Equation to Matlab code

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Lucca Martinelli
Lucca Martinelli on 5 Apr 2021
Hey, I need to write the following equations in MATLAB code. I can't find a mistake, but the displayed results are also not what I was expecting.
Could someone check for me maybe?
This is what I wrote:
y1 = (-W.*x1)./(384*E*I).*(16.*x1.^3-24*L.*x1.^2+9*L^2)+(M.*x1)./(6*E*I*L).*(x1.^2-3*L.*x1+2*L^2);
y2 = (-W*L)/(384*E*I).*(8.*x2.^3-24*L.*x2.^2+17*L^2.*x2-L^3)+(M.*x2)./(6*E*I*L).*(x2.^2-3*L.*x2+2*L^2);
  3 Comments
Lucca Martinelli
Lucca Martinelli on 5 Apr 2021
I am dealing with vectors but I believe that inser the dots in every important place, I will try putting some extra dots and check the difference. Thanks
Image Analyst
Image Analyst on 5 Apr 2021
  1. What were your input variable values?
  2. What did you get?
  3. What were you expecting?
Here is the pointing guideline FAQ again:
Also, I'll format your code as code by clicking on the Code icon, hopefully it's something you'll do yourself next time.

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Accepted Answer

John D'Errico
John D'Errico on 5 Apr 2021
Edited: John D'Errico on 5 Apr 2021
ALWAYS show what you wrote. Ony you know what you mean by dealing with vectors, etc. In the first case, do you have a vector of values for x, that varies from 0 to L/2?Are L, M, W, E, I all known constants with given values?
W = ???; % I've not filled these in because I have no idea what values you have
E = ???;
I = ???;
M = ???;
L = ???;
n = 100;
x1 = linspace(0,L/2,n);
First, you need to learn WHEN to use the dotted operators. You only need them when you are multiplying or dividing vectors or arrays of elements, or raising to powers. That tells MATLAB to perform element-wise operations.
However, you can multiply a vector by a scalar or multiply two scalars with no problem.
You can also divide a vector by a scalar using /, but you cannot divide a vector INTO a scalar using the / operator. That is if x is a vector, and a is a scalar, you can do these things using * or / :
x/a
x*a
a*x
because a is a scalar.
However to divide a scalar by a vector, you need to use ./ :
a./x
To raise elements to a power, you use .^ , thus you would do:
x.^a
a.^x
Raising a scalar to a scalar power is no problem. However
Addition and subtraction are always no problem. There is no .- or .+ operator to worry about.
Given all that, y1 should look like this:
y1 = (-W*x1)/(384*E*I).*(16*x1.^3-24*L*x1.^2+9*L^2)+(M*x1)/(6*E*I*L).*(x1.^2-3*L*x1+2*L^2);
  3 Comments
Walter Roberson
Walter Roberson on 5 Apr 2021
Well, it is a plot. We do not know what you were expecting.
If your constants are correct then y2 at L/2 is slightly negative.
L = 20;
E = 200e9;
I = 348e-6;
W = 5.4e3;
M = 200e3;
x1 = linspace(0,L/2);
x2 = linspace(L/2,L);
y1 = (-W.*x1)./(384*E*I).*(16.*x1.^3-24*L.*x1.^2+9*L^2)+(M.*x1)./(6*E*I*L).*(x1.^2-3*L.*x1+2*L^2);
y2 = (-W*L)/(384*E*I).*(8.*x2.^3-24*L.*x2.^2+17*L^2.*x2-L^3)+(M.*x2)./(6*E*I*L).*(x2.^2-3*L.*x2+2*L^2);
x = [x1 x2];
y = [y1 y2];
plot(x,y)
Lucca Martinelli
Lucca Martinelli on 5 Apr 2021
Thanks a lot!

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