In: Advanced Math
The data is in the nonlinear regime. The first column is plastic
strain values (e) and the second column is
corresponding true stress (s) values. In the nonlinear regime, the
relation between e and s is:
? = ???
Where K,n are material constants which need to be determined using
curve fitting.
a). Plot e vs s on a scatter plot
b). Find the constants n and K using curve fit. We
have not learned how to fit a power law in class but here
is a hint: if you take the log of the equation above, it becomes a
linear equation in log(e) and log(s).
Data
0.07913 | 400.1313 |
0.079274 | 400.394 |
0.079433 | 400.6575 |
0.079582 | 400.914 |
0.079731 | 401.1977 |
0.079875 | 401.4677 |
0.080019 | 401.7893 |
0.080168 | 402.084 |
0.080316 | 402.382 |
0.08046 | 402.6442 |
0.080609 | 402.8987 |
0.080753 | 403.1742 |
0.080907 | 403.4315 |
0.081056 | 403.7102 |
0.081205 | 403.9754 |
0.081359 | 404.258 |
0.081508 | 404.5485 |
0.081652 | 404.8328 |
0.081801 | 405.1216 |
0.081945 | 405.4012 |
0.082089 | 405.6714 |
0.082243 | 405.9478 |
0.082392 | 406.2127 |
0.082546 | 406.4651 |
0.082695 | 406.7232 |
0.082834 | 406.9823 |
0.082988 | 407.2513 |
0.083132 | 407.5583 |
0.08328 | 407.8404 |
0.083429 | 408.1223 |
0.083573 | 408.3959 |
0.083717 | 408.6451 |
0.083871 | 408.8914 |
0.084026 | 409.1303 |
0.084175 | 409.3806 |
0.084319 | 409.6619 |
0.084472 | 409.949 |
0.084621 | 410.2324 |
0.084765 | 410.4981 |
0.084914 | 410.7732 |
0.085068 | 411.0592 |
0.085212 | 411.3118 |
0.085357 | 411.515 |
0.085511 | 411.7665 |
0.085655 | 412.0262 |
0.085803 | 412.3185 |
0.085952 | 412.6021 |
0.086101 | 412.8792 |
0.08625 | 413.1566 |
0.086399 | 413.4088 |
USE MATLAB CODE ONLY!
USE MATLAB CODE ONLY!
THANK YOU
%%Matlab code for curve fitting
clear all
close all
%all data point
A=[0.07913 400.1313;...
0.079274 400.394;...
0.079433 400.6575;...
0.079582 400.914;...
0.079731 401.1977;...
0.079875 401.4677;...
0.080019 401.7893;...
0.080168 402.084;...
0.080316 402.382;...
0.08046 402.6442;...
0.080609 402.8987;...
0.080753 403.1742;...
0.080907 403.4315;...
0.081056 403.7102;...
0.081205 403.9754;...
0.081359 404.258;...
0.081508 404.5485;...
0.081652 404.8328;...
0.081801 405.1216;...
0.081945 405.4012;...
0.082089 405.6714;...
0.082243 405.9478;...
0.082392 406.2127;...
0.082546 406.4651;...
0.082695 406.7232;...
0.082834 406.9823;...
0.082988 407.2513;...
0.083132 407.5583;...
0.08328 407.8404;...
0.083429 408.1223;...
0.083573 408.3959;...
0.083717 408.6451;...
0.083871 408.8914;...
0.084026 409.1303;...
0.084175 409.3806;...
0.084319 409.6619;...
0.084472 409.949;...
0.084621 410.2324;...
0.084765 410.4981;...
0.084914 410.7732;...
0.085068 411.0592;...
0.085212 411.3118;...
0.085357 411.515;...
0.085511 411.7665;...
0.085655 412.0262;...
0.085803 412.3185;...
0.085952 412.6021;...
0.086101 412.8792;...
0.08625 413.1566;...
0.086399 413.4088];
%plotting the data
e=A(:,1); s=A(:,2);
scatter(log(e),log(s))
pp=polyfit(log(e),log(s),1);
yy=polyval(pp,log(e));
hold on
plot(log(e),yy)
title('log(s) vs. log(e) plot')
xlabel('log(e)')
ylabel('log(s)')
legend('scatter plot','linear fit')
fprintf('Value of log(k)= %f and n=%f\n', pp(2),pp(1))
fprintf('Hence K=%f\n',exp(pp(2)))
%%%%%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%%%%