# contour map for stress distribution

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Cem Eren Aslan on 31 May 2021
Commented: DGM on 31 May 2021
I want to plot a contour map. It relates to stress distribution of a circular area. My trial as follows:
X =linspace(-4.5,4.5);
R=linspace(-4.5,4.5) ;
[X,Y] = meshgrid(X,Y);
Z= 3426.8/63.3+660*X*4/3.14/4.5^4+1700*sqrt(R^2-X^2)*4/3.14/4.5^4;
contourf(X,Y,Z)
Q1) How can i obtain continuous contour and its scale

Sulaymon Eshkabilov on 31 May 2021
HI,
Here is a corrected code:
X =linspace(-4.5,4.5);
R=linspace(-4.5,4.5) ;
[X,Y] = meshgrid(X,R);
Z= 3426.8/63.3+660*X*4/3.14/4.5^4+1700*sqrt(Y.^2-X.^2)*4/3.14/4.5^4;
contour3(X,Y,abs(Z))

DGM on 31 May 2021
Edited: DGM on 31 May 2021
I don't know what you're doing, but this at least runs:
X = linspace(-4.5,4.5);
R = linspace(-4.5,4.5);
[X,Y] = meshgrid(X,R);
Z = 3426.8/63.3 + 660*X*4/3.14/4.5^4 + 1700*sqrt(Y.^2-X.^2)*4/3.14/4.5^4;
numlevels = 20; % set to whatever you want
contourf(X,Y,abs(Z),numlevels,'edgecolor','none') I doubt that's what you want, but it's what I could guess from what you wrote. If you figure out what you're trying to plot, you can use contourf() with however many levels you want to get more Z-resolution. At some point though, it stops being useful as a contour map. Otherwise, you can just display Z using imagesc().
EDIT:
I decided to adopt Sulaymon's interpretation of intent. That makes a bit more sense.
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DGM on 31 May 2021
That still doesn't make sense. You have a plot that's R on one axis and X on the other. As far as your coordinate setup indicates, you have a sectional plot. Like I said. I doubt it's what you want, but it's what you wrote. I don't know how you want to make this into something else.
If you can represent your information in polar coordinates, you could do something like this: