Area of the hysteresis loops

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Viswajit Talluru 2025-1-22

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评论： Star Strider 2025-2-5

stress_strain_cycle_count.csv

Hi Community,

I have a stress vs. strain plot with approximately 30 different loops, and I need to calculate the area enclosed by each loop. I've attached a file containing the stress vs. strain data.

Any help or guidance would be greatly appreciated!

Thank you!

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Answer 1

Star Strider 2025-1-22

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2173163-area-of-the-hysteresis-loops#answer_1557871

在 MATLAB Online 中打开

stress_strain_cycle_count.csv

These do not look like hysteresis curves to me. At least, I do not see any that form loops. (I am not certain what the ‘Cycle count’ does, so I used it to segregate the individual records.)

Try this —

T1 = readtable('stress_strain_cycle_count.csv', VariableNamingRule='preserve');

% disp(T1)

[Ucc,~,ix] = unique(T1.('Cycle count'));

cycles = accumarray(ix, (1:numel(ix)).', [], @(x){T1(x,:)})

cycles = 20x1 cell array

{6632x3 table} {1820x3 table} {1823x3 table} {1822x3 table} {1822x3 table} {1822x3 table} {1823x3 table} {1823x3 table} {1824x3 table} {1823x3 table} {1823x3 table} {1822x3 table} {1822x3 table} {1822x3 table} {1823x3 table} {1823x3 table}

% cycles{1}

figure

tiledlayout(5,4)

for k = 1:numel(cycles)

x = cycles{k}.Strain;

y = cycles{k}.Stress;

a(k) = trapz(x, y);

nexttile

plot(x, y)

grid

title(["Loop "+k "Area = "+round(a(k),3)])

end

●

Something is missing from these data that would form them into loops. Please provide it.

.

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Star Strider 2025-1-22

在 MATLAB Online 中打开

I plotted them correctly in my code, however there are no loops when I plot and integrate them individually (that appears to be the point). Each one is simply a zig-zag.

Somee of them have ‘pseudo-loops’ that aren’t actually loops. They consist of curves that intersect a straight line that then continues until it resets and does the same sort of thing at slightly different (x,y) values. It’s possible to calculate the area between the curve and the straight lilne, however it’s a lot of effort for (what appears to me to be) very little actual useful informaiton.

Example —

imshow(imread('output.png'))

T1 = readtable('stress_strain_cycle_count.csv', VariableNamingRule='preserve');

% disp(T1)

[Ucc,~,ix] = unique(T1.('Cycle count'));

cycles = accumarray(ix, (1:numel(ix)).', [], @(x){T1(x,:)})

cycles = 20x1 cell array

{6632x3 table} {1820x3 table} {1823x3 table} {1822x3 table} {1822x3 table} {1822x3 table} {1823x3 table} {1823x3 table} {1824x3 table} {1823x3 table} {1823x3 table} {1822x3 table} {1822x3 table} {1822x3 table} {1823x3 table} {1823x3 table}

% cycles{1}

figure

tiledlayout(2,2)

for k = 2:4:14

x = cycles{k}.Strain;

y = cycles{k}.Stress;

a(k) = trapz(x, y);

nexttile

plot(x, y)

grid

xlabel('Strain')

ylabel('Stress')

title(["Loop "+k "Area = "+round(a(k),3)])

end

●

figure

k = 2;

x = cycles{k}.Strain;

y = cycles{k}.Stress;

a(k) = trapz(x, y);

plot(x, y, '.-')

grid

xlabel('Strain')

ylabel('Stress')

title(["Loop "+k "Area = "+round(a(k),3)])

[pks,locs] = findpeaks(y, MinPeakProminence=0.8);

locs(1) = 1;

pks(1) = y(1);

Bline = [min(x) 1; max(x) 1] \ [min(y); max(y)]

Bline = 2×1

7.7090 -0.2116

idxrng = locs(1):locs(2);

sl = [x(idxrng) ones(numel(idxrng),1)] * Bline;

isxidx = find(diff(sign(sl-y(idxrng)))) % Intersection Indices

isxidx = 3×1

80 604 605

loopix = isxidx(1) : isxidx(end);

a = trapz(x(loopix),sl(loopix)) - trapz(x(loopix),y(loopix)) % Area Calculation

a = -0.0018

figure

plot(x(idxrng), y(idxrng), '.-', DisplayName='Segment Data')

hold on

plot(x(idxrng(1)), y(idxrng(1)), 'pc', MarkerSize=10, DisplayName='Segment Start')

plot(x(idxrng(end)), y(idxrng(end)), 'pm', MarkerSize=10, DisplayName='Segment End')

plot(x(idxrng), sl, '.-', DisplayName='Segment Straight Line')

plot(x(isxidx), y(isxidx), 'rs', DisplayName='Line & Curve Intersections')

hold off

grid

xlabel('Strain')

ylabel('Stress')

text(0.07, 0.25, sprintf('\\leftarrow Area = %.6f', abs(a)))

legend(Location='best')

t = linspace(0, 1, 500);

x = 2*cos(2*pi*t);

y = 2*sin(2*pi*t);

a(k) = trapz(x, y);

plot(x, y)

grid

axis('equal','padded')

title(["Circle With Radius = 2" "Area = "+round(-a(k),6) "Area = "+round(-a(k),2)/pi+" \times \pi"])

Note that this approach works with finding the area of a circle.

Under the circumstancees (noting that these aren’t actually hysteresis loops), i’ll stop here.

.

Viswajit Talluru 2025-1-22

output (1).png

Hello, thank you for your help! I’m approaching this from a testing perspective and often refer to these as hysteresis loops. Specifically, I’d like to focus on the first cycle (0.5 and 1 as the first set) to identify the intersection between these curves and calculate the area within the loops. This approach would yield 10 closed loops, including the one at 9.5 and the second 0 intersection. Across the entire plot, there are 30 such closed loops, and I’d like to determine the areas for all of them. Any assistance with this would be greatly appreciated!

Star Strider 2025-1-22

在 MATLAB Online 中打开

Many of the plot do not have those ’loops’, for example 1, 5, 9 annd 13, so it would be impossible to calculate the area of a loop that doesn’t exist.

Beyond that, for curves that cross, I outlined a way to determine the point of intersection and then calculate the enclosed area. (The indices are reversed, going from high to low rather than low to high, so the areas might appear to be negative. They aren’t. Just take the absolute values of the areas as I did to make them all positive.)

imshow(imread('output (1).png'))

T1 = readtable('stress_strain_cycle_count.csv', VariableNamingRule='preserve');

% disp(T1)

[Ucc,~,ix] = unique(T1.('Cycle count'));

cycles = accumarray(ix, (1:numel(ix)).', [], @(x){T1(x,:)})

cycles = 20x1 cell array

{6632x3 table} {1820x3 table} {1823x3 table} {1822x3 table} {1822x3 table} {1822x3 table} {1823x3 table} {1823x3 table} {1824x3 table} {1823x3 table} {1823x3 table} {1822x3 table} {1822x3 table} {1822x3 table} {1823x3 table} {1823x3 table}

% cycles{1}

figure

tiledlayout(2,2)

for k = 1:4:14

x = cycles{k}.Strain;

y = cycles{k}.Stress;

a(k) = trapz(x, y); % Area Between Vector And X-Axis

nexttile

plot(x, y)

grid

xlabel('Strain')

ylabel('Stress')

title(["Segment "+k "Area = "+round(a(k),3)])

end

sgtitle('These Vectors Don’t Display Any Loops')

figure

k = 2;

x = cycles{k}.Strain;

y = cycles{k}.Stress;

a(k) = trapz(x, y);

plot(x, y, '.-')

grid

xlabel('Strain')

ylabel('Stress')

title(["Loop "+k "Area = "+round(a(k),3)])

[pks,locs] = findpeaks(y, MinPeakProminence=0.8);

locs(1) = 1;

pks(1) = y(1);

Bline = [min(x) 1; max(x) 1] \ [min(y); max(y)] % Slope & Intercept Of The two-Point Linear Segment At The End

Bline = 2×1

7.7090 -0.2116

idxrng = locs(1):locs(2);

sl = [x(idxrng) ones(numel(idxrng),1)] * Bline; % Evaluate Straight Line

isxidx = find(diff(sign(sl-y(idxrng)))) % Intersection Indices

isxidx = 3×1

80 604 605

loopix = isxidx(1) : isxidx(end);

a = trapz(x(loopix),sl(loopix)) - trapz(x(loopix),y(loopix)) % Area Calculation

a = -0.0018

figure

plot(x(idxrng), y(idxrng), '.-', DisplayName='Segment Data')

hold on

plot(x(idxrng(1)), y(idxrng(1)), 'pc', MarkerSize=10, DisplayName='Segment Start')

plot(x(idxrng(end)), y(idxrng(end)), 'pm', MarkerSize=10, DisplayName='Segment End')

plot(x(idxrng), sl, '.-', DisplayName='Segment Straight Line')

plot(x(isxidx), y(isxidx), 'rs', DisplayName='Line & Curve Intersections')

hold off

grid

xlabel('Strain')

ylabel('Stress')

text(0.07, 0.25, sprintf('\\leftarrow Area = %.6f', abs(a)))

legend(Location='best')

t = linspace(0, 1, 500);

x = 2*cos(2*pi*t);

y = 2*sin(2*pi*t);

a(k) = trapz(x, y);

plot(x, y)

grid

axis('equal','padded')

title(["Circle With Radius = 2" "Area = "+round(-a(k),6) "Area = "+round(-a(k),2)/pi+" \times \pi"])

You will likely have to manually segment the data with ‘pseudo-loops’ and data, and then use the code I provided you to calculate the areas in the ‘pseudo-loops’. You can probably use either findpeaks or islocalmax for to detect the peaks separating them, and adding 1 and the end values (for example 1 and numel(x)) of each vector to provide the beginning and end reference indices. It would be easiest to visually determine which data have loops, rather than write rather complicated code to detect them. Then just use my code to find the areas. I

t might be best to put my area-finding code in a function, select the ‘pseudo-loops’ by their beginning and ending index values (from findpeaks or islocalmax), and then call the function for each set of stress-strain vectors that you previously identified.

Once you write it, I can probably help you get that code running if you have problems with it.

.

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Answer 2

Viswajit Talluru 2025-2-4

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2173163-area-of-the-hysteresis-loops#answer_1559033

area calculations.jpeg

Hello Star Strider,

I believe I have a potential solution for calculating the area. Our plan is to draw lines parallel to the y-axis from the first and last points of each half cycle. These lines, when connected to cycle 0, will form a closed loop whose area we need to compute. With approximately 30 loops present, this approach should yield about three distinct areas to be calculated. Any assistance or feedback on this method would be greatly appreciated.

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Star Strider 2025-2-5

在 MATLAB Online 中打开

area calculations.jpeg

II do not understand that approach, however if it works for you, then use it!

imshow(imread('area calculations.jpeg'))

●

.

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Area of the hysteresis loops

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Area of the hysteresis loops

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