Question

A straight bar 500mm long is 25 mm in diameter for the first 200mm length and 15mm diameter for the remaining length if the bar is subjected to an axial pull of 15 Kant what is the deformation of the bar

Answer 1

$\delta=\delta_1+\delta_2=\frac FE(\frac{L_1}{S_1}+\frac{L_2}{S_2})=0.66~mm.$