# ME 180 Sacramento State University Determining Material Properties Using MATLAB MATLAB for mechanical engineering. Homework Assignment 2: Complete Proble

ME 180 Sacramento State University Determining Material Properties Using MATLAB MATLAB for mechanical engineering.

Homework Assignment 2:

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ME 180 Sacramento State University Determining Material Properties Using MATLAB MATLAB for mechanical engineering. Homework Assignment 2: Complete Proble
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Complete Problem 2 (yes, only Problem 2) in the list of problems in the document found through this link:

Homework Assignment 2.pdf

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Find my solutions to all three problems (MATLAB code published to PDF) in the following:

HmwkAssgt2_Probs1_2_3_Solution_Homen.pdf

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Find my solutions to Problems 1 and 3 (m-files) in the following:

HmwkAssgt2_Prob1_Solution_Homen.m

HmwkAssgt2_Prob3_Solution_Homen.m

You can download and use my m-files as templates to save coding time with Problem 2.

Note that Problem 3 involves the same tasks you perform when completing your first lab experimental analysis: True Stress-True Strain and the power equation. Feel free to complete Lab 1 analysis using MATLAB in lieu of MS Excel. ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
Problem 1 (Expanded Prob 3.9 Dowling 5th ed.) ……………………………………………………………. 1
Problem 2 (Expanded Prob 3.10 Dowling 5th ed.) ………………………………………………………….. 9
Problem 3 (Expanded Prob 3.22 Dowling 5th ed.) …………………………………………………………. 18
Patrick Homen
Problem 1 (Expanded Prob 3.9 Dowling 5th ed.)
Stress-strain data from a tension test on AISI 4140 steel tempered at 427C (800F) are listed in Table 1.
The diameter before testing was 8.56 mm, and after fracture it was 6.74 mm. (a) Use the data to plot
an engineering stress-strain curve from intitial point to fracture. (b) Determine the following: E, 0.2%
offset yield strength, UTS, %Elongation (%El), & %Reduction in Area (%RA). (c) Use the data to plot an
engineering stress-strain curve from initial point to about 2% to show the initial part of the data plotted at
a sensitive strain scale. Use this plot to show the determination of the 0.2% Offset Yield Strength (show
offset line on this plot). (d) Calculate true stresses and strains for the data up to the UTS. Plot these values
and compare the resulting true stress-strain curve with the engineering stress-strain curve from (a). Do
this by showing both curves (data to UTS) on the same plot. (e) For the points from (d) that are beyond
yielding, calculate true plastic strains, & plot these with corresponding values of true stresses. Fit these
data to the power equation sigtru=H*epstruplast^n to obtain values of H & n. Show the data AND a “best
fit curve” on the plot. Note the “best fit curve” should be determined from the power equation and the
determined values of H & n, and NOT just a line connecting the data points. (f) Show the data and fit from
(e) on a log-log plot (g) Calculate the true fracture strength and true fracture strain, including the Mirone
correction for the former. Also calculate the true plastic strain at fracture and add the corresponding point
to an additional log-log plot as completed in (f). If your fitted line is extended (it should be a straight line
on this log-log plot), is it consistent with the fracture point?
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Stress & strain data from Table 1 (engineering stress-strain):
sigeng = engineering stress (MPa)
epsengpct = engineering strain (%)
epseng = engineering strain (unitless or mm/mm)
sigeng = [0 402 812 1198 1358 1403 1399 1406 1445 1466 1483 1492
1485 …
1462 1395 1325 1251 1171 1084]; %MPa
% Final value is fracture stress (fracture strength)
epsengpct = [0 0.197 0.405 0.595 0.681 0.732 0.781 1.010 1.386 …
1.823 2.50 3.28 3.99 4.56 5.52 6.49 7.46 8.46 9.40];
% Final value is fracture strain
% Calculate unitless engineering strain (don’t like %)
1
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
epseng = epsengpct/100;
% (a) Plot engineering stress vs engineering strain:
figure(1)
plot(epseng,sigeng,’k-o’)
title({‘AISI 4140 Steel (800F Temper)’;’Stress vs Strain’})
xlabel(‘Strain, mm/mm’)
ylabel(‘Stress, MPa’)
% (b) E, sigO, UTS, %El, %RA
% Elastic modulus, E:
% Regression on first 5 points for E (slope of first 5 points):
P = polyfit(epseng(1:5),sigeng(1:5),1);
% E = first element in “P”
E = P(1); % MPa
% Display E in output:
fprintf(‘Elastic Modulus = %5.0f MPan’,E)
% Given initial & final diameters Di & Df
Di = 8.56; % mm
Df = 6.74; % mm
% RA = Reduction in Area RA:
RA = (((Di^2)-(Df^2))/(Di^2))*100;
fprintf(‘%%Reduction of Area = %4.2f%%n’,RA)
% EL = %Elongation:
% %Elongation, EL, = strain at fracture – elastic recovered at
fracture:
% Assume “final point is fracture” means AT fracture
% Therefore must subtract elastic part from total strain at fracture
% for plastic strain after fracture to calculate %Elongation
% %EL = (epsfract-(sigfract/E))*100%
EL = (epseng(19)-sigeng(19)/E)*100;
fprintf(‘%%Elongation = %4.2f%%n’,EL)
% TS = tensile strength = ultimate tensile strength
% Assume UTS is highest stress in table (1492 MPa)
% UTS = Max Engrg Stress
TS = max(sigeng); %MPa
fprintf(‘Tensile Strength = %3.0f MPan’,TS)
% sigyield = 0.2% offset yield strength:
% Need sensitive stress-strain plot, part (c) for sigyield:
% (c) Plot first 10 data points for more sensitive strain scale (to
0.02):
% Open new plot window:
figure(2)
plot(epseng(1:10),sigeng(1:10),’k-o’);
title({‘AISI 4140 Steel (800F Temper)’;’Stress vs Strain for
Offset Yield Strength’})
xlabel(‘Strain’)
ylabel(‘Stress, MPa’)
hold on
% Create points for offset line:
% Top of Offset Line above yield point = 1500 MPa
sigoffsettop = 1500; %MPa
% Strain values for Offset Line:
2
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
epsoffset = [0.002 ((sigoffsettop/E)+0.002)];
sigoffset = [0 sigoffsettop];
plot(epsoffset,sigoffset,’k’);
% Show grid on plot to determine sigyield by inspection
grid on
% From plot, 0.2% Offset YS =~ avg stress pts 7 & 8
sigyield = (sigeng(7)+sigeng(8))/2; % MPa
% From plot, offset strain =~avg strain pts 7 & 8
epsyield = (epseng(7)+epseng(8))/2;
% Display 0.2% Offset Yield Strength in output:
fprintf(‘Yield Strength (0.2%% offset) = %3.0f MPan’,sigyield)
% (d) True stress & strains to UTS, plot & compare to Engr stressstrain
% Per Dowling (2019, 5th ed.) Figure 3.20 for true stress/strain:
% Determine 2 x yield strain formula limit:
formlimit = 2*epsyield; % limit = 0.0179
% for values below 2 x yield strain (formlimit):
%
sigtru = engineering stress
% for values above (beyond) 2 x yield strain:
%
sigtru = sigeng*(1+epseng)
% for all values of true strain up to UTS:
%
epstru = ln(1+epseng)
% Calculate true stress and true strain values
% Use formulas from Dowling (Fig3.20,5th ed; Fig4.19,4th ed):
% True stress (sigtru) = engnr stress (sigeng) up to 2 x yield strain
% 2 x yield strain =~ 0.0179 (first 9 points)
sigtru_1 = sigeng(1:9);
% True stress = engrg stress*(1+engrg strain) to UTS (points 10-12)
sigtru_2 = (ones(1,3)+ epseng(10:12)).*sigeng(10:12);
% These are true stress values to UTS
% Concatenate true stress values to form one vector of 12 elements:
sigtru = [sigtru_1,sigtru_2];
% Evaluate true strain values from epstru = ln(1+epseng)
epstru = log(ones(1,12)+epseng(1:12));
% Plot true stress-strain up to UTS:
figure(3)
plot(epstru,sigtru,’b-*’)
title({‘AISI 4140 Steel (800F Temper)’;’True Stress-True Strain to
UTS’})
xlabel(‘True Strain, mm/mm’)
ylabel(‘True Stress, MPa’)
grid on
% Plot True & Engr on one plot for comparison:
figure (4)
plot(epstru,sigtru,’b-‘)
title({‘AISI 4140 Steel (800F Temper)’;’True vs Engr Comparison’;’Data
up to UTS’})
xlabel(‘Strain (engr or true), mm/mm’)
ylabel(‘Stress (engr or true), MPa’)
grid on
hold on
plot(epseng(1:12),sigeng(1:12),’r-‘)
legend(‘true stress-strain’,’engrg stress-strain’,’Location’,’east’)
3
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
% (e) True stresses & true plastic strains beyond yielding to UTS
% (from part (d):
% Points beyond yielding to UTS are points 8 through 12 (5 points)
% Note still working only up to UTS, so true strain valid
% True plastic strain = true strain – true elastic strain
% True elastic strain = true stress/E
epstruelastic = sigtru/E;
epstruplastic = epstru-epstruelastic;
figure(5)
plot(epstruplastic(8:12),sigtru(8:12),’b*’)
grid on
title({‘AISI 4140 Steel (800F Temper)’;’Stress-True Plastic
Strain’;’Data from beyond yielding to UTS’})
xlabel(‘True Plastic Strain’)
ylabel(‘True Stress, MPa’)
hold on
% Show true stress vs true plastic strain on log-log plot:
% Calculate log of true plastic strain and log of true stress:
logepstruplastic = log10(epstruplastic);
logsigtru = log10(sigtru);
% Determine n & H:
Fit = polyfit(logepstruplastic(8:12),logsigtru(8:12),1);
% Display n & H:
n = Fit(1);
H = 10^Fit(2); %MPa
fprintf(‘The fitting constants are:ntStrain Hardening Exponent n =
%5.4fntStrength Coefficient H = %3.0f MPan’,n,H)
% Generate “fit” curve from true strain range and “n” & “H”, power
eqn:
% Create linearly spaced values of true plastic strain:
epstruplasticfit = linspace(epstruplastic(8),epstruplastic(12),1000);
% Use flow equation to calculate fitted true stress:
sigtrufit = H*(epstruplasticfit.^n);
% Add fit line to plot (figure 5)
plot(epstruplasticfit,sigtrufit,’r-‘)
legend(‘Data Points’,’Fit Line’,’Location’,’northwest’)
% (f) Now log-log plot, points after yield to UTS:
figure(6)
plot(logepstruplastic(8:12),logsigtru(8:12),’b*’)
title({‘AISI 4140 Steel (800F Temper)’;’Log(True Stress) vs Log(True
Plastic Strain)’})
xlabel(‘Log(True Plastic Strain)’)
ylabel(‘Log(True Stress), True Stress in MPa’)
grid on
hold on
% Show data and fit line on the plot:
plot(log10(epstruplasticfit),log10(sigtrufit),’r-‘)
legend(‘Data Points’,’Fit Line’,’Location’,’northwest’)
%(g) True fracture point:
% sigtrufrac = true fracture stress = fracture load/fracture area
% Pf = fracture load = fracture stress * initial area
4
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
Pf = sigeng(19)*(pi*(Di^2)/4); % Newtons
sigtrufrac = Pf/(pi*(Df^2)/4); % MPa
% Report true fracture stress:
fprintf(‘True Fracture Strength = %3.0f MPan’,sigtrufrac)
% Mirone correction:
% Need true strain beyond necking:
% epstrudiff = epstrufract – epsult (epstru>epsult)
% Need true strain at fracture, epstrufrac = 2*ln(Di/Df)
epstrufrac = 2*log(Di/Df);
% Report true fracture strain:
fprintf(‘True Fracture Strain = %5.5fn’,epstrufrac)
% for fracture point, use last point (pt 19) & ultimate point (pt 12)
epstrudiff = epstrufrac – epstru(12);
M = 1 – ((0.6058)*(epstrudiff^2))+((0.6317)*(epstrudiff^3))((0.2107)*(epstrudiff^4));
% Mirone corrected true fracture stress:
sigtrufracM = M*sigtrufrac;
% Last plot, log-log plot repeat w/added fracture point, fit line
extended:
figure(7)
plot(logepstruplastic(8:12),logsigtru(8:12),’b*’)
title({‘AISI 4140 Steel (800F Temper)’;’Log(True Stress) vs
Log(True Plastic Strain) to Fracture Point’})
xlabel(‘Log(True Plastic Strain)’)
ylabel(‘Log(True Stress), True Stress in MPa’)
grid on
hold on
plot(log10(epstrufrac),log10(sigtrufracM),’g*’)
% Show data and fit line on the plot:
epstruplasticfit2 = linspace(epstruplastic(8),epstrufrac,1000);
sigtrufit2 = H*(epstruplasticfit2.^n);
plot(log10(epstruplasticfit2),log10(sigtrufit2),’r-‘)
legend(‘Data Points’,’Fracture Point’,’Fit
Line’,’Location’,’northwest’)
% After inspection, fit line is NOT consistent with fracture point
fprintf(‘Fit line is NOT consistent with fracture pointn’)
Elastic Modulus = 199735 MPa
%Reduction of Area = 38.00%
%Elongation = 8.86%
Tensile Strength = 1492 MPa
Yield Strength (0.2% offset) = 1403 MPa
The fitting constants are:
Strain Hardening Exponent n = 0.0452
Strength Coefficient H = 1825 MPa
True Fracture Strength = 1748 MPa
True Fracture Strain = 0.47808
Fit line is NOT consistent with fracture point
5
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
6
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
7
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
8
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
Problem 2 (Expanded Prob 3.10 Dowling 5th
ed.)
Engineering stress-strain data from a tension test on 7075-T651 aluminum are given in Table 2. The diameter before testing was 9.07 mm, and after fracture the minimum diameter in the necked region was
7.78 mm. Analyze these data in terms of engineering and true stresses & strains as in Problem 1 (complete
all items (a) through (g)).
clc
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%
%
%
%
Stress & strain data from Table 2 (engineering stress-strain):
sigeng = engineering stress (MPa)
epsengpct = engineering strain (%)
epseng = engineering strain (unitless or mm/mm)
sigeng = [0 112 222 326 415 473 505 527 542 551 557 563 577 587 …
593 596 597 597 591 571]; %MPa
% Final value is fracture stress (fracture strength)
epsengpct = [0 0.165 0.322 0.474 0.605 0.703 0.797 0.953 1.209 …
1.498 1.819 2.3 4.02 5.98 8.02 9.52 10.97 12.5 13.9 15.33];
% Final value is fracture strain
% Calculate unitless engineering strain (don’t like %)
epseng = epsengpct/100;
9
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
% (a) Plot engineering stress vs engineering strain:
figure(8)
plot(epseng,sigeng,’k-o’)
title({‘7075-T651 Aluminum’;’Stress vs Strain’})
xlabel(‘Strain, mm/mm’)
ylabel(‘Stress, MPa’)
% (b) E, sigO, UTS, %El, %RA
% Elastic modulus, E:
% Regression on first 5 points for E (slope of first 5 points):
P = polyfit(epseng(1:5),sigeng(1:5),1);
% E = first element in “P”
E = P(1); % MPa
% Display E in output:
fprintf(‘Elastic Modulus = %5.0f MPan’,E)
% Given initial & final diameters Di & Df
Di = 9.07; % mm
Df = 7.78; % mm
% RA = Reduction in Area RA:
RA = (((Di^2)-(Df^2))/(Di^2))*100;
fprintf(‘%%Reduction of Area = %4.2f%%n’,RA)
% EL = %Elongation:
% %Elongation, EL, = strain at fracture – elastic recovered at
fracture:
% Assume “final point is fracture” means AT fracture
% Therefore must subtract elastic part from total strain at fracture
% for plastic strain after fracture to calculate %Elongation
% %EL = (epsfract-(sigfract/E))*100%
EL = (epseng(20)-sigeng(20)/E)*100;
fprintf(‘%%Elongation = %4.2f%%n’,EL)
% TS = tensile strength = ultimate tensile strength
% Assume UTS is highest stress in table (597 MPa)
% UTS = Max Engrg Stress
TS = max(sigeng); %MPa
fprintf(‘Tensile Strength = %3.0f MPan’,TS)
% Report Fracture Strength (Breaking Strength)
% Index last element in engineering stress vector
sigbreak = sigeng(end);
fprintf(‘Fracture (Breaking) Strength = %3.0f MPan’,sigbreak)
% sigyield = 0.2% offset yield strength:
% Need sensitive stress-strain plot, part (c) for sigyield:
% (c) Plot first 11 data points for more sensitive strain scale (to
~0.02):
% Open new plot window:
figure(9)
plot(epseng(1:11),sigeng(1:11),’k-o’);
title({‘7075-T651 Aluminum’;’Stress vs Strain for Offset Yield
Strength’})
xlabel(‘Strain’)
ylabel(‘Stress, MPa’)
hold on
% Create points for offset line:
10
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
% Top of Offset Line above yield point = 600 MPa
sigoffsettop = 600; %MPa
% Strain values for Offset Line:
epsoffset = [0.002 ((sigoffsettop/E)+0.002)];
sigoffset = [0 sigoffsettop];
plot(epsoffset,sigoffset,’k’);
% Show grid on plot to determine sigyield by inspection
grid on
% From plot, 0.2% Offset YS very close to 8th data point
sigyield = round(sigeng(8), -1); % MPa, rounded to 10’s place
% From plot, offset strain very close to 8th data point
epsyield = round(epseng(8), 3); %rounded to hundredths
% Display 0.2% Offset Yield Strength in output:
fprintf(‘Yield Strength (0.2%% offset) = %3.0f MPan’,sigyield)
% (d) True stress & strains to UTS, plot & compare to Engr stressstrain
% Per Dowling (2019, 5th ed.) Figure 3.20 for true stress/strain:
% Determine 2 x yield strain formula limit:
formlimit = 2*epsyield; % limit = 0.02
% for values below 2 x yield strain (formlimit):
%
sigtru = engineering stress
% for values above (beyond) 2 x yield strain:
%
sigtru = sigeng*(1+epseng)
% for all values of true strain up to UTS:
%
epstru = ln(1+epseng)
% Calculate true stress and true strain values
% Use formulas from Dowling (Fig3.20,5th ed; Fig4.19,4th ed):
% True stress (sigtru) = engnr stress (sigeng) up to 2 x yield strain
% 2 x yield strain =~ 0.02 (first 11 points)
sigtru_1 = sigeng(1:11);
% True stress = engrg stress*(1+engrg strain) to UTS (points 12-17)
sigtru_2 = (ones(1,6)+epseng(12:17)).*sigeng(12:17);
% These are true stress values to UTS
% Concatenate true stress values to form one vector of 12 elements:
sigtru = [sigtru_1,sigtru_2];
% Evaluate true strain values from epstru = ln(1+epseng)
epstru = log(ones(1,17)+epseng(1:17));
% Plot true stress-strain up to UTS:
figure(10)
plot(epstru,sigtru,’b*’)
title({‘7075-T651 Aluminum’;’7071-T651 Aluminum True Stress-True
Strain to UTS’})
xlabel(‘True Strain, mm/mm’)
ylabel(‘True Stress, MPa’)
grid on
% Plot True & Engr on one plot for comparison:
figure (11)
plot(epstru,sigtru,’b-‘)
title({‘7071-T651 Aluminum’;’True vs Engr Comparison’;’Data up to
UTS’})
xlabel(‘Strain (engr or true), mm/mm’)
ylabel(‘Stress (engr or true), MPa’)
grid on
11
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
hold on
plot(epseng(1:17),sigeng(1:17),’r-‘)
legend(‘true stress-strain’,’engrg stress-strain’,’Location’,’east’)
% (e) True stresses & true plastic strains beyond yielding to UTS
% (from part (d):
% Points beyond yielding to UTS are points 9 through 17 (9 points)
% Note still working only up to UTS, so true strain valid
% True plastic strain = true strain – true elastic strain
% True elastic strain = true stress/E
epstruelastic = sigtru/E;
epstruplastic = epstru-epstruelastic;
figure(12)
plot(epstruplastic(9:17),sigtru(9:17),’b*’)
grid on
title({‘7071-T651 Aluminum’;’Stress-True Plastic Strain’;’Data from
beyond yielding to UTS’})
xlabel(‘True Plastic Strain’)
ylabel(‘True Stress, MPa’)
hold on
% Show true stress vs true plastic strain on log-log plot:
% Calculate log of true plastic strain and log of true stress:
logepstruplastic = log10(epstruplastic);
logsigtru = log10(sigtru);
% Determine n & H:
Fit = polyfit(logepstruplastic(9:17),logsigtru(9:17),1);
% Display n & H:
n = Fit(1);
H = 10^Fit(2); %MPa
fprintf(‘The fitting constants are:ntStrain Hardening Exponent n =
%5.4fntStrength Coefficient H = %3.0f MPan’,n,H)
% Generate “fit” curve from true strain range and “n” & “H”, power
eqn:
% Create linearly spaced values of true plastic strain:
epstruplasticfit = linspace(epstruplastic(9),epstruplastic(17),1000);
% Use flow equation to calculate fitted true stress:
sigtrufit = H*(epstruplasticfit.^n);
% Add fit line to plot (figure 5)
plot(epstruplasticfit,sigtrufit,’r-‘)
legend(‘Data Points’,’Fit Line’,’Location’,’northwest’)
% (f) Now log-log plot, points after yield to UTS:
figure(13)
plot(logepstruplastic(9:17),logsigtru(9:17),’b*’)
title({‘7075-T651 Aluminum’;’Log(True Stress) vs Log(True Plastic
Strain)’})
xlabel(‘Log(True Plastic Strain)’)
ylabel(‘Log(True Stress), True Stress in MPa’)
grid on
hold on
% Show data and fit line on the plot:
plot(log10(epstruplasticfit),log10(sigtrufit),’r-‘)
legend(‘Data Points’,’Fit Line’,’Location’,’northwest’)
12
ME180 HMWK ASSGT 2
PROBS 1,2,3 SOLUTION
%(g) True fracture point:
% sigtrufrac = true fracture stress = fracture load/fracture area
% Pf = fracture load = fracture stress * initial area
Pf = sigeng(20)*(pi*(Di^2)/4); % Newtons
sigtrufrac = Pf/(pi*(Df^2)/4); % MPa
% Report true fracture stress:
fprintf(‘True Fracture Strength = %3.0f MPan’,sigtrufrac)
% Mirone correction:
% Need true strain beyond necking:
% epstrudiff = epstrufract – epsult (epstru>epsult)
% Need true strain at fracture, epstrufrac = 2*ln(Di/Df)
epstrufrac = 2*log(Di/Df);
% Report true fracture strain:
fprintf(‘True Fracture Strain = %5.5fn’,epstrufrac)
% for fracture point, use last point (pt 20) & ultimate point (pt 17)
epstrudiff = epstrufrac – epstru(17);
M = 1 – ((0.6058)*(epstrudiff^2))+((0.6317)*(epstrudiff^3))((0.2107)*(epstrudiff^4));
% Mirone corrected true fracture stress:
sigtrufracM = M*sigtrufrac;
% Last plot, log-log plot repeat w/added fracture point, fit line
extended:
figure(14)
plot(logepstruplastic(9:17),logsigtru(9:17),’b*’)
title({‘7075-T651 Aluminum’;’Log(True Stress) vs Log(True Plastic
Strain) to Fracture Point’})
xlabel(‘Log(True Plastic Strain)’)
ylabel(‘Log(True Stress), True Stress in MPa’)
grid on
hold on
plot(log10(epstrufrac),log10(sigtrufracM),’g*’)
% Show data and fit line on the plot:
epstruplasticfit2 = linspace(epstruplastic(9),epstrufrac,1000);
sigtrufit2 = H*(epstruplasticfit2.^n);
plot(log10(epstruplasticfit2),log10(sigtrufit2),’r-‘);
legend(‘Data Points’,’Fracture Point’,’Fit
Line’,’Location’,’northwest’);
% After inspection, fit line is NOT consistent with fracture point
fprintf(‘Fit line is NOT consistent with fracture pointnn’);
% (h) Tensile Test Properties Table:
fprintf(‘(h) Table of tensile test properties:nn’);
% Create array of values for fprintf to “print” from:
tableArray1 = [E sigyield TS sigbreak EL RA];
% Table Title:
fprintf(‘tttttttt
Tensile Test Propertiesn’);
% Underline Title:
fprintf(‘t—————————————————–n’);
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