Solution to Problem 218 Axial Deformation

$\delta = \dfrac{PL}{AE}$
 

 

from the frigure:
$d\delta = \dfrac{dP \, x}{AE}$
 

Where:
dP = centrifugal force of differential mass
dP = dM ω² x = (ρA dx)ω² x
dP = ρAω² x dx
 

$d\delta = \dfrac{(\rho A \omega^2 x \, dx) x}{AE}$

$\delta = \dfrac{\rho \omega^2}{E} \displaystyle \int_0^L x^2 \, dx = \dfrac {\rho \omega^2}{E} \left[ \dfrac{x^3}{3} \right]_0^L$

$\delta = \dfrac{\rho \omega^2}{E}\, [ \, L^3 - 0^3 \, ]$

$\delta = \rho \omega^2 L^3 / 3E$       okay!