Why are my variables saving only Nan entries?

Cal = struct;
Cal.data = readtable('Lab_4_Data.txt');

Warning: The DATETIME data was created using format 'MM/dd/uuuu' but also matched 'dd/MM/uuuu'.
To avoid ambiguity, supply a datetime format using SETVAROPTS, e.g.
opts = setvaropts(opts,varname,'InputFormat','MM/dd/uuuu');

Cal.data_uE1 = Cal.data.Var3;
Cal.data_uE2 = Cal.data.Var4;
Cal.data_uE3 = Cal.data.Var5;

Cal.data_E1 = Cal.data_uE1 * 1000000;
Cal.data_E2 = Cal.data_uE2 * 1000000;
Cal.data_E3 = Cal.data_uE3 * 1000000;

masses = [0 50 100 200 300 500]; % grams
xN = (masses ./ 1000) * 9.81;  % in newtons

% Calculate shear strain (𝛾xy)
Cal.data_gamma_xy = 2 * Cal.data_E2 - (Cal.data_E1 + Cal.data_E3);

% Calculate principal strains (Ɛ1, Ɛ2, Ɛ3)
Cal.data_epsilon_1 = Cal.data_E1;
Cal.data_epsilon_2 = 0.5 * Cal.data_gamma_xy;
Cal.data_epsilon_3 = Cal.data_E3;

% Calculate Poisson's ratio (v)
Cal.data_nu = abs(Cal.data_epsilon_2) ./ Cal.data_epsilon_1;

% Calculate angle of rotation (𝜃𝜃p)
Cal.data_theta_p = atand(Cal.data_gamma_xy ./ (Cal.data_epsilon_1 - Cal.data_epsilon_3));

% Calculate principal stresses (Txx, Tyy)
E = 69000; % Young's Modulus for aluminum in MPa
Cal.data_Txx = ((E) / (1 - Cal.data_nu.^2)) .* (Cal.data_epsilon_1 + Cal.data_nu .* Cal.data_epsilon_2);
Cal.data_Tyy = ((E) / (1 - Cal.data_nu.^2)) .* (Cal.data_epsilon_2 + Cal.data_nu .* Cal.data_epsilon_1);

% Calculate longitudinal stress from the beam flexure equation
P = xN; % applied load in Newtons
L = 0.241; % length of the beam in meters (replace with your actual value)
b = 0.025; % width of the beam in meters (replace with your actual value)
h = 0.00379; % thickness of the beam in meters (replace with your actual value)
c = h / 2;
% distance between gage's mounting line and dent at the end of the beam in meters
L_adjusted = 0.051;
I = (b * h^3) / 12;
Cal.data_sigma_L = (6 * P * L_adjusted) / (b * h^2);

% Calculate load cell calibration factor
Calibration_Factor = max(abs(Cal.data_epsilon_1)) / max(P);

% Plot 1: Principal Strains vs. Applied Load
figure;
plot(xN, Cal.data_epsilon_1(1:6), '-o', 'DisplayName', 'Ɛ1');
hold on;
plot(xN, Cal.data_epsilon_2(1:6), '-o', 'DisplayName', 'Ɛ2');
plot(xN, Cal.data_epsilon_3(1:6), '-o', 'DisplayName', 'Ɛ3');
xlabel('Applied Load (N)');
ylabel('Principal Strains (\mu\epsilon)');
title('Principal Strains vs. Applied Load');
legend('Location', 'Best');
grid on;

% Plot 2: Principal Stresses vs. Applied Load
figure;
plot(xN, Cal.data_Txx(1:6), '-o', 'DisplayName', 'Txx');
hold on;
plot(xN, Cal.data_Tyy(1:6), '-o', 'DisplayName', 'Tyy');
xlabel('Applied Load (N)');
ylabel('Principal Stresses (MPa)');
title('Principal Stresses vs. Applied Load');
legend('Location', 'Best');
grid on;

% Plot 3: Comparison of Measured and Theoretical Longitudinal Stress
figure;
plot(xN, Cal.data_sigma_L(1:6), '-o', 'DisplayName', 'Measured Longitudinal Stress');
hold on;
plot(xN, Cal.data_Txx(1:6), '-o', 'DisplayName', 'Theoretical Longitudinal Stress');
xlabel('Applied Load (N)');
ylabel('Stress (MPa)');
title('Comparison of Measured and Theoretical Longitudinal Stress');
legend('Location', 'Best');
grid on;

% Display Load Cell Calibration Factor
disp('Load Cell Calibration Factor (N/Ɛ):');

Load Cell Calibration Factor (N/Ɛ):

disp(Calibration_Factor);

   1.1621e+07

The .txt file is attached.

This code runs, however the variables data_nu, data_theta_p, data_Txx, and data_Tyy within the Cal struct all have Nan values. All other created variables contain the correct data pulled from the .txt file. Why is this happening?

[Stephen23]

"Why is this happening?"

Because the first eighty-five rows of channel data are zeros. What do you expect to get when you divide zero by zero?

Date    Time	200245-Ch 1 ue	200245-Ch 2 ue	200245-Ch 3 ue
11/11/21 04:09:57:5492	0	0	0
11/11/21 04:09:57:6792	0	0	0
11/11/21 04:09:57:7992	0	0	0
11/11/21 04:09:57:9293	0	0	0
11/11/21 04:09:58:0593	0	0	0
11/11/21 04:09:58:1894	0	0	0
11/11/21 04:09:58:3194	0	0	0
11/11/21 04:09:58:4494	0	0	0
11/11/21 04:09:58:5795	0	0	0
11/11/21 04:09:58:7095	0	0	0
11/11/21 04:09:58:8395	0	0	0
11/11/21 04:09:58:9594	0	0	0
11/11/21 04:09:59:0896	0	0	0
11/11/21 04:09:59:2197	0	0	0
11/11/21 04:09:59:3596	0	0	0
11/11/21 04:09:59:4897	0	0	0
11/11/21 04:09:59:6197	0	0	0
11/11/21 04:09:59:7498	0	0	0
11/11/21 04:09:59:8799	0	0	0
... etc

[hyu34gyd5]

My apologies, I came to this platform for help because I'm new to MATLAB and coding in general. I get it, you're the smart guy, but can you actually help?

I'm well aware that the first 85 entries are zero indicating that zero load had been applied. I still need the rest of the data (not all zeros) to plot my figures.

Answer 1

[Stephen23]

Cal.data_Txx = ((E) ./ (1 - Cal.data_nu.^2)) .* (Cal.data_epsilon_1 + Cal.data_nu .* Cal.data_epsilon_2);
Cal.data_Tyy = ((E) ./ (1 - Cal.data_nu.^2)) .* (Cal.data_epsilon_2 + Cal.data_nu .* Cal.data_epsilon_1);
%                   ^^ you used the wrong operator here

The first 85 NaN are expected, as I explained in my comment.

[hyu34gyd5]

Changing the operator made no difference in my figures.

[Stephen23]

"Changing the operator made no difference in my figures."

Because in all of your figures you only plot the first six elements of those vectors, like this:

Cal.data_epsilon_1(1:6)

Once again: the first 85 elements will be NaN because your data are zeros. You are plotting the first 6 NaN elements. Your channel data appear to correspond to be sampled at particular times (total 212 samples), but you are trying to plot them as if they correspond to a vector of six masses. How do those masses correspond to your sampled data?

As far as I can guess, your channel data has six distinct groups, does each group correspond to one mass? In that case you will need map each group to each mass, e.g. using known limits to the channel values, KMEANS, or similar.

Cal = struct;
Cal.data = readtable('Lab_4_Data.txt');

Warning: The DATETIME data was created using format 'MM/dd/uuuu' but also matched 'dd/MM/uuuu'.
To avoid ambiguity, supply a datetime format using SETVAROPTS, e.g.
opts = setvaropts(opts,varname,'InputFormat','MM/dd/uuuu');

Cal.data_uE1 = Cal.data.Var3;
Cal.data_uE2 = Cal.data.Var4;
Cal.data_uE3 = Cal.data.Var5;

Cal.data_E1 = Cal.data_uE1 * 1000000;
Cal.data_E2 = Cal.data_uE2 * 1000000;
Cal.data_E3 = Cal.data_uE3 * 1000000;

masses = [0 50 100 200 300 500]; % grams
xN = (masses ./ 1000) * 9.81;  % in newtons

% Calculate shear strain (𝛾xy)
Cal.data_gamma_xy = 2 * Cal.data_E2 - (Cal.data_E1 + Cal.data_E3);

% Calculate principal strains (Ɛ1, Ɛ2, Ɛ3)
Cal.data_epsilon_1 = Cal.data_E1;
Cal.data_epsilon_2 = 0.5 * Cal.data_gamma_xy;
Cal.data_epsilon_3 = Cal.data_E3;

% Calculate Poisson's ratio (v)
Cal.data_nu = abs(Cal.data_epsilon_2) ./ Cal.data_epsilon_1;

% Calculate angle of rotation (𝜃𝜃p)
Cal.data_theta_p = atand(Cal.data_gamma_xy ./ (Cal.data_epsilon_1 - Cal.data_epsilon_3));

% Calculate principal stresses (Txx, Tyy)
E = 69000; % Young's Modulus for aluminum in MPa
Cal.data_Txx = ((E) ./ (1 - Cal.data_nu.^2)) .* (Cal.data_epsilon_1 + Cal.data_nu .* Cal.data_epsilon_2);
Cal.data_Tyy = ((E) ./ (1 - Cal.data_nu.^2)) .* (Cal.data_epsilon_2 + Cal.data_nu .* Cal.data_epsilon_1);

% Calculate longitudinal stress from the beam flexure equation
P = xN; % applied load in Newtons
L = 0.241; % length of the beam in meters (replace with your actual value)
b = 0.025; % width of the beam in meters (replace with your actual value)
h = 0.00379; % thickness of the beam in meters (replace with your actual value)
c = h / 2;
% distance between gage's mounting line and dent at the end of the beam in meters
L_adjusted = 0.051;
I = (b * h^3) / 12;
Cal.data_sigma_L = (6 * P * L_adjusted) / (b * h^2);

% Calculate load cell calibration factor
Calibration_Factor = max(abs(Cal.data_epsilon_1)) / max(P);

% Plot 1: Principal Strains vs. Applied Load
figure;
plot(Cal.data_epsilon_1, '-o', 'DisplayName', 'Ɛ1');
hold on;
plot(Cal.data_epsilon_2, '-o', 'DisplayName', 'Ɛ2');
plot(Cal.data_epsilon_3, '-o', 'DisplayName', 'Ɛ3');
xlabel('Applied Load (N)');
ylabel('Principal Strains (\mu\epsilon)');
title('Principal Strains vs. Applied Load');
legend('Location', 'Best');
grid on;

% Plot 2: Principal Stresses vs. Applied Load
figure;
plot(Cal.data_Txx, '-o', 'DisplayName', 'Txx');
hold on;
plot(Cal.data_Tyy, '-o', 'DisplayName', 'Tyy');
xlabel('Applied Load (N)');
ylabel('Principal Stresses (MPa)');
title('Principal Stresses vs. Applied Load');
legend('Location', 'Best');
grid on;

% Plot 3: Comparison of Measured and Theoretical Longitudinal Stress
figure;
plot(Cal.data_sigma_L, '-o', 'DisplayName', 'Measured Longitudinal Stress');
hold on;
plot(Cal.data_Txx, '-o', 'DisplayName', 'Theoretical Longitudinal Stress');
xlabel('Applied Load (N)');
ylabel('Stress (MPa)');
title('Comparison of Measured and Theoretical Longitudinal Stress');
legend('Location', 'Best');
grid on;

% Display Load Cell Calibration Factor
disp('Load Cell Calibration Factor (N/Ɛ):');

Load Cell Calibration Factor (N/Ɛ):

disp(Calibration_Factor);

   1.1621e+07

[hyu34gyd5]

Yes, the masses are the applied loads that correspond to each jump in the data. I was plotting 1:6 because I want to irradicate the extra noise. I only want to plot one data point for each of the applied load settings.

I'm not familar with the methods you suggested using.

[Stephen23]

"I only want to plot one data point for each of the applied load settings."

But your data file has multiple data points for each group. You need to decide how to obtain one data point from each group: take the mean, the max, the min, the mode, or some other value. Which do you want?

Another important question: how do you distinguish between the groups in the data file? Is there a particular value or parameter that specifies each group? Or a duration that each group was sampled for? Or do we need to identify the groups by a change in the data values?

"I'm not familar with the methods you suggested using."

We can look at the that once we know how you want to calculate one value from each group.

[hyu34gyd5]

I think I'd like to take the mean of each group, but the data was collected using random start/stop so the intervals between groups are not the same and I can't seem to distinguish between them inside my script. Each group of data represent the microstrains corresponding to applied loads 0 g, 50 g, 100 g, 200 g, 300 g, and 500 g respectively. Columns 3, 4 and 5 contain these strain values collected from three separate gages.

So I want to plot one point for each of the applied load conditions for all three gages. Six points per gage, 18 points total per plot.

[Sam Chak]

Hi @hyu34gyd5

Since the code is provided, most of the time, we can only advise on what went wrong in the code. If you are very new to MATLAB and coding, you may have difficulty understanding some technical aspects that we explain.

Therefore, I suggest that you provide a very clear picture (graphically) of what you intend to do with the data. Also, provide a sample outcome or plot that you expected.

Furthermore, some of us are not experts in the stress and strain field. Thus, we cannot verify if the formulas in the code are correct or not. You are advised to attach some images of the formulas so that we can check them.

[Stephen23]

You could use K-means clustering to detect the groups:

https://en.wikipedia.org/wiki/K-means_clustering

after which it is very simple to take the mean of each group. First import the data from file:

fnm = 'Lab_4_Data.txt';
opt = detectImportOptions(fnm, 'Delimiter','\t', 'VariableNamingRule','preserve');
tbl = readtable(fnm,opt)

tbl = 212×4 table
           Date    Time           200245-Ch 1 ue    200245-Ch 2 ue    200245-Ch 3 ue
    __________________________    ______________    ______________    ______________

    {'11/11/21 04:09:57:5492'}          0                 0                 0       
    {'11/11/21 04:09:57:6792'}          0                 0                 0       
    {'11/11/21 04:09:57:7992'}          0                 0                 0       
    {'11/11/21 04:09:57:9293'}          0                 0                 0       
    {'11/11/21 04:09:58:0593'}          0                 0                 0       
    {'11/11/21 04:09:58:1894'}          0                 0                 0       
    {'11/11/21 04:09:58:3194'}          0                 0                 0       
    {'11/11/21 04:09:58:4494'}          0                 0                 0       
    {'11/11/21 04:09:58:5795'}          0                 0                 0       
    {'11/11/21 04:09:58:7095'}          0                 0                 0       
    {'11/11/21 04:09:58:8395'}          0                 0                 0       
    {'11/11/21 04:09:58:9594'}          0                 0                 0       
    {'11/11/21 04:09:59:0896'}          0                 0                 0       
    {'11/11/21 04:09:59:2197'}          0                 0                 0       
    {'11/11/21 04:09:59:3596'}          0                 0                 0       
    {'11/11/21 04:09:59:4897'}          0                 0                 0       

Then identify each group using K-means algorithm and take the mean of each channel in each group:

masses = [0,50,100,200,300,500]; % grams
mat = tbl{:,digitsPattern+wildcardPattern};
idx = kmeans(mat,numel(masses));
[~,~,idy] = unique(idx,'stable');
tmp = splitapply(@mean,mat,idy)

tmp = 6×3

0 0 0 -5.6364 21.8485 26.9394 -11.0000 44.0000 54.0000 -22.7097 87.3226 107.1290 -34.0000 131.0000 160.8824 -56.7826 218.4783 267.5652

Cal = struct;
Cal.data_uE1 = tmp(:,1);
Cal.data_uE2 = tmp(:,2);
Cal.data_uE3 = tmp(:,3);

After that comes the rest of your code:

Cal.data_E1 = Cal.data_uE1 * 1000000;
Cal.data_E2 = Cal.data_uE2 * 1000000;
Cal.data_E3 = Cal.data_uE3 * 1000000;
xN = (masses ./ 1000) * 9.81;  % in newtons
% Calculate shear strain (𝛾xy)
Cal.data_gamma_xy = 2 * Cal.data_E2 - (Cal.data_E1 + Cal.data_E3);
% Calculate principal strains (Ɛ1, Ɛ2, Ɛ3)
Cal.data_epsilon_1 = Cal.data_E1;
Cal.data_epsilon_2 = 0.5 * Cal.data_gamma_xy;
Cal.data_epsilon_3 = Cal.data_E3;
% Calculate Poisson's ratio (v)
Cal.data_nu = abs(Cal.data_epsilon_2) ./ Cal.data_epsilon_1;
% Calculate angle of rotation (𝜃𝜃p)
Cal.data_theta_p = atand(Cal.data_gamma_xy ./ (Cal.data_epsilon_1 - Cal.data_epsilon_3));
% Calculate principal stresses (Txx, Tyy)
E = 69000; % Young's Modulus for aluminum in MPa
Cal.data_Txx = ((E) ./ (1 - Cal.data_nu.^2)) .* (Cal.data_epsilon_1 + Cal.data_nu .* Cal.data_epsilon_2);
Cal.data_Tyy = ((E) ./ (1 - Cal.data_nu.^2)) .* (Cal.data_epsilon_2 + Cal.data_nu .* Cal.data_epsilon_1);
% Calculate longitudinal stress from the beam flexure equation
P = xN; % applied load in Newtons
L = 0.241; % length of the beam in meters (replace with your actual value)
b = 0.025; % width of the beam in meters (replace with your actual value)
h = 0.00379; % thickness of the beam in meters (replace with your actual value)
c = h / 2;
% distance between gage's mounting line and dent at the end of the beam in meters
L_adjusted = 0.051;
I = (b * h^3) / 12;
Cal.data_sigma_L = (6 * P * L_adjusted) / (b * h^2);
% Calculate load cell calibration factor
Calibration_Factor = max(abs(Cal.data_epsilon_1)) / max(P);
% Plot 1: Principal Strains vs. Applied Load
figure;
plot(xN,Cal.data_epsilon_1, '-o', 'DisplayName', 'Ɛ1');
hold on;
plot(xN,Cal.data_epsilon_2, '-o', 'DisplayName', 'Ɛ2');
plot(xN,Cal.data_epsilon_3, '-o', 'DisplayName', 'Ɛ3');
xlabel('Applied Load (N)');
ylabel('Principal Strains (\mu\epsilon)');
title('Principal Strains vs. Applied Load');
legend('Location', 'Best');
grid on;

% Plot 2: Principal Stresses vs. Applied Load
figure;
plot(xN,Cal.data_Txx, '-o', 'DisplayName', 'Txx');
hold on;
plot(xN,Cal.data_Tyy, '-o', 'DisplayName', 'Tyy');
xlabel('Applied Load (N)');
ylabel('Principal Stresses (MPa)');
title('Principal Stresses vs. Applied Load');
legend('Location', 'Best');
grid on;

% Plot 3: Comparison of Measured and Theoretical Longitudinal Stress
figure;
plot(xN,Cal.data_sigma_L, '-o', 'DisplayName', 'Measured Longitudinal Stress');
hold on;
plot(xN,Cal.data_Txx, '-o', 'DisplayName', 'Theoretical Longitudinal Stress');
xlabel('Applied Load (N)');
ylabel('Stress (MPa)');
title('Comparison of Measured and Theoretical Longitudinal Stress');
legend('Location', 'Best');
grid on;

% Display Load Cell Calibration Factor
disp('Load Cell Calibration Factor (N/Ɛ):');

Load Cell Calibration Factor (N/Ɛ):

disp(Calibration_Factor);

   1.1576e+07