Help executing calculation in for loop

Hello,

I am pretty new to matlab but have to use it for a class. I am currently trying to calculate the maximum force in a member using a moment equation for a range of angles.

I have created a for loop to run through my range of angles to calculate the component of the force in the y direction. When alpha (angle) is 0 the component will just be 1 instead of multiplied by the cosine of the angle because it will be entirely in the y direction.

I created a table to display the alpha angle and what the force should be muitiplied by to find the y component. So it has two columns and a bunch of rows.

I now need to be able to calculate the Force variable out of a moment equation using these values that I calculated. I wrote a program like this alreday that only calculates the component and force for one alpha angle instead of many using this:

ME = ((W*dW)-(FDC*cosd(myalpha)*dCB))==0;
FDC = round((solve(ME,FDC)),3);

Where W, dW, dCB, and myalpha are all known values.

Now I am trying to do that same thing but for a range. This is what my loop looks like:

for alpha = -43.43:1:43.43
    if alpha == 0
        FDCm = 1;
    elseif alpha > 0
        FDCm = cosd(alpha);
    elseif alpha < 0
        FDCm = cosd(alpha);
    end
    T = [T; alpha FDCm];
end
T

I need to do what I did for one angle for many angles and have it output into a table as well. I have tried many different things and keep getting errors no matter what I try. Any help is appreciated!

[Jan]

Whenever you mention an error in the forum, post a copy of the error message.

I do not understand, what you are asking for. The loop can be simplified to:

alpha = (-43.43:43:43).';
T     = [alpha, cosd(alpha)];

There is no need for the IF branchs.

[Serina Robbins]

I tried doing that but it didn't give me what I want. I am sorry I am trying to figure out how to explain what I need to someone not in my class.

This is my entire code:

clc

lBG = 13.5;
dG = 1.75;
lBH = 34;
dH = 3;
alpha = -43.43:1:43.43;
dCB = 5;
lBE = 32;
W = 450;
dW = 16;

lGH = sqrt((((lBG-dH).^2)+((lBH-dG).^2)-(2*(lBG-dH)*(lBH-dG)*cosd(90+alpha))));
d = alpha;

plot(lGH,d);

lGHmin = min(lGH);
lGHmax = max(lGH);
fprintf('Fully extended hydraulic cylinder:%.2f\n ',lGHmax);
fprintf('Fully retracted hydraulic cylinder:%.2f\n ',lGHmin);

syms W dW FDC dCB Ey Ez By Bz FGH FDCm ME

T = [];
for alpha = -43.43:1:43.43
    if alpha == 0
        FDCm = 1;
    elseif alpha > 0
       FDCm = cosd(alpha);
    elseif alpha < 0
     FDCm = cosd(alpha);
    end
    T = [T; alpha FDCm];
end

And this is what it outputs:

Fully extended hydraulic cylinder:40.11
 Fully retracted hydraulic cylinder:26.17
 
T =

  -43.4300    0.7262
  -42.4300    0.7381
  -41.4300    0.7498
  -40.4300    0.7612
  -39.4300    0.7724
  -38.4300    0.7834
  -37.4300    0.7941
  -36.4300    0.8046
  -35.4300    0.8148
  -34.4300    0.8248
  -33.4300    0.8346
  -32.4300    0.8440
  -31.4300    0.8533
  -30.4300    0.8622
  -29.4300    0.8710
  -28.4300    0.8794
  -27.4300    0.8876
  -26.4300    0.8955
  -25.4300    0.9031
  -24.4300    0.9105
  -23.4300    0.9175
  -22.4300    0.9243
  -21.4300    0.9309
  -20.4300    0.9371
  -19.4300    0.9430
  -18.4300    0.9487
  -17.4300    0.9541
  -16.4300    0.9592
  -15.4300    0.9640
  -14.4300    0.9685
  -13.4300    0.9727
  -12.4300    0.9766
  -11.4300    0.9802
  -10.4300    0.9835
   -9.4300    0.9865
   -8.4300    0.9892
   -7.4300    0.9916
   -6.4300    0.9937
   -5.4300    0.9955
   -4.4300    0.9970
   -3.4300    0.9982
   -2.4300    0.9991
   -1.4300    0.9997
   -0.4300    1.0000
    0.5700    1.0000
    1.5700    0.9996
    2.5700    0.9990
    3.5700    0.9981
    4.5700    0.9968
    5.5700    0.9953
    6.5700    0.9934
    7.5700    0.9913
    8.5700    0.9888
    9.5700    0.9861
   10.5700    0.9830
   11.5700    0.9797
   12.5700    0.9760
   13.5700    0.9721
   14.5700    0.9678
   15.5700    0.9633
   16.5700    0.9585
   17.5700    0.9533
   18.5700    0.9479
   19.5700    0.9422
   20.5700    0.9362
   21.5700    0.9300
   22.5700    0.9234
   23.5700    0.9166
   24.5700    0.9095
   25.5700    0.9021
   26.5700    0.8944
   27.5700    0.8864
   28.5700    0.8782
   29.5700    0.8698
   30.5700    0.8610
   31.5700    0.8520
   32.5700    0.8427
   33.5700    0.8332
   34.5700    0.8234
   35.5700    0.8134
   36.5700    0.8031
   37.5700    0.7926
   38.5700    0.7818
   39.5700    0.7708
   40.5700    0.7596
   41.5700    0.7481
   42.5700    0.7365

So I add my moment equation to be able to solve for FCD and my loop becomes:

   for alpha = -43.43:1:43.43
    if alpha == 0
        FDCm = 1;
    elseif alpha > 0
       FDCm = cosd(alpha);
    elseif alpha < 0
     FDCm = cosd(alpha);
    end
    T = [T; alpha FDCm];
    ME = ((W*dW)-(FDC*FDCm*dCB))==0;
    FDC = solve(ME,FDC);

end

But I get the eorror message:

 Operator '*' is not supported for operands of type 'struct'.

Error in trial (line 35)
    ME = ((W*dW)-(FDC*FDCm*dCB))==0;

That is the equation I need to use and I need to be able to solve for FCD

[Jan]

I mention it again:

    if alpha == 0
        FDCm = 1;
    elseif alpha > 0
       FDCm = cosd(alpha);
    elseif alpha < 0
     FDCm = cosd(alpha);
    end
    

is exactly the same as:

    FDCm = cosd(alpha);

Find out, which of the variables is a struct, e.g. by uisng the debugger and the command whos

[Serina Robbins]

Okay thank you, I got that to work realizing I forgot to delete part of it the first time I tried to run it. Now that I have cosd(alpha) for all the angles I need in my range, how can I make a calculation using the new variable?

I tried doing this:

for alpha = -43.43:1:43.43
    FDCm = cosd(alpha);
    FDC = ((W*dW)/(FDCm*dCB));
end

But I keep getting this:

FDC =
 
(2251799813685248*W*dW)/(1658341119971255*dCB)

When I actually need FCD for every alpha value in my range.

Is there a way to do that calculation and put it as another column in my table? I can't get it to calculate anything but that one line.

[David Hill]

Look at my answer below. You should not use symbolics. No loop is needed.

Answer 1

[David Hill]

myalpha=-43.43:1:43.43;
for k=1:length(myalpha)
  FDC(k)=round((W.*dW)/cosd(myalpha(k))./dCB,3);%I am assuming that W, dW, and dCB are all the same size or are scalar
end

[Serina Robbins]

I made that my loop but all I get is

ans =
 
0.001*round((1125899906842624000*W*dW)/(817645190541711*dCB))

But I need it as a number instead of variables and I need a different answer for every alpha value

[David Hill]

No loop is needed. Solve function and symbolics are not needed. Solving for FDC algebraically is straight forward.

lBG = 13.5;
dG = 1.75;
lBH = 34;
dH = 3;
alpha = -43.43:1:43.43;
dCB = 5;
lBE = 32;
W = 450;
dW = 16;
lGH = sqrt((((lBG-dH).^2)+((lBH-dG).^2)-(2*(lBG-dH)*(lBH-dG)*cosd(90+alpha))));
d = alpha;
plot(lGH,d);
lGHmin = min(lGH);
lGHmax = max(lGH);
FDC=round(W*dW./cosd(alpha)/dCB,3);
T=[alpha',cosd(alpha)',FDC'];

[Serina Robbins]

Thank you, that worked for me. When I display the table it is inscientific notation so its displaying like -0.0434.. How can I stop this? And I will probably be back with more questions as this is my first step sadly.

[David Hill]

help format

[Serina Robbins]

That worked. Now I believe this will be my last question.. I need to find the maximum FDC, which I did using max(FDC) but I need to find the alpha that corresponds to that max(FDC) value. How can I do that? There is more than one so I need a way that can output more than one answer.

I tried using the indexOfMaxFDC but it is giving me the max cosd(alpha) instead of alpha.

Thank you for all your help!

[David Hill]

[~,idx]=max(FDC);
alpha(idx)

[Serina Robbins]

That gives me on value but not the other, is there a way to give me the other?