Using this definition, the axial elongation corresponding to the measured $1,200 \mathrm{\mu \epsilon }$ strain can be computed for rod $\left(1\right)$:

${\mathrm{\delta }}_1={\mathrm{\epsilon }}_1{\mathrm{L}}_1=(1,200 \mathrm{\mu \epsilon })\frac{1 \mathrm{mm}/\mathrm{mm}}{1\times {10}^6\mathrm{\mu \epsilon }}(1,500 \mathrm{mm})=1.8 \mathrm{mm}$