Solution
Before starting this example, please read a short Theoretical Introduction.
We draw the moment diagrams for the beam in the state of initial loading and separately for unit loading.
External loading state - state P + moment diagram Mp
Unit state + moment diagram M
Maximum deflection at the end of the beam:
\begin{aligned}&w_{max}=\frac{1}{EI}\int(Mp(x)\cdot M(x)) dS=\\ \end{aligned} \begin{aligned}&=\frac{1}{EI}(\frac{1}{3}\cdot 180\cdot 6\cdot 6-\frac{1}{3}\cdot 45\cdot 6\cdot 6)=\frac{1620}{EI}\\ \end{aligned}
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