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Nataly
2 days ago
13

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

his material elongate when a true stress of 414 MPa (60050 psi) is applied if the original length is 520 mm (20.47 in.)? Assume a value of 0.22 for the strain-hardening exponent, n.

Physics
1 answer:
inna [2.9K]2 days ago
8 0

Answer:

24.348 mm

Explanation:

NB: I'll be uploading images to represent missing mathematical expressions or unique characters that aren’t easily typed

K = d / €^n

Note: d refers to the Greek letter epsilon.

K = 345 / 0.02⁰.²² = 816 mPa

The true strain based on the stress of 414 mPa =

€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576

The actual relationship between true strain and length is given by

€ = ln(Li/Lo)

To isolate Li by rearranging,

Li = Lo.e^€

Li = 520e⁰.⁰⁴⁵⁷⁶

Li = 544.348 mm

The elongation can be calculated from

Change in L = Li - Lo = 544.348 - 520 change in L = 24.348 mm.

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