Compatibility equation.
$\frac{{\mathrm{F}}_1{\mathrm{L}}_1}{{\mathrm{A}}_1{\mathrm{E}}_1}+{\mathrm{\alpha }}_1∆{\mathrm{TL}}_1+\frac{{\mathrm{F}}_2{\mathrm{L}}_2}{{\mathrm{A}}_2{\mathrm{E}}_2}+{\mathrm{\alpha }}_2∆{\mathrm{TL}}_2=0$

Solve for ${\mathrm{F}}_1$.
${\mathrm{F}}_1=\frac{-∆\mathrm{T}({\mathrm{\alpha }}_1{\mathrm{L}}_1+{\mathrm{\alpha }}_2{\mathrm{L}}_2)}{{\displaystyle \frac{{\mathrm{L}}_1}{{\mathrm{A}}_1{\mathrm{E}}_1}}+{\displaystyle \frac{{\mathrm{L}}_2}{{\mathrm{A}}_2{\mathrm{E}}_2}}}$

$=\frac{-({75}^{\circ }\mathrm{C}-{20}^{\circ }\mathrm{C})\left[(23.6\times {10}^{-6}/{}^{\circ }\mathrm{C})(400 \mathrm{mm})+(12.0\times {10}^{-6}/{}^{\circ }\mathrm{C})(200 \mathrm{mm})\right]}{{\displaystyle \frac{400 \mathrm{mm}}{(200 {\mathrm{mm}}^2)(70,000 \mathrm{N}/{\mathrm{mm}}^2)}+\frac{200 \mathrm{mm}}{(450 {\mathrm{mm}}^2)(200,000 \mathrm{N}/{\mathrm{mm}}^2)}}}$

$=\frac{-0.6512 \mathrm{mm}}{2.8571429\times {10}^{-5} \mathrm{mm}/\mathrm{N}+2.2222222\times {10}^{-6} \mathrm{mm}/\mathrm{N}}$

$=-21,147.216 \mathrm{N}=-21.1 \mathrm{kN}$

since ${\mathrm{F}}_1={\mathrm{F}}_2$.

The negative value indicates that these internal forces are actually compression, not tension as originally assumed.