## Materials Science and Engineering: An Introduction

$Δl = 23.54 mm$
Given: metal alloy with true stress of 345 MPa that produces true plastic strain of 0.02 original length is 500 mm value for n = 0.22 Required: elongation when true stress of 415 MPa is applied Solution: Solve value for K by rearranging Equation 16.9: $K = \frac{σ_{T}}{(ε_{T})^{n}} =\frac{345 MPa}{(0.02)^{0.22}} = 815.82 MPa$ Solving for $ε_{T}$ using Equation 6.19 when 415 MPa is applied: $ε_{T} = (\frac{σ_{T}}{K})^{1/n} = (\frac{415 MPa}{815.82 MPa})^{1/0.22} = 0.046 = ln (\frac{l_{i}}{l_{0}})$ $l_{i} = l_{0}e^{0.046} =(500 mm)e^{0.046} = 523.54 mm$ Solving for elongation: $Δl = l_{i}-l_{0} = 523.54 mm - 500 mm = 23.54 mm$