Use the internal force to calculate the normal stresses in members $\left(1\right)$ and $\left(2\right)$. The normal stress in aluminum rod $\left(1\right)$ is

$$\begin{gathered} {\mathrm{\sigma }}_1=\frac{{\mathrm{F}}_1}{{\mathrm{A}}_1}=\frac{-21,147.216 \mathrm{N}}{200 {\mathrm{mm}}^2} \\ =\mathbf{105}\mathbf{.}\mathbf{7}\mathbf{ }\mathbf{MPa}\mathbf{ }\mathbf{\left(}\mathbf{C}\mathbf{\right)} \end{gathered}$$