From the force deformation relationship, compute the deformation in member $\left(1\right)$:

$$\begin{gathered} {\mathrm{\delta }}_1=\frac{{\mathrm{F}}_1{\mathrm{L}}_1}{{\mathrm{A}}_1{\mathrm{E}}_1} \\ =\frac{(19,466 \mathrm{N})(900 \mathrm{mm})}{(160 {\mathrm{mm}}^2)(70,000 \mathrm{N}/{\mathrm{mm}}^2)} \\ =1.564 \mathrm{mm} \end{gathered}$$

Therefore,

${\mathrm{v}}_{\mathrm{D}}=\mathbf{3}\mathbf{.}\mathbf{38} \mathrm{mm}$