Solution
\begin{aligned}
&L=5 \quad m \\
&EI=40000 \quad k N m^{2} \\
&p=12 \quad \frac{k N}{m} \\
&h=1 \quad m \\
&v_{1}=0.0012375 \quad m \\
&v_{2}=0.0036 \quad m \\
&v_{3}=0.0060375 \quad m \\
&M=-E I \cdot \frac{v_{1}-2 \cdot v_{2}+v_{3}}{h^{2}}=-3 \quad k N m
\end{aligned}
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