One end of a uniform rod of mass M and cross sectional area A is suspended from a rigid support and equal mass M is suspended from the other end. What is the stress at the mid point of the rod.

Since the rod is uniform, half of its weight can be taken to act at its mid point. For mid point of the rod the total weight suspended,

$\text{F}\text{=}$ weight of suspended mass + weight of half the rod $\text{= Mg}\text{+}\frac{\text{Mg}}{2}=\frac{3}{2}\text{Mg}$

Stress at mid point $=\frac{\text{F}}{\text{A}}=\frac{3\text{Mg}}{2\text{A}}$