hi, it says "Unable to perform assignment because the size of the left side is 1-by-3 and the size of the right side is 1-by-2" and the error is at line 96 in my code

% Units used: 
LUnit = 'ft';			% Length = ft or m
FUnit = 'lb';			% Force = lb or kip or N or kN
%.. Input problem data
nNodes = 6;              % Number of nodes
nMembers = 9;           % Number of members
nReactions = 2;         % Number of vector force equations
nLoads = 3;             % Number of applied loads
nDim = 2;               % Dimension 2=2D, 3=3D
nDOF = nDim*nNodes;      % Number of degrees of freedom
tol  = 1.e-8;           % Remainder tolerance on norm of residual force

%.. Input coordinates of each node
Node(1,:) = [0,0];
Node(2,:) = [3,3];
Node(3,:) = [6,6];
Node(4,:) = [9,9];
Node(5,:) = [18,0];
Node(6,:) = [9,0];

%.. Input connections/members
%.. First input is the start node; Second input is the end node
Member(1,:) = [1,2];
Member(2,:) = [1,6];
Member(3,:) = [2,6];
Member(4,:) = [2,3];
Member(5,:) = [3,6];
Member(6,:) = [3,4];
Member(7,:) = [4,6];
Member(8,:) = [4,5];
Member(9,:) = [5,6];

%.. Input support reactions    
%.. First input is the node number, second input is resistance in the e1
%.. direction, third input is resistance in the e2 direction
%.. 1 for support in that direction, 0 for no support
Reaction(1,:) = [1,1,1];
Reaction(2,:) = [5,0,1];

%.. Input loads  
%.. First input is the node number, remaining inputs are the magnitude of
%.. the force in the e1 or e2 direction.
Load(1,:) = [2,500,-866.025];
Load(2,:) = [3,866.025,-500];
Load(3,:) = [4,1000,0];
%%  Compute the unit vector (direction) for every member
connect=zeros(nMembers,nDim);
Length=zeros(nMembers,1);
PosVec = zeros(nMembers, 3);		%Initialize Position Vector matrix
UnVec = zeros(nMembers, 3);		%Initialize Unit Vector matrix
OppUnVec = zeros(nMembers, 3);	%Initialize opposite direction matrix
SupportReaction = zeros(3, nNodes);
for i = 1 : nMembers
    SN = Member(i,1); 
    EN = Member(i,2); 
    % *****  This section is where the core of your CP3 code should go.  *****
    
    connect(i,:) = Node(EN,:) -Node(SN,:);
    Length(i) = norm(connect(i,:));
    PosVec(i,: ) = connect(i,:);
    UnVec(i,:) = connect(i,:) / Length(i); 
    OppUnVec(i,:) = -PosVec(i,:) / norm(PosVec(i,:));
end

Unable to perform assignment because the size of the left side is 1-by-3 and the size of the right side is 1-by-2.

%% The coefficient matrix, C, contains the unit vectors (direction vectors)
% for each member force associated with each node.  The unit vectors are
% then separated into e1, e2, e3 components and placed in separate rows. 
% Therefore, each node has 2 rows (for 2-D) or 3 rows (for 3-D) associated
% with it; one for each of the e1, e2, e3 components.
% This corresponds with the Degrees of Freedom, nDOF, which is 2*nNodes
% for 2-D or 3*nNodes for 3-D.
% So the coefficient matrix has 2 or 3 rows for each node and one column
% for each member force.
C = zeros(3*nNodes, nMembers);	%Initialize coefficient matrix
% Loop through all nodes, create the vector force equation for that node
% and store it in the proper row of the coefficient matrix, C.
for i = 1 : nNodes
    for j = 1 : nMembers
        SN = Member(j,1); 
        EN = Member(j,2);
        if (SN == i)  % If member j starts at node i
            C(3*i-2, j) = UnVec(j,1);
            C(3*i-1, j) = UnVec(j,2);
            C(3*i, j) = UnVec(j,3);
        elseif (EN == i)  % If member j ends at node i
            C(3*i-2, j) = OppUnVec(j,1);
            C(3*i-1, j) = OppUnVec(j,2);
            C(3*i, j) = OppUnVec(j,3);
        end  % if SN
    end  % for nMembers
end  % for nNodes
%% The External Force matrix, F, contains the magnitude of all externally
% applied loads, (input as e1, e2, e3 components), stored in the
% proper rows to corespond with the node it is applied at.
% Therefore, the F matrix has 2 or 3 rows for each node and one column.
F = zeros(3*nNodes, 1);		%Initialize External Force matrix
% Loop through all externally applied loads and place each component of the
% load (e1, e2, e3) in the proper rows in the external load matrix, F.
for i = 1 : nLoads
    j = Load(i,1);		% Which node, j is the load i on
    F(3*j-2) = Load(i,2);
    F(3*j-1) = Load(i,3);
    F(3*j) = Load(i,4);
end  % for i,nLoads
%% The Restraint matrix, Crest, contains the unit vectors (directions)
% for each member force associated with those nodes that are restrained by
% external supports.
% The Crest matrix has one row for each reaction component and one column
% for each member.
%nReactions=1;
%.... Loop through all nodes and determine if there is a reaction at that
%.... node and seperate the reaction node equations from the force load
%.... equations.
nReaction = 1;       %Used to count the number of reaction equations
%.... The nReactionEquation vector determines if a row should be put in the
%.... Crest matrix or the Cfree matrix
nReactionEquation = zeros(1, 3*nNodes);
for j = 1 : nReactions
    for i = 1 : nNodes
        if (Reaction(j,1) == i)
            if (Reaction(j,2) == 1)
                Crest(nReaction,:) = C(3*i-2,:);
                nReactionEquation(3*i-2) = 1;       
                % If nReactionEquation is 1 it should be in the Crest
                % matrix and not the Cfree matrix.
                nReaction = nReaction+1;
            end
            if (Reaction(j,3) == 1)
                Crest(nReaction,:) = C(3*i-1,:);
                nReactionEquation(3*i-1) = 1;
                % If nReactionEquation is 1 it should be in the Crest
                % matrix and not the Cfree matrix.
                nReaction = nReaction+1;
            end
            if (Reaction(j,4) == 1)
                Crest(nReaction,:) = C(3*i,:);
                nReactionEquation(3*i) = 1;
                % If nReactionEquation is 1 it should be in the Crest
                % matrix and not the Cfree matrix.
                nReaction = nReaction+1;
            end
        end
    end
end
nForce = 1;             %Used to count the number of Cfree equations
for (i = 1 : 3*nNodes)
    if (nReactionEquation(i) == 1)
    else
        Cfree(nForce,:) = C(i,:);
        Ffree(nForce) = F(i);
        nForce = nForce + 1;
    end
end
%% Solve first for the member forces and then solve for the support
% reactions.
% All of the values in the Cfree and Ffree matrices are known so you can
% now solve for the T matrix, which is the tension force in each member.
T = Cfree \ -Ffree';      % finish this equation
% Since all values in the Crest and T matrices are now known, you can solve
% for the support reactions.
Reactions = -1*Crest * T;       % finish this equation
% Calculate the residual to find the amount of error and ensure that
% equilibrium was satisfied.
Residual1 = (Cfree * T + Ffree');
Residual2 = (Crest * T + Reactions);
Res = norm(Residual1) + norm(Residual2);
% Create the support reaction matrix.
nReaction = 1;
for i = 1 : nNodes
    if (nReactionEquation(3*i-2) == 1)
        SupportReaction(i,1) = i;
        SupportReaction(i,2) = Reactions(nReaction);
        nReaction = nReaction+1;   
    end
    if (nReactionEquation(3*i-1) == 1)
        SupportReaction(i,1) = i;
        SupportReaction(i,3) = Reactions(nReaction);
        nReaction = nReaction+1;   
    end
    if (nReactionEquation(3*i) == 1)
        SupportReaction(i,1) = i;
        SupportReaction(i,4) = Reactions(nReaction);
        nReaction = nReaction+1;   
    end
end
%% Create output for command window 
fprintf('%s\n'     ,  '----------------------------------------')
fprintf('%s\n'     ,  '-------------    Truss    --------------')
fprintf('%s\n'     ,  '----------------------------------------')
fprintf('%s',' Node Coordinates (', LUnit, ')')
fprintf ('\n')
for i = 1 : nNodes
    fprintf('%s%4i%8.3f%8.3f%8.3f\n','   Node: ', i, Node(i,:)')
end % for i, nNodes
fprintf ('\n')
fprintf('%s',' Member Forces (', FUnit, ')')
fprintf ('\n')
for i = 1 : nMembers
    fprintf('%s%4i%12.3f\n','   Member Force: ',i, T(i)) 
end % for i, nMembers
fprintf ('\n')
for i = 1 : nReactions
    for j = 1 : nNodes
        if (Reaction(i,1) == j)
            fprintf('%s%s%s%4i%12.3f%12.3f%12.3f\n',...
                '   Support Reaction (',FUnit,') at Node ',...
                SupportReaction(j,:)) 
            fprintf ('\n')
        end % if Reaction
    end % for j, nNodes
end % for i, nReactions
fprintf('\n%s%8.3f\n','   Residual Error: ', Res)
fprintf('\n\n\n')


%% Create the plot............................................
% Once your code is modified the plot will be 3-D and so you may need the
% "Rotate 3D" command to view the truss figure properly.
%.. Plot members with color indicating tension or compression 
Marker = ceil(25 / sqrt(nNodes));
fig1 = figure(1); clf; grid on; axis equal; 
hold on;
xlabel(cat(2, 'X  (',LUnit,')' )); 
ylabel(cat(2, 'Y  (',LUnit,')' )); 
zlabel(cat(2, 'Z  (',LUnit,')' ));
title('Truss Analysis');
small = max(T)*tol;
for m = 1 : nMembers
    if (T(m) < -small)
        MColor = 'b';       % Color compression members blue
    elseif (T(m) > small)
        MColor = 'r';       % Color tension members red
    else
        MColor = 'k';       % Color "zero force members" black
    end % if T
    SN = Member(m,1); 
    EN = Member(m,2); 
    X = [Node(SN,1); Node(EN,1)];
    Y = [Node(SN,2); Node(EN,2)];
    Z = [Node(SN,3); Node(EN,3)];
    p = plot3(X,Y,Z);
    set(p, 'Color', MColor, 'LineWidth', 6/nDim);
end % for m, nMembers

%.. Plot external reaction forces
Rlength = 0.1 * max(max(Node));  % establish size of line
e = [ 1, 0, 0 ; 0, 1, 0 ; 0, 0, 1];
for i = 1 : nReactions
    for j = 1 : nNodes
        if (Reaction(i,1) == j)
            for k = 1 : 3
                if(Reaction(i, k+1) == 1)
                    xR(1,:) = Node(j,:);
                    xR(2,:) = Node(j,:) - Rlength * e(k, 1:nDim);
                    xR(3,:) = Node(j,:) - Rlength * e(k, 1:nDim);
                    p = plot3(xR(:,1),xR(:,2),xR(:,3));
                    set(p,'Color','g','LineWidth',8/nDim);
                end % if Reaction == 1
            end % for k
        end % if Reaction == j
    end % for j, nNodes
end % for i, nReactions
%.. Plot loads
Llength = 0.1 * max(max(Node));  % establish size of line
Fmax = max(max(abs(Load)));     % establish maximum force level
e = [ 1, 0, 0 ; 0, 1, 0 ; 0, 0, 1];      
for i = 1 : nLoads
    for j = 1 : nNodes
        if (Load(i,1) == j)
            for k = 1 : 3
                F = Load(i, k+1) / Fmax;
                xL(1,:) = Node(j,:);
                xL(2,:) = Node(j,:) + Llength * F * e(k, 1:nDim);
                xL(3,:) = Node(j,:) + Llength * F * e(k, 1:nDim); 
                p = plot3(xL(:,1), xL(:,2),xL(:,3));
    			set(p,'Color','c','LineWidth',8/nDim);
            end % for k
        end % if Load == j
    end % for j, nNodes
end % for i, nLoads

%.. Plot nodes      
plot3(Node(:,1), Node(:,2), Node(:,3),'o', 'LineWidth', 2,...
    'MarkerFaceColor','w',...
    'MarkerEdgeColor','k',...
    'MarkerSize',Marker);
view(3);

% End of Program

Answer 1

[Walter Roberson]

Node(1,:) = [0,0];
Node(2,:) = [3,3];
Node(3,:) = [6,6];
Node(4,:) = [9,9];
Node(5,:) = [18,0];
Node(6,:) = [9,0];

That creates Node as a 6 x 2 matrix.

nMembers = 9;           % Number of members
nDim = 2;               % Dimension 2=2D, 3=3D
connect=zeros(nMembers,nDim);

That creates connect as a 9 x 2 matrix.

    connect(i,:) = Node(EN,:) -Node(SN,:);

With Node(En,:) being a 1 x 2 matrix and Node(SN,:) being a 1 x 2 matrix, the size of the right hand side is 1 x 2, and it is being stored in a left hand side that is 1 x 2, so that part is fine.

    PosVec = zeros(nMembers, 3);		%Initialize Position Vector matrix

That creates PosVec as a 9 x 3 matrix.

    PosVec(i,: ) = connect(i,:);

We showed above that connect(i,:) is a 1 x 2 matrix. You are trying to store that 1 x 2 matrix into PosVec(i,:) which is a 1 x 3 matrix. There are not enough values on the right hand side to match the size of the left hand side.

    UnVec(i,:) = connect(i,:) / Length(i); 

That line has the same problem.