A cylindrical metal specimen 12.0 mm in diameter and 120 mm long is to be subjec

Question

A cylindrical metal specimen 12.0 mm in diameter and 120 mm long is to be subjected to a tensile stress of 40 MPa . at this stress level the resulting deformation is totally elastic. a) if the elongation must be less than 0.065 mm, what is the maximum strain the metal specimen can sustain? (use 2 decimal places) b) using the maximum strain you found in a , calculate the minimum stifness or young's modulus required of the specimen so that it does not elongate past 0.065 mm under the applied stress of 40 MPa

Explanation / Answer

a) strain = delta L / L = 0.065/120 = 5.4 * 10^-4 ...... b) E = stress/strain = 40*10^6/(5.4 * 10^-4) = 74.07 GPa

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