binwidth is not working
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Hi;
%statistical parameters
STD_Strain = 25;
Mean_strain = 150;
AN=1000;
Stress = 210000*(STD_Strain.*randn(AN,1) + Mean_strain)*10^-6;
%Stress range-rainflow counting
[c,hist,edges,rmm,idx]=rainflow(Stress)
BinWidth=2;
h=histogram('BinEdges',edges','BinCounts',sum(hist,2))
xlabel('Stress Range')
ylabel('Cycle Counts')
The target of this code is to conduct fatigue analysis.
I tried by many ways to specify the bin width. Still, the result is always bin width = 3.
h =
Histogram with properties:
Data: []
Values: [64 54.5000 67 47.5000 36 27 19 6.5000 5 1 1.5000]
NumBins: 11
BinEdges: [0 3 6 9 12 15 18 21 24 27 30 33]
BinWidth: 3
BinLimits: [0 33]
Normalization: 'count'
FaceColor: 'auto'
EdgeColor: [0 0 0]
What is wrong with my code?
Thank you in advance
3 comentarios
Voss
el 3 de En. de 2023
It seems that the rainflow function does the histogram-binning of the Stress values you give it. (I say that because rainflow returns hist and edges.) Therefore, to modify how the binning is done (e.g., by using a specific bin width), you'll have to modify rainflow.
If you share the code for the rainflow function, then someone can probably show you how to be able to specify your own bin width.
Rik
el 4 de En. de 2023
It is an internal function: rainflow doc page, so modifying it should be avoided. I can't see from the documentation how you can extract the actual values or force it to use a specific binning.
Voss
el 4 de En. de 2023
My mistake. Thanks for pointing that out.
Respuestas (1)
You're just creating a variable. You do not use that variable as an input argument, or to modify the histogram object after you create it. How is Matlab supposed to know you meant that?
edit:
Below are the two possible ways you can change the code. However, neither will work, because you manually set the bin counts.
%statistical parameters
STD_Strain = 25;
Mean_strain = 150;
AN=1000;
Stress = 210000*(STD_Strain.*randn(AN,1) + Mean_strain)*10^-6;
%Stress range-rainflow counting
[c,hist,edges,rmm,idx]=rainflow(Stress);
BinWidth=2;
h=histogram('BinEdges',edges','BinCounts',sum(hist,2));
xlabel('Stress Range')
ylabel('Cycle Counts')
if false
% option 1:
h.BinWidth=BinWidth;
% option 2:
h=histogram('BinEdges',edges','BinCounts',sum(hist,2),'BinWidth',BinWidth);
end
3 comentarios
Mohamed Zied
el 3 de En. de 2023
Rik
el 3 de En. de 2023
The fundamental problem is that you don't provide the original source data to the histogram function, so it can't split bins for you.
You need to explain what exactly the source data is. Then you can use the histogram function as it is intended to be used, instead of a fancier version of the bar function.
Mohamed Zied
el 4 de En. de 2023
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