For Problem 2: E 0 0 0 □ (MPa) 220 200 180 160 140 120 100 80 60 40 20 0.001 0.002 0.003 0.005 0.01 0.02 0.04 0.06 0.08

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answerhappygod
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For Problem 2: E 0 0 0 □ (MPa) 220 200 180 160 140 120 100 80 60 40 20 0.001 0.002 0.003 0.005 0.01 0.02 0.04 0.06 0.08

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For Problem 2 E 0 0 0 Mpa 220 200 180 160 140 120 100 80 60 40 20 0 001 0 002 0 003 0 005 0 01 0 02 0 04 0 06 0 08 1
For Problem 2 E 0 0 0 Mpa 220 200 180 160 140 120 100 80 60 40 20 0 001 0 002 0 003 0 005 0 01 0 02 0 04 0 06 0 08 1 (115.24 KiB) Viewed 16 times
For Problem 2: E 0 0 0 □ (MPa) 220 200 180 160 140 120 100 80 60 40 20 0.001 0.002 0.003 0.005 0.01 0.02 0.04 0.06 0.08 0.10 0.15 0.20 0.25 0.3* 100 50 150 175 185 192 186 175 193 198 200 201 200 197 0.04, 198 0.15, 197 0.2, 192 0.3, 175 0.2 0.3 0.02, 193 0.01, 185 0.005, 175 0.003, 150 0.002, 100 0.001, 50 0 0,0 0 0.02 0.04 0.08, 201 0.06, 200 0.06 0.08 0.1, 200 0.1 0.12 0.14 0.16 € (m/m) 0.18 0.25, 186 0.22 0.24 0.26 0.28
2. 15 pts. Using the engineering stress-strain data on the last page for a newly discovered metal: a. Estimate the modulus of elasticity 3 pts. 3 pts. b. Estimate the industry-standard yield strength of the metal c. Determine the ultimate tensile strength of the alloy 3 pts. d. Determine the percent elongation at fracture 3 pts. e. Estimate the modulus of resilience 3 pts.
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