stiffness matrix for each member
% 12*m
Ei(:,i)=e;
end
% Deflections in global coordinate syste 6*n
V=1-(D.Re|D.St);
% f: A vector that indicates the number of degree of freedom ndof*1
f=find(V);
V(f)=S(f,f)\(D.Load(f)-Pf(f));
% Supports reactions in global coordinate system 6*n
R=reshape(S*V(:)+Pf,6,n);
R(f)=0;V=V+D.St;
for i=1:m
% internal forces and moments in local coordinate system 12*m
Q(:,i)=Ni(:,:,i)*V(Ei(:,i))+Qfi(:,i);
end
所以这是问题 如何将负载更改为动态表单？现在代码是静态的。这意味着现在我们将一个集中的负载放到一个节点上，我们得到了答案，例如基于增加的集中负载的其他节点的位移。但是我们如何将其更改为动态加载形式，这意味着我们将一个具有不同时间值（例如5秒）的负载添加到节点，并在不同的时间步骤中获得其他节点的答案。
您可以使用以下命令运行第二个代码：
d = DataT3D; [Q，V，R] = MSA（d）; **
需要帮助。
答案 0 :(得分：0)
试试这个：
<强> DataT3D
function D=DataT3D
%m number of elements
%n number of nodes
m=25;n=10;
%coordinates of nodes [(X Y Z) for each node]
Coord=[-37.5 0 200;37.5 0 200;-37.5 37.5 100;37.5 37.5 100;37.5 -37.5 100;-37.5 -37.5 100;
-100 100 0;100 100 0;100 -100 0;-100 -100 0];
%conection of the nodes [first in coordinates is the first node and ...]
Con=[1 2;1 4;2 3;1 5;2 6;2 4;2 5;1 3;1 6;3 6;4 5;3 4;5 6;3 10;6 7;4 9;5 8;4 7;3 8;5 10;6 9;
6 10;3 7;4 8;5 9];
Con(:,3:4)=0;
%Re degrees of freedom for each node (free=0 & fixed=1)
Re=ones(n,6);
Re(1:6,1:3)=zeros(6,3);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% concentrated loads on nodes
record = [1 2 3 4 5 6 7 8 9 10]; % Example forces on node 1 on different time intervals
% Create storage for Q, V and R
allQ = cell(2,1);
allV = cell(2,1);
allR = cell(2,1);
for t=1:length(record)
Load = [record(t) 0 0 0 0 0; zeros(n*1,6)]; % Load has only a Fx and all other forces and moments are zero
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% uniform loads in local coordinate system
w=zeros(m,3);
% E: material elastic modules G:shear elastic modules J:torsional constant
E=ones(1,m)*1e4;nu=0.3;G=E/(2*(1+nu));
% A:cross sectional area and Iy Iz: moment of inertia
A=ones(1,m)*0.5;Iz=ones(1,m);Iy=ones(1,m);J=ones(1,m);
%St: settlement of supports & displacements of free nodes
St=zeros(n,6); be=zeros(1,m);
% All of the variables are transposed and stored in a structure array in the
%name of D
D=struct('m',m,'n',n,'Coord',Coord','Con',Con','Re',Re',...
'Load',Load','w',w','E',E','G',G','A',A','Iz',Iz','Iy',...
Iy','J',J','St',St','be',be');
[allQ{t},allV{t},allR{t}]=MSA(D); % Save the results for Q, V and R
end
allQ = cell2mat(allQ)
allV = cell2mat(allV)
allR = cell2mat(allR)
end
<强> MSA
function [Q,V,R]=MSA(D)
m=D.m;
n=D.n;
% the matrix to store K*T for each member 12*12*m
Ni=zeros(12,12,m);
% global stiffness matrix of the structure 6n*6n
S=zeros(6*n);
% element fixed end forces in global coordinate 6n*1
Pf=S(:,1);
% internal forces and moments in local coordinate system for each member
% 12*m
Q=zeros(12,m);
% element fixed end forces in local coordinate for each member 12*m
Qfi=Q;
% member code numbers* (mcn) in global stiffness matrix for each member
% 12*m
Ei=Q;
for i=1:m
% connectivity and release of the both member ends 4*1
H=D.Con(:,i);
% difference of beginning and end nodes coordinates 3*1
C=D.Coord(:,H(2))-D.Coord(:,H(1));
% member code numbers (mcn) in global stiffness matrix for a member
% 1*12
e=[6*H(1)-5:6*H(1),6*H(2)-5:6*H(2)];
c=D.be(i);
[a,b,L]=cart2sph(C(1),C(3),C(2));
ca=cos(a); sa=sin(a); cb=cos(b); sb=sin(b); cc=cos(c); sc=sin(c);
r=[1 0 0;0 cc sc;0 -sc cc]*[cb sb 0;-sb cb 0;0 0 1]*[ca 0 sa;0 1 0;-sa 0 ca];
% transformation matrix related to the
% coordinate transformation which in considering member orientation
% 12*12
T=kron(eye(4),r);
co=2*L*[6/L 3 2*L L];
x=D.A(i)*L^2; y=D.Iy(i)*co; z=D.Iz(i)*co;
g=D.G(i)*D.J(i)*L^2/D.E(i);
% local stiffness matrix for each member
K1=diag([x,z(1),y(1)]);
K2=[0 0 0;0 0 z(2);0 -y(2) 0];
K3=diag([g,y(3),z(3)]);
K4=diag([-g,y(4),z(4)]);
K=D.E(i)/L^3*[K1 K2 -K1 K2;K2' K3 -K2' K4;-K1 -K2 K1 -K2;K2' K4 -K2' K3];
% uniform loads in local coordinate system for each member 1*3
w=D.w(:,i)';
% local fixed-end force vector for a member, corresponding to external
% loads 12*1
Qf=-L^2/12*[6*w/L 0 -w(3) w(2) 6*w/L 0 w(3) -w(2)]';
% local fixed-end force vector for a member, corresponding to support
% displacements 12*1
Qfs=K*T*D.St(e)';
A=diag([0 -0.5 -0.5]);
B(2,3)=1.5/L;
B(3,2)=-1.5/L;
W=diag([1,0,0]);
Z=zeros(3);
M=eye(12);
p=4:6;
q=10:12;
% type of member release* 0 1 2 3
% M: A matrix for modifying stiffness matrix and local fixed-end force
%vector of a released member ends such K=M*K , Qf=M*Qf and Qfs=M*Qfs
switch 2*H(3)+H(4)
case 0;B=2*B/3; % released at both ends
M(:,[p,q])=[-B -B;W Z;B B;Z W];
case 1; % released at the beginning
M(:,p)=[-B;W;B;A];
case 2; % released at the end
M(:,q)=[-B;A;B;W];
end
K=M*K;Ni(:,:,i)=K*T;
% global stiffness matrix of the structure 6n*6n
S(e,e)=S(e,e)+T'*Ni(:,:,i);
Qfi(:,i)=M*Qf;
% element fixed end forces in global coordinate 6n*1
Pf(e)=Pf(e)+T'*M*(Qf+Qfs);
% member code numbers* (mcn) in global stiffness matrix for each member
% 12*m
Ei(:,i)=e;
end
% Deflections in global coordinate syste 6*n
V=1-(D.Re|D.St);
% f: A vector that indicates the number of degree of freedom ndof*1
f=find(V);
D.Load(f);
V(f)=S(f,f)\(D.Load(f)-Pf(f));
% Supports reactions in global coordinate system 6*n
R=reshape(S*V(:)+Pf,6,n);
R(f)=0;V=V+D.St;
for i=1:m
% internal forces and moments in local coordinate system 12*m
Q(:,i)=Ni(:,:,i)*V(Ei(:,i))+Qfi(:,i);
end