and the compatibility equation:

$\frac{1}{760 \mathrm{mm}}\frac{{\mathrm{F}}_1{\mathrm{L}}_1}{{\mathrm{A}}_1{\mathrm{E}}_1}=-\frac{1}{1,520 \mathrm{mm}}\frac{{\mathrm{F}}_2{\mathrm{L}}_2}{{\mathrm{A}}_2{\mathrm{E}}_2}$

can be solved simultaneously to determine ${\mathrm{F}}_1$ and ${\mathrm{F}}_2$.

${\mathrm{F}}_1=+19.466 \mathrm{kN} {\mathrm{F}}_2=-28.030 \mathrm{kN}$

To compute the normal stress in each member, divide the internal axial force by the cross-sectional area ${\mathrm{A}}_1={\mathrm{A}}_2=160 {\mathrm{mm}}^2$:

${\mathrm{\sigma }}_1=\mathbf{121}\mathbf{.}\mathbf{7} \mathrm{MPa} \left(\mathrm{T}\right) {\mathrm{\sigma }}_2=\mathbf{175}\mathbf{.}\mathbf{2} \mathrm{MPa} \left(\mathrm{C}\right)$