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December 12, 2016 14:05
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% Calculations for ME 241 Lab 6: Tensile Testing | |
% Group 2, December 12, Fall 2016 | |
clear; | |
clc; | |
F_al = [0 | |
181.1 | |
362.2 | |
543.3 | |
724.4 | |
769.6 | |
814.9 | |
860.2 | |
905.5 | |
916.6 | |
927.8 | |
938.9 | |
950.1 | |
961.2 | |
972.4 | |
983.5 | |
994.7 | |
1005.8 | |
1017.0]; | |
d_al = [0.0000 | |
0.0008 | |
0.0016 | |
0.0024 | |
0.0032 | |
0.0034 | |
0.0037 | |
0.0042 | |
0.0060 | |
0.0071 | |
0.0086 | |
0.0108 | |
0.0141 | |
0.0189 | |
0.0258 | |
0.0357 | |
0.0500 | |
0.0706 | |
0.1000]; | |
F_steel = [0 | |
242.8 | |
485.5 | |
728.3 | |
971.1 | |
1031.8 | |
1092.4 | |
1153.1 | |
1213.8 | |
1236.8 | |
1259.8 | |
1282.7 | |
1305.7 | |
1328.7 | |
1351.6 | |
1374.6 | |
1397.5 | |
1420.5 | |
1443.5]; | |
d_steel = [0 | |
0.0004 | |
0.0007 | |
0.0011 | |
0.0015 | |
0.0016 | |
0.0018 | |
0.0023 | |
0.0039 | |
0.0051 | |
0.0069 | |
0.0098 | |
0.0142 | |
0.0209 | |
0.0309 | |
0.0460 | |
0.0683 | |
0.1014 | |
0.1500]; | |
% calculate stress at dogbone area (smallest area) | |
% stress = force / area (psi) | |
A = 0.25 * 0.100; | |
stress_al = F_al / A; | |
stress_steel = F_steel / A; | |
% calculate strain = (L - L0) / L0 (in/in) | |
L0 = 1; % original length | |
strain_al = d_al / L0; | |
strain_steel = d_steel / L0; | |
% calculate modulus of elasticity | |
% E = stress / strain | |
E_al = stress_al ./ strain_al; | |
E_steel = stress_steel ./ strain_steel; | |
% Ran code and hand picked the linear regions of modulus of elasticity: | |
% Also need to get the appropriate X values (displacements.) | |
E_linear_al = E_al(2:6); | |
E_linear_steel = E_steel(2:7); | |
d_linear_al = d_al(2:6); | |
d_linear_steel = d_steel(2:7); | |
% Use polyfit to get the slope and intercept for linear fit: | |
p_al_coeffs = polyfit(d_linear_al, E_linear_al, 1); | |
p_steel_coeffs = polyfit(d_linear_steel, E_linear_steel, 1); | |
p_al = p_al_coeffs(1) * d_linear_al + p_al_coeffs(2); | |
p_steel = p_steel_coeffs(1) * d_linear_steel + p_steel_coeffs(2); | |
modulus_elasticity_aluminum = p_al_coeffs(2) | |
modulus_elasticity_steel = p_steel_coeffs(2) | |
% Determine tensile strength... that's just stress when it broke. | |
tensile_strength_aluminum = stress_al(end) | |
tensile_strength_steel = stress_steel(end) | |
hold on | |
plot(d_al, F_al) | |
plot(d_steel, F_steel) | |
title('Force vs Displacement'); | |
xlabel('Displacement (in)'); | |
ylabel('Force (lb)'); | |
legend('6061-T6 Aluminum', '1018 Cold Rolled Steel', 'location', 'best') | |
figure(); | |
plot(d_linear_steel, E_linear_steel, 'b', d_linear_steel, p_steel, 'r') | |
title('Modulus of Elasticity (Steel)') | |
xlabel('Strain (in/in)') | |
ylabel('Stress (psi)') | |
label1 = ['E (steel): ' num2str(modulus_elasticity_steel)] | |
legend('Experimental Data', label1, 'location', 'best') | |
figure(); | |
plot(d_linear_al, E_linear_al, 'b', d_linear_al, p_al, 'r') | |
title('Modulus of Elasticity (Aluminum)') | |
xlabel('Strain (in/in)') | |
ylabel('Stress (psi)') | |
label2 = ['E (aluminum): ' num2str(modulus_elasticity_aluminum)] | |
legend('Experimental Data', label2, 'location', 'best') | |
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