Answer
$Δl = 23.54 mm$
Work Step by Step
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:
$Δl = l_{i}-l_{0} = 523.54 mm - 500 mm = 23.54 mm$
