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me3255_finalproject_group29/problem_3.m
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% 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') |