Question

2. A load of 50 kg is applied to the lower end of a steel rod 80 cm long and 0.6 cm in diameter. How much will the rod stretch? Y = 190 GPa for steel.

Answer 1

$\sigma=E\epsilon\to F/A=E\Delta l/l\to \Delta l=\frac{Fl}{EA}=$

$=\frac{50\cdot9.81\cdot 0.8\cdot4}{190\cdot10^9\cdot 3.14\cdot 0.006^2}=7.3\cdot10^{-5}m$ Answer