Elongation of uniform bar under its self weight

Consider a uniform vertical bar of length $L$, area $A$, Young's modulus $E$ and unit weight (weight density) $\gamma = \rho g$, hanging from a fixed support at the top and carrying only its own weight.

 ═══════  fixed support
   │ ↑ x measured from the free (bottom) end
   │
   │  dx
   │
   ○  free end

Derivation

Take a section at distance $x$ from the free end. The load it carries is the weight of the bar below it:

$$P_x = \gamma A x$$

The stress at the section is $\sigma_x = \dfrac{P_x}{A} = \gamma x$, which varies linearly from zero at the free end to $\gamma L$ at the support.

The elongation of a small element of length $dx$:

$$d\delta = \frac{\sigma_x}{E}dx = \frac{\gamma x}{E}dx$$

Total elongation:

$$\delta = \int_0^L\frac{\gamma x}{E}dx = \frac{\gamma L^2}{2E}$$

With the total weight $W = \gamma A L$:

$$\boxed{\delta = \frac{\gamma L^2}{2E} = \frac{WL}{2AE}}$$

Meaning: the elongation due to self-weight is half the elongation that the same weight $W$ would cause if it were applied as a point load at the free end ($\delta = WL/AE$). Maximum stress $\sigma_{max} = \gamma L$ occurs at the support.

Example: a steel bar of $L = 100$ m with $\gamma = 77$ kN/m³ and $E = 200$ GPa: $\delta = \dfrac{77\times10^{-6}\ \text{N/mm}^3\times(10^5)^2}{2\times200\times10^3} = 1.925$ mm, with $\sigma_{max} = 7.7$ N/mm².