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A tensile test was conducted on a specimen that had an original diameter of 0.5 in. It fractured with an applied load of 71100lb. Determine the true stress and true strain at fracture if measured diam...

A tensile test was conducted on a specimen that had an original diameter of 0.5 in. It fractured with an applied load of 71100lb. Determine the true stress and true strain at fracture if measured diameter of the fracture part was 0.3in.

I believe for fracture, true stress and strain are:

True strain = ln(original area/final area)

True stress = applied load/final area

However, Im not quite sure

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Answer #1

Original diameter d,5 duue stress, 11100 午1005859.24 psi true strain. Ini 印1り1슷:1 , n.hty! Ao ET 102165

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