Answer
$ε= 1.39 \times 10^{-3}$
Work Step by Step
Given:
copper with 15.2 mm x 19.1 mm (0.60 in. x 0.75 in.) cross section
tension of 44,500 N (10,000 lbf)
Required:
strain
Solution:
Using the combination of Equations 6.1 and 6.5:
$ε = \frac{σ}{E} = \frac{F}{A_{0}E}$
$A_{0} = (15.2 mm x 19.1 mm) = 290.32 mm^{2} = 2.90 \times 10^{-4} m^{2}$ and $E = 110 GPa = 110 \times 10^{9} N/m^{2}$
Substituting, it follows:
$ε= \frac{44,500 N}{(2.90 \times 10^{-4} m^{2})(110 \times 10^{9} N/m^{2})} = 1.39 \times 10^{-3}$
