Rotation and deflection at free end, BMD

Assumptions (figure partly unclear): AB: $L=5$ m, stiffness $2EI$, UDL taken as $w=10$ kN/m; BC: $L=5$ m, stiffness $EI$; A fixed, B roller; at the free end C a 50 kN downward load and a 20 kN m clockwise moment. Results are in terms of $EI$ (kN m$^2$). Nodes: 1 = A, 2 = B, 3 = C. Units: kN, m.

Element stiffness

$$[k]=\frac{EI}{L^3}\begin{bmatrix}12&6L&-12&6L\\6L&4L^2&-6L&2L^2\\-12&-6L&12&-6L\\6L&2L^2&-6L&4L^2\end{bmatrix},\quad \{d\}=\{v_i,\theta_i,v_j,\theta_j\}^T$$

(Upward forces and displacements, and counter-clockwise moments and rotations, are positive.) with $EI\to2EI$ for AB. Equivalent loads for AB: $-25$ and $-20.833$ at A; $-25$ and $+20.833$ at B. At C: $F_{v3}=-50$ kN, $F_{\theta3}=-20$ kN m (clockwise).

Assembled equation

Boundary conditions: $v_1=\theta_1=0$, $v_2=0$. Free DOFs: $\theta_2,v_3,\theta_3$.

$$EI\,\begin{bmatrix}2.4000 & -0.2400 & 0.4000 \\ -0.2400 & 0.0960 & -0.2400 \\ 0.4000 & -0.2400 & 0.8000\end{bmatrix}\begin{Bmatrix}\theta_2\\v_3\\\theta_3\end{Bmatrix}=\begin{Bmatrix}20.833 \\ -50.000 \\ -20.000\end{Bmatrix}$$

Solving:

$$\theta_B=-\frac{155.73}{EI}\ \text{rad},\qquad v_C=-\frac{3112.0}{EI}\ \text{m},\qquad \theta_C=-\frac{880.73}{EI}\ \text{rad}$$

Free end C: deflection 3112.0/EI m downward; rotation 880.73/EI rad (clockwise).

End forces

Reactions: $R_A=-49.75$ kN (downward, since the overhang hogs the beam), $M_A=-103.75$ kN m (clockwise), $R_B=149.75$ kN. Check: $-49.75+149.75=100$ kN $=50+10\times5$.

Bending moment diagram (sagging +)

The moment at B is $-(50\times5+20)=-270$ kN m. In BC the moment varies linearly from $-270$ to $-20$.

 A(fixed)                  B(roller)                C(free)
 |-------------------------|------------------------|
 BMD: +103.75 (x=1.77 m: 0)   -270                   -20   (kN m)

Answer: $\theta_C=-880.73/EI$ rad, $v_C=-3112.0/EI$ m (downward).