Como ya tenemos todos los vales necesarios sustituimos en la formula.
\(\delta ac=\frac{PL}{AE}=\left(-18.34\times10^3N\right)\left(0.4\ m\right)\div\left(\pi\left(0.01m\right)^2\right)\left(200\times10^9Pa\right)=-1.16\times10^{-4}\approx-0.116\ mm\)
\(\delta bd=\left(-91.66\times10^3\ N\right)\left(0.4\ m\right)\div\left(\pi\left(0.01m\right)^2\right)\left(200\times10^9Pa\right)=-5.83\times10^{-4}\approx-0.583\ mm\)