Question P.205: A uniform bar of length L, cross-sectional area A, and unit ...

\delta=\frac{P L}{A E}

From the figure:

δ = dδ
P = Wy = (ρAy)g
L = dy

\begin{aligned}& d\delta={\frac{(\rho A y)g\,d y}{A E}}\\& \delta=\frac{\rho g}{E}\int_{0}^{L}y\,d y=\frac{\rho g}{E}\left[\frac{y^{2}}{2}\right]_{0}^{L}\\& \delta=\frac{\rho g}{2{ E}}\left[\,L^{2}-0^{2}\,\right]={\rho g L}^{2}/2E\;{ok !}\end{aligned}

Given the total mass M:

\begin{aligned}& \rho=M/V=M/A L\\&\delta={\rho g L^{2}}/{2E}= ({M}/{AL})({gL^2}/{2E})\\& \delta={M g L}/{2A E}\,\ o k!\end{aligned}

Another Solution:

The weight will act at the center of gravity of the bar.

\delta=\frac{P L}{A E}

Where:

P = W = (ρAL)g
L = L/2

\begin{aligned}& \delta=\frac{[(\rho A L)g](L/2)}{A E}\\& \delta=\frac{\rho g L^{2}}{2E}\;o k!\end{aligned}

For you to feel the situation, position yourself in pull-up exercise with your hands on the bar and your body hang freely above the ground. Notice that your arms suffer all your weight and your lower body fells no stress (center of weight is approximately just below the chest). If your body is the bar, the elongation will occur at the upper half of it.