Question

A brass rod is 15cm long with a circular cross-section of 2cm. A force of 10kN is applied to the end of the rod parallel to its length. The young's modulus for the rod under compression is 90x10^9 N/m^2. The rod shortens by how many millimetres?

Answer 1

Young's modulus is

$$E=\frac{\sigma}{\epsilon},$$

Strain is

$$\epsilon=\frac{\Delta l}{l}=\frac{\sigma}{E}.$$

The stress is

$$\sigma=\frac{F}{A}=\frac{4F}{\pi d^2}.$$

Finally,

$$\Delta l=l\frac{\sigma}{E}=l\frac{F}{AE}=0.083\text{ mm}.$$