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

The stretch of a rod is given by the following expression:

$$\Delta L = \dfrac{FL}{AY}$$

where $F = mg$ is the force applied to the lower end of the steel rod, $L = 80cm = 0.8m$ is the length of the rod, $A = \pi d^2 /4$ is a cross-section area of the rod and $Y = 190\times 10^{9}Pa$ is the Young modulus. Thus, obtain:

$$\Delta L = \dfrac{FL}{AY} = \dfrac{4mgL}{\pi d^2Y} = \dfrac{4\cdot 50\cdot 9.8\cdot 0.8}{\pi\cdot 0.006^2\cdot 190\cdot 10^9} \approx 0.073mm$$

Answer. 0.073mm.