problem_3.m

% Problem 3
clear
clc
setdefaults
% Plug in different values for q to find corresponding deflections.
q = [1,10,20,30,50];

%Plug in different values for P to find corresponding deflections.
P= [0,100,200,300];

% Young's Modulus for aluminum [Pascals]:
E = 70e9;
% Width of beam [meters]:
b = 0.1;
% Height of beam [meters]:
h = 0.01;
% Second moment of inertia [meters^4]:
I = (b*h.^3)/12;

% Create matrix for plotting for each segment length
w6_max_plot=zeros(4,5);
w10_max_plot=zeros(4,5);
w20_max_plot=zeros(4,5);

for s= 1:4

% Plotting q vs. max deflection:
for i = 1:5
    %----------------------------------------------------------------------
   % Define the 6 segment matrix:


 % Segment length for 6 segments
    h_6 = 1/6;
    A_6 = zeros(5);
    b_6 = ((h_6^4*q(i))/(E*I))*ones(5,1);

    % Diagonal vectors of the A matrix:
    ones_vect_6 = ones(3,1);
    fours_vect_6 = -4*ones(4,1);
    sixes_vect_6 = 6*ones(5,1);
    % Create A matrix by summing the diagonal vectors.
    A_6 = diag(ones_vect_6,-2) + diag(fours_vect_6,-1) + diag(sixes_vect_6) + diag(ones_vect_6,2) + diag(fours_vect_6,1);
    A_6(1,1) = 5;
    A_6(5,5) = 5;

    %define Matrix due to force P>0;
    P_6=zeros(5);

    %Diagonal vectors of the P matrix:
    twos_vect_6=-2*ones(5,1);
    ones_vect_6=ones(4,1);
    % Creating diagonal P matrix by summing diagonal vectors.
    P_6=diag(twos_vect_6,0)+diag(ones_vect_6,-1)+diag(ones_vect_6,1);

    %multiply matrix by (Ph^2)/(EI)
    P_6=((P(s)*(h_6)^2)/(E*I))*P_6;

    %define final matrix as A_6 - P_6
    A_6=A_6-P_6;

    %find unkown w for each segment
    w_6 = inv(A_6) * b_6;
    w6_max(i) = max(w_6);
 %----------------------------------------------------------------------
    % Define the 10 segment matrix:
    % Segment length for 10 segments
    h_10 = 1/10;
    A_10 = zeros(9);
    b_10 = ((h_10^4*q(i))/(E*I))*ones(9,1);

    % Diagonal vectors of the A matrix:
    ones_vect_10 = ones(7,1);
    fours_vect_10 = -4*ones(8,1);
    sixes_vect_10 = 6*ones(9,1);
    % Create A matrix by summing the diagonal vectors.
    A_10 = diag(sixes_vect_10)+diag(fours_vect_10,-1)+diag(fours_vect_10,1)+diag(ones_vect_10,-2)+diag(ones_vect_10,2);
    A_10(1,1) = 5;
    A_10(9,9) = 5;

    % define Matrix due to force P>0;
    P_10=zeros(9);

    %Diagonal vectors of the P matrix:
    twos_vect_10=-2*ones(9,1);
    ones_vect_10=ones(8,1);
    % Creating diagonal P matrix by summing diagonal vectors.
    P_10=diag(twos_vect_10,0)+diag(ones_vect_10,-1)+diag(ones_vect_10,1);

    %multiply matrix by (Ph^2)/(EI)
    P_10=((P(s)*(h_10)^2)/(E*I))*P_10;

    %define final matrix as A_10 - P_10
    A_10=A_10-P_10;

    %find unkown w for each segment

    w_10 = inv(A_10) * b_10;
    w10_max(i) = max(w_10);

%----------------------------------------------------------------------
    % Define the 20 segment matrix:
    % Segment length for 20 segments
    h_20=1/20;

    A_20 = zeros(19);
    b_20 = ((h_20^4*q(i))/(E*I))*ones(19,1);
    % Diagonal vectors of the A matrix:
    sixes_vect_20 = 6*ones(1,19);
    fours_vect_20 = -4*ones(1,18);
    ones_vect_20 = ones(1,17);
    % Create A matrix by summing the diagonal vectors.
    A_20 = diag(sixes_vect_20)+diag(fours_vect_20,-1)+diag(fours_vect_20,1)+diag(ones_vect_20,-2)+diag(ones_vect_20,2);
    A_20(1,1) = 5;
    A_20(19,19) = 5;

    %define Matrix due to force P>0;
    P_20=zeros(19);

    %Diagonal vectors of the P matrix:
    twos_vect_20=-2*ones(19,1);
    ones_vect_20=ones(18,1);
    % Creating diagonal P matrix by summing diagonal vectors.
    P_20=diag(twos_vect_20,0)+diag(ones_vect_20,-1)+diag(ones_vect_20,1);

    %multiply matrix by (Ph^2)/(EI)
    P_20=((P(s)*(h_20)^2)/(E*I))*P_20;

    %define final matrix as A_20 - P_20
    A_20=A_20-P_20;

    %find unkown w for each segment
    w_20 = inv(A_20) * b_20;
    w20_max(i) = max(w_20);
    %----------------------------------------------------------------------
end
%define max deformation matrix values for the current load
w6_max_plot(s,:)=w6_max;
w10_max_plot(s,:)=w10_max;
w20_max_plot(s,:)=w20_max;

end

% Plot q vs. max deflection for each segment size.  There will be three
% different graphs, each corresponding to a different segment size.
% each graph  will show 4 different lines representing the four different
% applied loads.


figure
plot(w6_max_plot(1,:),q,'-o',w6_max_plot(2,:),q,'-o',w6_max_plot(3,:),q,'-o',w6_max_plot(4,:),q,'-o');
title(['Maximum Deflection for 6 segments \newline      at Varying Distributed Loadings'])
xlabel('\delta_{x} [m]')
ylabel('q [N/m]')
legend('\delta_{x} for P=0', '\delta_{x} for P=100','\delta_{x} for P=200','\delta_{x} for P=300','Location','southeast')

figure
plot(w10_max_plot(1,:),q,'-o',w10_max_plot(2,:),q,'-o',w10_max_plot(3,:),q,'-o',w10_max_plot(4,:),q,'-o');
title(['Maximum Deflection for 10 segments \newline     at Varying Distributed Loadings']);
xlabel('\delta_{x} [m]')
ylabel('q [N/m]')
legend('\delta_{x} for P=0', '\delta_{x} for P=100','\delta_{x} for P=200','\delta_{x} for P=300','Location','southeast')

figure
plot(w20_max_plot(1,:),q,'-o',w20_max_plot(2,:),q,'-o',w20_max_plot(3,:),q,'-o',w20_max_plot(4,:),q,'-o');
title(['Maximum Deflection for 20 segments \newline     at Varying Distributed Loadings'])
xlabel('\delta_{x} [m]')
ylabel('q [N/m]')
legend('\delta_{x} for P=0', '\delta_{x} for P=100','\delta_{x} for P=200','\delta_{x} for P=300','Location','southeast')