$$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{B}}}{480 \mathrm{mm}}=\frac{2.58 \mathrm{mm}}{600 \mathrm{mm}} \\ \therefore {\mathrm{V}}_{\mathrm{B}}=2.064 \mathrm{mm} \end{gathered}$$

Since member $\left(1\right)$ is connected to the rigid plate at $\mathrm{B}$, it elongates by $2.064 \mathrm{mm}$. Use this elongation to determine the strain in member $\left(1\right)$.

$$\begin{gathered} {\mathrm{\epsilon }}_1=\frac{2.064 \mathrm{mm}}{1,200 \mathrm{mm}} \\ =0.00172 \mathrm{mm}/\mathrm{mm} \\ =\mathbf{1}\mathbf{,}\mathbf{720}\mathbf{ }\mathbf{\mu \epsilon } \end{gathered}$$