Equation to Matlab code
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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 个评论
Walter Roberson
2021-4-5
just in case you deal with vectors, change all * to .* and all / to ./
Also it can be rewriten in terms of a new variable that is x/L in which case the pieces become 0 to 1/2 and 1/2 to 1
Lucca Martinelli
2021-4-5
Image Analyst
2021-4-5
- What were your input variable values?
- What did you get?
- 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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John D'Errico
2021-4-5
编辑：John D'Errico
2021-4-5
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 个评论
Lucca Martinelli
2021-4-5
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
2021-4-5
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