Equation to Matlab code

2 次查看（过去 30 天）
Lucca Martinelli
Lucca Martinelli 2021-4-5
评论： Lucca Martinelli 2021-4-5
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
显示 1更早的评论
Lucca Martinelli
Lucca Martinelli 2021-4-5
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 2021-4-5
  1. What were your input variable values?
  2. What did you get?
  3. What were you expecting?
Here is the pointing guideline FAQ again:
TUTORIAL: How to ask a question (on Answers) and get a fast answer
Also, I'll format your code as code by clicking on the Code icon, hopefully it's something you'll do yourself next time.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

John D'Errico
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
显示 1更早的评论
Walter Roberson
Walter Roberson 2021-4-5
Ran in:
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)

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Mathematics 的更多信息

标签

产品


版本

R2020b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by