To find the rigid bar deflection at $\mathrm{A}$, recall that
${\mathrm{v}}_{\mathrm{A}}={\mathrm{\delta }}_1$

$$\begin{gathered} {\mathrm{v}}_{\mathrm{A}}={\mathrm{\delta }}_1=\frac{(11.062 \mathrm{kN})(600 \mathrm{mm})(1,000 \mathrm{N}/\mathrm{kN})}{(70.0 {\mathrm{mm}}^2)(200,000 \mathrm{N}/{\mathrm{mm}}^2)} \\ {\mathrm{v}}_{\mathrm{A}}=\mathbf{0}\mathbf{.}\mathbf{474}\mathbf{ }\mathbf{mm} \left(\mathrm{downward}\right) \end{gathered}$$