Materials Science and Engineering: An Introduction

Published by Wiley
ISBN 10: 1118324579
ISBN 13: 978-1-11832-457-8

Chapter 6 - Mechanical Properties of Metals - Questions and Problems - Page 207: 6.7b


$l_{f} = 76.25 mm$

Work Step by Step

Given: brass alloy- plastic deformation begins at 345 MPa stress modulus of elasticity - 103 GPa = $103 \times 10^{3} MPa$ Required: maximum length stretched without deformation if original length is 76 mm Solution: Using a combination of Equation 6.2 and 6.5, it follows: $l_{f} = l_{0}(1 + \frac{σ}{E}) = (76 mm)(1 + \frac{345 MPa}{103 \times 10^{3} MPa} )= 76.25 mm$
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