Application of For Next Statement?

Question

Sagar Trehan 2015-2-3

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https://ww2.mathworks.cn/matlabcentral/answers/174940-application-of-for-next-statement

评论： Michael Haderlein 2015-2-4

Hi Guys,

I am unable plot the graphs for the Stress Intensity Functions using For Next Loop. Hers is the script, can anybody explain me the fallacy for the code.

Code:

    %------------Stress Plot---------------%
    theta=0:0.1:360
    n=length(theta)
    m=1/3
    for x=1:n
        a=((0.25).*(1+cos(x)+(1.5).*(sin(x)).^2));
        b=((0.25).*((1-(2.*m.^2)).*(1+cos(x))+(1.5.*(sin(x)).^2)));    
    end 
    figure
    polar(theta,a,'b')
    hold on
    polar(theta,b,'r')

Thanks & Regards

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Answer 1

Michael Haderlein 2015-2-3

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https://ww2.mathworks.cn/matlabcentral/answers/174940-application-of-for-next-statement#answer_166847

编辑：Michael Haderlein 2015-2-3

Please make your code readable using the {}Code button on top of the edit window in future.

You are overwriting a in every loop iteration. Make it a(x) =... and it will work (the same with b).

Edit: I just realized you span theta from 0 to 360. sin and cos are radian based, so either span theta from 0 to 2*pi or use sind and cosd functions.

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Sagar Trehan 2015-2-4

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Hi Michael,

Thanks for your response. I am posting two different codes for the same output. With first code I am getting correct output but I am trying the same with For Loop but output is not as expected.

if true
%-----------------------------------------------------------------%

%------Von Mises Stress Plot for Crack Tip Plasticity-------------%

%-----------------------------------------------------------------%

theta=0:0.001:2*pi; ms=(1+cos(theta)+1.5.*(sin(theta)).^2).*(0.25) vs=(((1-(2.*(1/3)))^2).*(1+cos(theta))+((3/2).*((sin(theta)).^2)))*(0.25) ts=(((cos(theta./2)).^2).*((1+sin(theta./2)).^2)) rs=(((cos(theta./2)).^2).*((1-(2.*(1/3))+(sin(theta./2)))).^2) figure polar(theta,ms,'r') hold on polar(theta,vs,'b') hold off figure polar(theta,ts,'b') hold on polar(theta,rs,'g')

    % code
  end

2nd Code:

if true%------------Stress Plot---------------%

theta=0:0.001:2*pi n=length(theta) m=1/3

for x=1:n a(x)=((0.25).*(1+cosd(x)+(1.5).*(sind(x)).^2)); b(x)=((0.25).*((1-(2.*m.^2)).*(1+cosd(x))+(1.5.*(sind(x)).^2))); end figure polar(theta,a,'b') hold on polar(theta,b,'r') % code end

Pls refer the second code for the plot generation. I want to understand where I have done the mistake in understanding.

Regards, Sagar Trehan

Michael Haderlein 2015-2-4

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Really, the code is barely readable without the code formatting. That's most likely the reason why I didn't see one important point in your code: You use x as argument of the trigonometric functions. It must be theta(x), of course. Anyway, you don't even need a loop:

a=((0.25).*(1+cos(theta)+(1.5).*(sin(theta)).^2));

will work for all theta at once.

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