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The data shown in the following table are from a tensile test of high-strength steel. TENSILE-TEST DATA Load (lb) Elongation (in.) 1.000 2.000 6.000 10.000 12.000 12.900 13.400 13.600 13.800 14.000 14,400 15.200 16.800 18.400 20.000 22,400 22.600 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 ultimate stress percent elongation in 2.00 in percent reduction in area 0.0336 0.0507 0.1108 Fracture The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure). Gage length @ At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in. Plot the conventional stress-strain curve for the steel. (Submit a file with a maximum size of 1 MB.) Choose File No fie chosen This answer has not been graded yet ultimate stress, percer percent elongation in 2.00 in. (at fracture), and percent reduction in area (at fracture). (Enter the proportional limit, modulus of elasticity, yield stress, and ultimate stress in psi.) Determine the proportional limit, modulus of elasticity (the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, proportional limit 59.91 x modulus of elasticity yield stress 30046 Xp 68.897 X P 112.8307539 X psi 30.83%

The data shown in the following table are from a tensile test of high-strength steel. TENSILE-TEST DATA Load (lb) Elongation (in.) 1.000 2.000 6.000 10.000 12.000 12.900 13.400 13.600 13.800 14.000 14,400 15.200 16.800 18.400 20.000 22,400 22.600 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 ultimate stress percent elongation in 2.00 in percent reduction in area 0.0336 0.0507 0.1108 Fracture The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure). Gage length @ At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in. Plot the conventional stress-strain curve for the steel. (Submit a file with a maximum size of 1 MB.) Choose File No fie chosen This answer has not been graded yet ultimate stress, percer percent elongation in 2.00 in. (at fracture), and percent reduction in area (at fracture). (Enter the proportional limit, modulus of elasticity, yield stress, and ultimate stress in psi.) Determine the proportional limit, modulus of elasticity (the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, proportional limit 59.91 x modulus of elasticity yield stress 30046 Xp 68.897 X P 112.8307539 X psi 30.83%

Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
5th Edition
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Saeed Moaveni
Chapter7: Length And Length-related Variables In Engineering
Section: Chapter Questions
Problem 43P: A 10 cm long rectangular bar (when subjected to a tensile load) deforms by 0.1 mm. Calculate the...
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The data shown in the following table are from a tensile test of high-strength steel. TENSILE-TEST DATA Load (lb) Elongation (in.) 1,000 2,000 6,000 10,000 12,000 12,900 13,400 13,600 13,800 14,000 14,400 15,200 16,800 18,400 20,000 22,400 22,600 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 Fracture The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure). Gage length Ⓡ At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in. Plot the conventional stress-strain curve for the steel. (Submit a file with maximum size of 1 MB.) Choose File No file chosen This answer has not been graded yet. Determine the proportional limit, modulus of elasticity (the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in. (at fracture), and percent reduction in area (at fracture). (Enter the proportional limit, modulus of elasticity, yield stress, and ultimate stress in psi.) proportional limit 59.91 X psi modulus of elasticity 30046 X psi 68.897 X psi yield stress ultimate stress 112.8307539 X psi percent elongation in 2.00 in. percent reduction in area 30.83✔ %
Transcribed Image Text:The data shown in the following table are from a tensile test of high-strength steel. TENSILE-TEST DATA Load (lb) Elongation (in.) 1,000 2,000 6,000 10,000 12,000 12,900 13,400 13,600 13,800 14,000 14,400 15,200 16,800 18,400 20,000 22,400 22,600 0.0002 0.0006 0.0019 0.0033 0.0039 0.0043 0.0047 0.0054 0.0063 0.0090 0.0102 0.0130 0.0230 0.0336 0.0507 0.1108 Fracture The test specimen has a diameter of 0.505 in. and a gage length of 2.00 in. (see figure). Gage length Ⓡ At fracture, the elongation between the gage marks is 0.12 in. and the minimum diameter is 0.42 in. Plot the conventional stress-strain curve for the steel. (Submit a file with maximum size of 1 MB.) Choose File No file chosen This answer has not been graded yet. Determine the proportional limit, modulus of elasticity (the slope of the initial part of the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent elongation in 2.00 in. (at fracture), and percent reduction in area (at fracture). (Enter the proportional limit, modulus of elasticity, yield stress, and ultimate stress in psi.) proportional limit 59.91 X psi modulus of elasticity 30046 X psi 68.897 X psi yield stress ultimate stress 112.8307539 X psi percent elongation in 2.00 in. percent reduction in area 30.83✔ %
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