Error Using Rainflow fonction

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JoeTeg
JoeTeg 2024-7-29
评论： Star Strider 2024-7-29
Hello everyone,
I'm not really good at Matlab, but I need to use it to process some load cycles result using the rainflow function (please see attachment)
I have written the code as below
clear all,
clc,
[filename,pathname] = uigetfile({'*.xlsx;*.xlsm','Excel Worksheets (*.xlsx,*.xlsm)'},'Select File');
[file,sheets] = xlsfinfo([pathname,filename]);
num_1 = xlsread(filename);
[num_3, txt]= xlsread(filename);
%read only row and column with number
Data_1 = num_3(4:end,:); % 4:end, --> choose all the row start from 4 % : --> choose all the column % start at row 4 and read all column
X = num_3(4:end,7); % 4,--> Choose only row 4 seread oly row 4 and all column.
Y = num_3(4:end,6);
Z = num_3(4:end,8);
figure(1)
plot3(X,Y,Z);
t = linspace(min(Y),max(Y),length(Z)/10);
d = linspace(min(Y),max(Y),length(Z)/5);
figure(2)
Z = [X Y]
rainflow(Z,t);
and unfortunately have the errors below after the code:
Error using rainflow
Expected X to be a vector.
Error in rainflow>parseInputs (line 184)
validateattributes(x,{'single','double'},...
Error in rainflow (line 81)
[x,t,ext,td] = parseInputs(x,varargin{:});
Error in KW31_BMW_Load_Cycle_Ju (line 39)
rainflow(Z,t)
Can someone please tell me were I´making the mytake by using the Rainflow function?
Thank in advance

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回答（2 个）

Animesh
Animesh 2024-7-29
编辑：Animesh 2024-7-29
Hey @JoeTeg
The "rainflow" function is typically used for fatigue analysis and expects a time history of stress or strain as input. Thus the provide input should be a vector.
The current code is passing a matrix "Z" composed of "X" and "Y" values, making "Z" a 51 x 2 matrix, which is unsuitable for the "rainflow" function.
If you wish to concatenate data of "X" and "Y", try using "vertcat" function to concatenate them vertically.
Z = vertcat (X,Y)
You can refer the following MathWorks documentation for more information :
https://www.mathworks.com/help/signal/ref/rainflow.html
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Animesh
Animesh 2024-7-29
Hey @JoeTeg
From what I can gather, you want to evaluate three load cycles together. The "rainflow" function in MATLAB is designed to process a single vector of data, typically a time history of stress or strain. It cannot directly handle three separate vectors like "X", "Y", and "Z" simultaneously.
Here's a way to approach it: combine the data. If you need to evaluate the load cycles for the combined data, you need to create a single vector that represents the load history. This could be done by concatenating the vectors "X","Y", and "Z" in a meaningful way.
% Read only row and column with number
Data_1 = num_3(4:end, :); % 4:end, --> choose all the rows starting from 4 % : --> choose all the columns
X = num_3(4:end, 7); % Column 7
Y = num_3(4:end, 6); % Column 6
Z = num_3(4:end, 8); % Column 8
% Combine X, Y, Z into a single vector
combinedData = [X; Y; Z];
Also make sure that the time vector "t" has same dimensions as the transpose of "combinedData".
t=linspace(0, length(combinedData)-1, length(combinedData));
Now, you can evaluate "rainflow" on this data:
rainflow(combinedData, t);
JoeTeg
JoeTeg 2024-7-29
thank a lot @Animesh for taking from your time to help me. I really appreciate it. I will try the code a let you know if i stil have some questions.

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Star Strider
Star Strider 2024-7-29
Ran in:
The rainflow function only takes vector arguments.
What result do you want?
In the interim,try this —
T1 = readtable('Data_for_Rainflow.xlsx', 'VariableNamingRule','preserve')
T1 = 52x8 table
Zeit Pressure_1 Pressure_2 Masse Flow Pressure_3 adiabatic power Normlize Load Cycle 1 to 3 ____ __________ __________ __________ __________ _______________ _____________ ____________ NaN NaN NaN NaN NaN NaN NaN NaN 0 102.01 97.432 5.0249 1.0747 0.031449 0.023871 1 0.5 140.42 111.94 36.53 1.486 1.3174 1 2 1 140.41 111.94 36.511 1.4859 1.3167 0.99942 3 1.5 139.69 111.58 35.947 1.478 1.2779 0.97 1 2 138.72 111.1 35.198 1.4675 1.2271 0.93142 2 2.5 137.88 110.68 34.546 1.4584 1.1836 0.89844 3 3 137.17 110.33 33.991 1.4506 1.1472 0.8708 1 3.5 136.5 110.03 33.446 1.4433 1.1127 0.84463 2 4 135.89 109.78 32.948 1.4368 1.082 0.82129 3 4.5 135.18 109.48 32.375 1.4292 1.0468 0.79459 1 5 134.48 109.18 31.809 1.4217 1.0127 0.76866 2 5.5 133.86 108.91 31.301 1.415 0.98236 0.74566 3 6 133.26 108.66 30.817 1.4086 0.95389 0.72405 1 6.5 132.94 108.52 30.556 1.4052 0.9388 0.7126 2 7 132.7 108.42 30.361 1.4026 0.92756 0.70406 3
VN = T1.Properties.VariableNames;
num_3 = T1{2:end,:};
X = num_3(:,7); % 4,--> Choose only row 4 seread oly row 4 and all column.
Y = num_3(:,6);
Z = num_3(:,8);
t = num_3(:,1);
figure(1)
plot3(X,Y,Z);
grid on
xlabel(VN{7})
ylabel(VN{6})
zlabel(VN{8})
% t = linspace(min(Y),max(Y),length(Z)/10);
d = linspace(min(Y),max(Y),length(Z)/5);
figure(2)
Z = [X Y]
Z = 51x2
0.0239 0.0314 1.0000 1.3174 0.9994 1.3167 0.9700 1.2779 0.9314 1.2271 0.8984 1.1836 0.8708 1.1472 0.8446 1.1127 0.8213 1.0820 0.7946 1.0468
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
figure
rainflow(Z(:,1),t)
sgtitle(VN{7})
figure
rainflow(Z(:,2),t);
sgtitle(VN{6})
.
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JoeTeg
JoeTeg 2024-7-29
Thx i will try it and come back to you with more detaisl. It is also the first time for me to process such kind of data
i wish you a nice evening
Star Strider
Star Strider 2024-7-29
Thank you!
I have never needed to use that function either, ono my own data. I am learning as well.

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