A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.20 is produced when a true stress of 575 MPa (83,500 psi) is applied; for the same metal, the value of K in Equation 6.19 is 860 MPa (125,000 psi). Calculate the true strain that results from the application of a true stress of 600 MPa (87,000 psi). Part 1 To solve this problem, calculate the value of n using the known relationship between true stress, true strain, the strain- hardening coefficient, and K (Equation 6.19). Or = Ke, What is n? n = i

A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.20 is produced when a true stress of 575 MPa (83,500 psi) is applied; for the same metal, the value of K in Equation 6.19 is 860 MPa (125,000 psi). Calculate the true strain that results from the application of a true stress of 600 MPa (87,000 psi). Part 1 To solve this problem, calculate the value of n using the known relationship between true stress, true strain, the strain- hardening coefficient, and K (Equation 6.19). Or = Ke, What is n? n = i

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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A tensile test is performed on a metal specimen, and it is found that a true plastic strain of 0.20 is produced when a true stress of
575 MPa (83,500 psi) is applied; for the same metal, the value of K in Equation 6.19 is 860 MPa (125,000 psi). Calculate the true
strain that results from the application of a true stress of 600 MPa (87,000 psi).
Part 1
To solve this problem, calculate the value of n using the known relationship between true stress, true strain, the strain-
hardening coefficient, and K (Equation 6.19).
or = Ke,
What is n?
n =
i

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Question 4 Save Answer You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as ✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is ✓ MPa for 0.1 plastic strain, proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k= MPa. Hence, we can predict that we need a prior plastic strain of approximately True Stress (MPa) True Stress (MPa) True Stress (MPa) 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 Ep = 0.1 0.01 Ep=0.2 0.01 Ep=0.3 0.01 0.02 True Strain 0.02 0.03 0.02 0.03 True Strain 0.03 0.04 0.04 0.04 True Stress (MPa) 0.05 0.000 0.001 True Stress (MPa) Ep=0.1 True…
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A cylindrical specimen of a steel alloy 10 mm in diameter and 100mm long is pulled in tension at a load of 23500N and the strain found to be 0.0015, the elongation will be: Select one: 0.15mm Ob. 0.023mm 0.100mm Od. 0.0015mm
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I am not reallyunderstanding what this problem wants with the wording, coud you help me clarify that please? Thank you so much!!