Consider a cylindrical titanium wire 3.0 mm in diameter and 2.5x\(10^4\) mm long. Calculate its elongation when a load of 500 N is applied. Assume that the deformation is totally elastic.

Titanium

\(l_{0} = \) 3.0mm

\(d_{0} =\) \(2.5 x 10^4\)mm

Load: F = 500 N

Elongation \(\implies\) \(\Delta l\) = ?

Ti \(\implies\) E = 107GPA

\(A = A_{0} = {\pi \over 4} d_{0}^2 \) \(\quad\) \(\epsilon = {\Delta l \over l_{0}}\) \(\quad\) \(\sigma = {F \over A_{0}}\) \(\quad\) \(\sigma = E \epsilon\)

Take the assumptions and solve for elongation:

\(\Delta l = \epsilon l_{0}\)

\(\Delta l = {\sigma \over E} l_0\)

\(\Delta l = {F \over A_0} {l_0 \over E}\)

\(\Delta l = {F l_0 \over {\pi \over 4} {d_0}^2 E}\)

\( \Delta l = {(2.5x10^{4}mm)(500 N) \over ({\pi \over 4})(3x10^{-3}m)^2(107x10^9 N/m^2)} \) = 0.0165 m

\( \Delta l\) = 16.5 mm