Formula 5

\begin{array}{l} σ_{l} = \frac{2 \cdot 103 . 25^{2}}{109 . 55^{2} - 103 . 25^{2}} = 15.9 N / mm ² \\ {\bar{σ}}_{t} = \frac{2 \cdot 103.25}{109.55 - 103.25} = 32.8 N / mm ² \\ {\bar{σ}}_{r} = \frac{- 2 \cdot 103.25}{109.55 103.25} = - 1, 0 N / mm ² \end{array}

Do you have any questions?