## Materials Science and Engineering: An Introduction

$l_{f} = 76.25 mm$
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: $l_{f} = l_{0}(1 + \frac{σ}{E}) = (76 mm)(1 + \frac{345 MPa}{103 \times 10^{3} MPa} )= 76.25 mm$