How do I save the output array (y2) for each loop?

How do I save the output array (y2) for each loop?

At the end, I would like a 3x10 array stored in the variable ysum. Where each row is the unique calculation based on the "c" count.

I've been going at this for 6 hours and can't seem to get it.

Thanks in advanced.

Fmultiple = input('Enter multiple concentrated forces [N] in array format: ');
amultiple = input(sprintf('Enter multiple locations (0 - %0.0f) [m]: ', L));

z = length(amultiple); % this is the length of the amultiple user vector input

% a new blank vector must be created to which outputs in the loops will be
% appended
ysum = [];% this will store all the values of the deflection calculations
c = 0;
for i = 1:z; % this loop from spot 1 of the array inputed all the way to the end
    a = amultiple(i) % indexes i for amultiple array
    F = Fmultiple(i) % indexes i for Fmultiple array
    c = c + 1
    y = [] % reset y each round
    I = (w .* (h .^3)) ./ 12;
    R = (F ./ L) .* (L - a);
    theta = ((F .* a) ./ (6 .* E .* I .* L)) .* ((2 .* L) - a) .* (L - a);
    if c > 0
        y1 = [];
        y2 = [];
        x = 0:L;
        for n = 1:length(x);
            if x(n) <= a
                def = (-1 .* theta .* x(n)) + ((R * x(n) .^3) ./ (6 .* E .* I));
                y1 = [y1, def];
            else
                def = (-1 .* theta .* x(n)) + ((R * x(n) .^3) ./ (6 .* E .* I)) - ((F ./ (6 .* E .* I)) .* ((x(n) - a) .^3));
                y2 = [y1, def];
                ysum %<===       %MARKED
            end
        end
    end
end

Answer 1

[Lauren Michaud]

I was able to get help that lead to a solution.

For completeness, i'm posting it here.

loop = 1;
while loop == 1
% Task 4

% what is the objective of this for loop
% for multiple forces, each element of the array is it's own point
% for for F = [800, 350, 200] a = [3, 5, 8]; 800 is the F, 3 is the a
% the output of this array will be concatenated with the next round which
% will be F = 350, a = 5
% lastly F = 200, a = 8
% this will yield 3 rows
% two for loops will be needed

Fmultiple = input('Enter multiple concentrated forces [N] in array format: ');
amultiple = input(sprintf('Enter multiple locations (0 - %0.0f) [m]: ', L));

z = length(amultiple); % this is the length of the amultiple user vector input

% a new blank vector must be created to which outputs in the loops will be
% appended
%c = 0;
%ysum = zeros(1,L) % create a zero table that is the size of the expected array
Msum = 0;
for i = 1:z; % this loop from spot 1 of the array inputed all the way to the end
    a = amultiple(i); % indexes i for amultiple array
    F = Fmultiple(i); % indexes i for Fmultiple array
    %c = c + 1
    M = []; % reset y each round
    I = (w .* (h .^3)) ./ 12;
    R = (F ./ L) .* (L - a);
    theta = ((F .* a) ./ (6 .* E .* I .* L)) .* ((2 .* L) - a) .* (L - a);
    %if c > 0
     %   y = []
        x = linspace(0,L,1000);
        for n = 1:length(x)
            if x(n) <= a
                def = (-1 .* theta .* x(n)) + ((R * x(n) .^3) ./ (6 .* E .* I));
                M = [M, def];
            else
                def = (-1 .* theta .* x(n)) + ((R * x(n) .^3) ./ (6 .* E .* I)) - ((F ./ (6 .* E .* I)) .* ((x(n) - a) .^3));
                M = [M, def];
            end
        end
        Msum = (Msum + M);
       % ysum(c,:) = y 
       %   end
end
MaxMin = abs(min(Msum))*1000; % finds the greatest minimum value and takes the absolute value of it
fprintf('The maximum deflection of the beam is %0.3f [mm].', MaxMin);
   
figure(2)
plot(x,(Msum.*1000));
x = linspace(0,L);
grid on;
xlabel('Length of the Beam [m]')
ylabel('Deflection [mm]')
title(sprintf('Deflection of %s undermultiple concentrated loads',Material(material_task1)))

loop = menu('Do you wish to repeat multiple loads calculation?', "Yes", "No");
end

Answer 2

[Walter Roberson]

Under what circumstance could c be <= 0 such that the if would fail ?

amultiple = input(sprintf('Enter multiple locations (0 - %0.0f) [m]: ', L));

That tells us that L is a length in meters.

        x = 0:L;

That tells us that L is an integer count, not a fractional length.

       for n = 1:length(x);

The way you constructed x, x is always going to be integers, and n is always going to be x+1

         if x(n) <= a

Why bother with x there when you could just use n? If n<=a+1 ?

                 ysum %<===       %MARKED

could you confirm that should only be assigned into if the if fails?

I would like a 3x10 array stored in the variable ysum

Where does the 3 come from? Where does the 10 come from?

Your a values are obviously expected to be different. Why would you expect the same number of columns of output for the different a values ?

[Lauren Michaud]

Thanks for your quick response.

I used

c > 0

just as a trigger to make sure that the if statement runs. It's not meant to fail - but always run

"L" is indeed a length in meters. This length can change and it's not hard coded, so this code permits "L" to vary.

for n = 1:length(x)

performs the loop for

x = 0:L

this was simplified from the original

x = 0:0.1:L

So it evaluates the equations in the loop at 0.1 increments the entire distance of L.

I chose a 3x10 array because the loop will output a 1x10 array, but it will do it a total of 3 times as it moves along user inputed array which was 1x3. I made it simple so I can debug a 1x10 array vs a 1x1000 array

After some thought, I have somewhat figure out a solution

ysum(c,:) = y2

The problem is the above code runs when y2 is a 1x5 array and not yet a 1x10 array, and so i'm getting an error

Error: Unable to perform assignment because the size of the left side is 1-by-10 and the size of the right side is 1-by-5.

The goal is for the output in the for loop to be saved each time and not overwrite the previous loop.

There should be 3 unique arrays of 1 x 10 dimension.

Here is the full code.

loop = 1;
while loop == 1
% Task 4

% what is the objective of this for loop
% for multiple forces, each element of the array is it's own point
% for for F = [800, 350, 200] a = [3, 5, 8]; 800 is the F, 3 is the a
% the output of this array will be concatenated with the next round which
% will be F = 350, a = 5
% lastly F = 200, a = 8
% this will yield 3 rows
% two for loops will be needed

Fmultiple = input('Enter multiple concentrated forces [N] in array format: ');
amultiple = input(sprintf('Enter multiple locations (0 - %0.0f) [m]: ', L));

z = length(amultiple); % this is the length of the amultiple user vector input

% a new blank vector must be created to which outputs in the loops will be
% appended
c = 0;
ysum = zeros(z,L) % create a zero table that is the size of the expected array

for i = 1:z; % this loop from spot 1 of the array inputed all the way to the end
    a = amultiple(i) % indexes i for amultiple array
    F = Fmultiple(i) % indexes i for Fmultiple array
    c = c + 1
    y = [] % reset y each round
    I = (w .* (h .^3)) ./ 12;
    R = (F ./ L) .* (L - a);
    theta = ((F .* a) ./ (6 .* E .* I .* L)) .* ((2 .* L) - a) .* (L - a);
    if c > 0
        y1 = [];
        y2 = [];
        x = 0:L;
        for n = 1:length(x);
            if x(n) <= a
                def = (-1 .* theta .* x(n)) + ((R * x(n) .^3) ./ (6 .* E .* I));
                y1 = [y1, def];
            else
                def = (-1 .* theta .* x(n)) + ((R * x(n) .^3) ./ (6 .* E .* I)) - ((F ./ (6 .* E .* I)) .* ((x(n) - a) .^3));
                y2 = [y1, def];
             
            end
            
        end
        
    end
    ysum(c,:) = y2
end
loop = menu('Do you wish to repeat multiple loads calculation?', "Yes", "No");
end

Thanks again for your help. Hope I made this clear.

I've been working on this little piece of script for 12 hours now. 1 hour per character. lol

ysum(c,:) = y2