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Task solution:

1. Drawing a diagram of moments against external load (Mp) and unit moment diagrams M1 - beam loaded with concentrated force (to calculate vertical displacement) and M2 - beam loaded with concentrated moment (to calculate rotation angle).

\begin{array}{lcc} \sum M_{A}=0 & 20-10-6 V_{C}+20 \cdot 4=0 & V_{C}=15 k N \\ \sum M_{C}=0 & 20-10-20 \cdot 2+6 V_{A}=0 & V_{A}=5 k N \\ \sum Y=0 & V_{A}+V_{C}-20=0 & \mathrm{~L}=P \end{array}

2. Displacement of point B vertically

\begin{aligned} \Delta_{B}=\int \frac{M_{P} \cdot M_{1}}{E J} d x= \end{aligned} \begin{aligned} =\frac{1}{E J}\left[\frac{1}{6} \cdot 20 \cdot \frac{4}{3} \cdot 4+\frac{1}{3} \cdot 40 \cdot \frac{4}{3} \cdot 4+\frac{1}{3} \cdot 40 \cdot \frac{4}{3} \cdot 2+\frac{1}{6} \cdot 10 \cdot \frac{4}{3} \cdot 2\right]=\frac{128.89}{E J} \end{aligned}

3. Rotation angle at point B

\begin{aligned} \varphi_{B}=\int \frac{M_{P} \cdot M_{2}}{E J} d x= \end{aligned} \begin{aligned} =\frac{1}{E J}\left[-\frac{1}{6} \cdot 20 \cdot \frac{2}{3} \cdot 4-\frac{1}{3} \cdot 40 \cdot \frac{2}{3} \cdot 4+\frac{1}{3} \cdot 40 \cdot \frac{1}{3} \cdot 2+\frac{1}{6} \cdot 10 \cdot \frac{1}{3} \cdot 2\right]=-\frac{34.44}{E J} \end{aligned}