Answer
$Δl = 13.8 mm$
Work Step by Step
Given:
cylindrical nickel wire 2.0 mm in diameter and $3 \times 10^{4}$ mm long
Required:
elongation when 300 N load is applied assuming the deformation is totally elastic
Solution:
Using combination of Equations 6.1, 6.2, and 6.5, and from Table 6.1 and that $E = 207 GPa = 207 \times 10^{9} N/m^{2}$:
$Δl = l_{0}ε = l_{0} \frac{σ}{E} = \frac{l_{0}F}{Eπ(\frac{d_{0}}{2})^{2}} = \frac{4l_{0}F}{Eπd_{0}^{2}} = \frac{(4)(30 m)(300 N)}{(207 \times 10^{9} N/m^{2})(π)(2 \times 10^{-3} m)^{2}} = 0.0138 m = 13.8 mm$
