contour plot real numbers

3 views (last 30 days)
Derek Reitnouer
Derek Reitnouer on 19 Feb 2019
Commented: Derek Reitnouer on 25 Feb 2019
I wrote a program that just takes an input of x and y values and calculates the principal stresses. It keeps giving me an error about input arguments for contour must be real. All the variables are squared so I am not sure how I would be getting this problem. Thanks.
clc
clear all
% Load
P = 1;
% Length of Beam
L = 1;
% Height of Beam from Neutral Axis
c = 0.1;
x = linspace(0,L);
y = linspace(-c,c);
[X,Y] = meshgrid(x,y);
sigmaP1 = ((3*P/(4*(c^3)))*[X*Y+[((c^2-Y^2)^2)+(X^2)*(Y^2)]^(1/2)]);
contourf(X,Y,sigmaP1,'ShowText','on')

Accepted Answer

Adam Danz
Adam Danz on 19 Feb 2019
Edited: Adam Danz on 19 Feb 2019
The values of 'sigmaP1 ' are imaginary. You can identify this just by looking at those values.
sigmaP1 (1:5)
ans =
Columns 1 through 5
1282.8 + 1.9693e-08i 1282.8 + 8.1557e-09i 1282.8 + 2.4166e-08i 1282.8 - 2.1095e-07i 1282.8 - 2.3987e-08i
Or by using isreal()
isreal(sigmaP1)
ans =
logical
0 %which means that your data contains at least 1 imaginary number
In fact, none of the elements of sigmaP1 are real numbers as confirmed by
any(imag(sigmaP1(:))==0)
ans =
logical
0 %no real numbers found
  12 Comments
Adam Danz
Adam Danz on 21 Feb 2019
This is telling. Since my break-down produces the same exact results as the full equation, that confirms that the break-down is correct. But since the break-down results to not match your expected results, that tells us that the original equation is wrong (or your expected results are wrong).
I think it's a great idea to reduce the number of inputs to ~5 so you can manage the comparison. I haven't read through your code above and I didn't see a specific follow-up question but if you get stuck in this trouble-shooting process, let me know.
Derek Reitnouer
Derek Reitnouer on 25 Feb 2019
Thats correct, the breakdown does match the full equation. Maybe I'm not understanding the way meshgrid works. If I go down to 3 variables, the big X equals a 3x3 matrix consisting of 0, 0.5, 1. The big Y is the same with variables -0.1,00.1. The output for sigmP1(1,1) should be calculated using the values X=0 and Y = -0.1. The equation you wrote down is the correct equations and it should return 0, not 75.

Sign in to comment.

More Answers (0)

Products


Release

R2017b

Community Treasure Hunt

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

Start Hunting!