Using this definition, the axial elongation corresponding to the measured $+910 \mathrm{\mu \epsilon }$ strain can be computed for rod $\left(1\right)$:

$$\begin{gathered} {\mathrm{\delta }}_1={\mathrm{\epsilon }}_1{\mathrm{L}}_1 \\ =(+910 \mathrm{\mu \epsilon })\frac{1 \mathrm{mm}/\mathrm{mm}}{1\times {10}^6\mathrm{\mu \epsilon }}(2,000 \mathrm{mm}) \\ =+1.82 \mathrm{mm} \end{gathered}$$