Slope deflection: beam with settlement of A and C

Data and assumptions: A is fixed; AB = 5 m (EI) with 60 kN at 2 m from A; BC = 6 m (2EI) with 2 t/m $=2\times9.81=19.62$ kN/m; CD = 2 m (EI) is an unloaded overhang (it carries no moment). B and C are supports. Support A sinks 1 cm (without rotating) and C sinks 1.5 cm. $E=2\times10^5$ MPa $=2\times10^8$ kN/m², $I=10\,000$ cm⁴ $=10^{-4}$ m⁴, so $EI=20000$ kNm².

Chord rotations (clockwise +)

• $\psi_{AB}=\dfrac{\delta_B-\delta_A}{L}=\dfrac{0-(-0.010)}{5}\ \text{(B is higher than A by 10 mm)}\Rightarrow\psi_{AB}=-0.00200$ (anticlockwise)
• $\psi_{BC}=\dfrac{0.015}{6}=0.00250$ (C sinks relative to B, clockwise)
• Overhang CD: no moment.

Fixed-end moments (kNm, clockwise +)

• AB ($a=2$, $b=3$): $FEM_{AB}=-\dfrac{60\times2\times3^2}{5^2}=-43.20$, $FEM_{BA}=+\dfrac{60\times2^2\times3}{5^2}=28.80$
• BC: $\mp\dfrac{19.62\times6^2}{12}=\mp58.86$

Slope-deflection equations ($\theta_A=0$)

• $M_{AB}=8000.00\left(\theta_{B}\right) + 4.80$
• $M_{BA}=8000.00\left(2\theta_{B}\right) + 76.80$
• $M_{BC}=13333.33\left(2\theta_{B} + \theta_{C}\right) - 158.86$
• $M_{CB}=13333.33\left(2\theta_{C} + \theta_{B}\right) - 41.14$

Equilibrium

Joint B: $M_{BA}+M_{BC}=0$; joint C: $M_{CB}=0$ (the overhang CD is unloaded, so there is no moment from it).

$$\begin{bmatrix}42666.6667 & 13333.3333 \\ 13333.3333 & 26666.6667\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}82.060 \\ 41.140\end{Bmatrix}$$ $$\theta_{B}=0.00170806,\quad \theta_{C}=0.000688722$$

Final end moments (kNm) and BMD

Checks: Joint B: $104.13+(-104.13)=0.00$ ✓; joint C: $0.00=0.00$ ✓.

Reactions (kN, kNm): A: $H=0.00$, $V=11.48$, $M=-18.46$; B: $H=0.00$, $V=124.73$, $M=0.00$; C: $H=0.00$, $V=41.51$, $M=0.00$.

Answer: $M_{AB}=18.46$, $M_{BA}=104.13$, $M_{BC}=-104.13$ kNm and $M_{CB}=0$ (support moment at C is zero since CD carries no load); BC sagging maximum 43.90 kNm.