${\mathrm{\delta }}_1={\mathrm{u}}_{\mathrm{B}}-\cancel{{{\mathrm{u}}_{\mathrm{A}}}^{}}={\mathrm{u}}_{\mathrm{B}}$

To find the deflection at joint $\mathrm{B}$, we need only compute the deformation in member $\left(1\right)$ using the force-deformation relationship:

${\mathrm{\delta }}_1=\frac{{\mathrm{F}}_1{\mathrm{L}}_1}{{\mathrm{A}}_1{\mathrm{E}}_1}+{\mathrm{\alpha }}_1∆{\mathrm{TL}}_1$

$$\begin{gathered} {\mathrm{\delta }}_1=\frac{(-21,147.216 \mathrm{N})(400 \mathrm{mm})}{(200 {\mathrm{mm}}^2)(70,000 \mathrm{N}/{\mathrm{mm}}^2)} \\ +\left(23.6\times {10}^{-6}{/}^{\circ }\mathrm{C}\right)\left({55}^{\circ }\mathrm{C}\right)(400 \mathrm{mm}) \\ =-0.0850 \mathrm{mm} \mathbf{(}\mathbf{i}\mathbf{.}\mathbf{e}\mathbf{.}\mathbf{,}\mathbf{ }\mathbf{contraction}\mathbf{)} \end{gathered}$$

$\therefore {\mathrm{u}}_{\mathrm{B}}={\mathrm{\delta }}_1=\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{0850}\mathbf{ }\mathbf{mm}\mathbf{ }\mathbf{(}\mathbf{i}\mathbf{.}\mathbf{e}\mathbf{.}\mathbf{,}\mathbf{ }\mathbf{upward}\mathbf{)}$