how can I calculate rate of compression?

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Manav Divekar
Manav Divekar 2022-1-30
回答： Vinayak 2024-1-11
I want to test rate of compression (mm/s), average Elastic moduli and Plot force-displacement. from the given given text file. Since there are other text files whoes block size are different what changes can i make in my code to make it work for all without manually changing values. For this particula set of value 10% strain rate was applied.
Data = inportdata('DHD_10%_sec_T1.txt'); % read text file: DHD_10%_sec_T1.txt
dL = Data(:,1)'; % relative displacement [mm] data
F = Data(:,2)'; % load = force [N] data
ADF = 10; % adjustment factor for zeroing
L = 39.96; % lenght if block [mm]
w = 25.8666667; % width if block[mm]
th = 25.9; % thickness if block[mm]
% get the zero point of force
NF = numel(F); % number of elements of F
PosF_L = F > 0; % logical positive
IndV = find(PosF_L); % index of positives
NIndV = numel(IndV); % number of elements of IndV
Nneg = NF-NIndV; % number of elements of negatives
Diff = zeros(1,NIndV-ADF);
for c1 = 1:NIndV-ADF
Diff(c1) = IndV(c1+ADF)-IndV(c1); % difference between points separated by ADF
end
Diff2 = Diff(2:end)-Diff(1:end-1); % difference between 2 adjacent points
for c2 = 1:c1-1
if Diff2(c2) ~= 0
Diff2(c2) = c2; % replaces the elements with increasing counter
end
end
IndZero = max(Diff2)+Nneg+1; % index of zero F
% zeroed dl and F
dL = dL(IndZero:end)-dL(IndZero); % relative displacement [mm]
F = F(IndZero:end)-F(IndZero); % force [N]
% intensive properties
A = w*L*th; % area [mm^2]
A = A/1e6; % area [m^2]
strain = dL/L; % strain [mm/mm]
stress = F/A; % stress [N/m^2 = Pa]
% adjust units
strain = strain*100; % strain [%]
stress = stress/1e6; % stress [MPa]
% locate the ultimate stress
dsigma = diff(stress); % derivative of stress
dsigma = round(dsigma*1000)/1000; % round off dsigma
Ndsig = numel(dsigma); % number of elements of dsigma
MSp = zeros(1,Ndsig); MSpi = MSp; UsigInd = Ndsig; % initial values
for c3 = 1:Ndsig
[MSp(c3),MSpi(c3)] = max(stress(1:c3)); % maximum stress and its index vectors
if c3 > 2
% condition: max is established when the next two are decreasing
% and the derivative must be negative
if (MSp(c3) == MSp(c3-1)) && (MSp(c3) == MSp(c3-2)) && (dsigma(c3-1) < 0) && (dsigma(c3-2) < 0)
UsigInd = c3-2; % index of maximum point
break;
end
end
end
% trim data
cutoff = round(UsigInd*1.15); % maximum index
if cutoff <= Ndsig
strain = strain(1:cutoff); % strain [%]
stress = stress(1:cutoff); % stress [MPa]
end
% readjust the zero
Ustr = max(stress); % maximum stress [MPa]
limStr = Ustr*0.01; % 1% of maximum stress [MPa]
Lsig = stress >= limStr; % logical true
LsigInd = find(Lsig); % index of true
zInd = LsigInd(1); % new index
strain = strain(zInd:end)-strain(zInd); % strain [%]
stress = stress(zInd:end)-stress(zInd); % stress [MPa]
% true values
strainT = -log(1-strain/100)*100; % true strain [%]
stressT = stress.*(1-strain/100); % true stress [MPa]
[UCS,UCSi] = max(stressT); % ultimate compressive strength [MPa]
UCstrain = strainT(UCSi); % ultimate compressive strain [%]
% linear regression to determine the compressive modulus
ND = numel(stressT); % number of elements
% initial values
RSQ = zeros(1,ND-2); x = NaN(1,ND); y = x; yf = y; m = RSQ; b = m; dy = b;
for c4 = 1:ND-2
x = strainT(1:c4+2); % x observed [%]
y = stressT(1:c4+2); % y observed [MPa]
m(c4) = (((c4+2)*sum(x.*y))-(sum(x)*sum(y)))/(((c4+2)*sum(x.^2))-(sum(x)^2)) ; % slope of line fit [MPa/% = 100*MPa]
b(c4) = ((sum(y))-(m(c4)*(sum(x))))/(c4+2); % y-intercept of line fit [MPa]
yf = m(c4)*x + b(c4); % y fit [MPa]
SSE = sum((y-yf).^2); % sum of squares error
SST = sum((y-mean(y)).^2); % sum of squares total
RSQ(c4) = 1 - (SSE/SST); % coefficient of determination vector
%xi(c4) = -b/m(c4);
dy(c4) = abs(y(c4+2) - (m(c4)*x(c4+2)+b(c4)))/UCS; % change in y over UCS
end
[~,ECi] = max(dy >= 0.01); % index of the compressive modulus, 10% cutoff
rsq = RSQ(ECi); % r^2 = coefficient of determination
EC = m(ECi)*100; % compressive modulus [MPa]
% yield values
Ystrain = strainT(ECi+2); % yield strain [%]
Ystress = stressT(ECi+2); % yield stress [%]

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回答（1 个）

Vinayak
Vinayak 2024-1-11
Hi Manav
From what I noticed in the file you shared, it contains three columns of data. I presume that the other files are in similar format. You could write a custom function that determines the value of strain rate and block sizes based on file name. You can also write a function to go through all files in an existing directory and process them accordingly. Here is a sample code for the given file to calculate the strain rate:
% Get a list of all files in the directory
directoryPath = 'path/to/your/directory'; % Replace with your directory path
files = dir(directoryPath);
fileNames = {files.name};
for i = 1:length(fileNames)
fileName = fileNames{i};
% Ignore directories and hidden files
if ~files(i).isdir && ~startsWith(fileName, '.')
% Extract the strain rate from the file name
strainRate = getStrainRateFromFileName(fileName);
% Display the strain rate
fprintf('Processing file: %s with strain rate: %d%%\n', fileName, strainRate);
% Process the stress/strain data from the file
processData(fullfile(directoryPath, fileName), strainRate);
end
end
function strainRate = getStrainRateFromFileName(fileName)
% Regex pattern to match digits followed by a percentage sign
pattern = '_(\d+)%';
tokens = regexp(fileName, pattern, 'tokens');
if ~isempty(tokens)
strainRate = str2double(tokens{1}{1});
else
error('Strain rate could not be extracted from the file name.');
end
end
function processData(filePath, strainRate)
Data = importdata(filePath);
ADF = strainRate;
% ... (rest of the old code)
end
You may read more about “regexp” and “dir” in the following documentation:

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