Question

For some metal alloy, a true stress of 345 MPa (50,000 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 415 MPa (60,000 psi) is applied if the original length is 500 mm (20 in.)? Assume a value of 0.22 for the strain-hardening exponent, n.

Answer 1

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- Given Data: | true stress (0) = 375 MPa true strain (&p= 0.02 Case 1 Specimen Original length = 500mm = 0:5m. Case dit true Stress (62) = 415 MPa Strain hardening exponent (n)-0.22 Solution: By Hollomon's equation reken ro applied Stress K> strength coefficient (constant for a given material) E> Strain. raen. Stress a (strain (since kis constant

E2 = 0.031703 ... Strain induced in case a. Starin e elongation original dength 0:031703. elongation Soo mm elongation = 500x0·031703 - = 15.85 mm • the elongation for 415 MPa stress is 13.85 onn.