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: 6.7a


$F_{y} = 44,850 N$

Work Step by Step

Given: brass alloy- plastic deformation begins at 345 MPa stress modulus of elasticity - 103 GPa = $103 \times 10^{9} N/m^{2}$ Required: maximum load to be applied to specimen with cross sectional area of $130 mm^{2}$ without plastic deformation ($F_{y}$) Solution: Using Equation 6.1, it follows: $F_{y} = σ_{y}A_{0} = (345 \times 10^{6} N/m^{2})(130 \times 10^{-6} m^{2}) = 44,850 N$
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