## Materials Science and Engineering: An Introduction

$F_{y} = 44,850 N$
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: $F_{y} = σ_{y}A_{0} = (345 \times 10^{6} N/m^{2})(130 \times 10^{-6} m^{2}) = 44,850 N$