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tatiyna
2 years ago
8

For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.

On the basis of this information, what engineering stress (in MPa) is necessary to produce an engineering plastic strain of 0.250
Engineering stress (MPa) Engineering strain
203 0.160
224 0.289
Engineering
1 answer:
kirza4 [7]2 years ago
7 0

Answer:

203.0160

Explanation:

Because you add then subtract then multiply buy 7 the subtract then divide then you add that to the other numbers you got than boom

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A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength
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a) Ef = 0.755

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