Slope deflection: continuous beam, 90 kN and 30 kN

Data: A is fixed; AB = 15 m (2EI) carries 15 kN/m; B and C are simple supports; BC = 10 m (EI) carries 90 kN at 4 m from B; CD is a 2 m overhang with 30 kN at its free end D.

Step 1: Overhang CD

$M_C=30\times2=60$ kNm (hogging), a known moment at joint C.

Step 2: Fixed-end moments (clockwise +)

• AB: $\mp\dfrac{15\times15^2}{12}=\mp281.25$ kNm
• BC ($a=4$, $b=6$): $FEM_{BC}=-\dfrac{90\times4\times6^2}{10^2}=-129.60$, $FEM_{CB}=+\dfrac{90\times4^2\times6}{10^2}=86.40$ kNm

Step 3: Slope-deflection equations ($\theta_A=0$, $\psi=0$)

• $M_{AB}=\frac{2(2EI)}{15}\left(\theta_{B}\right) - 281.25$
• $M_{BA}=\frac{2(2EI)}{15}\left(2\theta_{B}\right) + 281.25$
• $M_{BC}=\frac{2EI}{10}\left(2\theta_{B} + \theta_{C}\right) - 129.60$
• $M_{CB}=\frac{2EI}{10}\left(2\theta_{C} + \theta_{B}\right) + 86.40$

Step 4: Joint equilibrium

• Joint B: $M_{BA}+M_{BC}=0$
• Joint C: $M_{CB}+M_{CD}=0$ with $M_{CD}=-60$ kNm, i.e. $M_{CB}=60$ kNm

$$EI\begin{bmatrix}0.9333 & 0.2000 \\ 0.2000 & 0.4000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-151.650 \\ -26.400\end{Bmatrix}$$ $$EI\theta_{B}=-166.1400,\quad EI\theta_{C}=17.0700$$

Step 5: Final end moments (kNm)

Checks:

• Joint B: (192.64) + (-192.64) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (60.00) + (-60.00) = 0.00 kNm (applied clockwise moment 0.00) ✓

Bending moment diagram ordinates (sagging +, kNm)

Reactions: A: $H=0.00$, $V=121.36$, $M=325.55$; B: $H=0.00$, $V=170.90$, $M=0.00$; C: $H=0.00$, $V=52.74$, $M=0.00$.

Answer: $M_{AB}=-325.55$, $M_{BA}=192.64$, $M_{BC}=-192.64$, $M_{CB}=60.00$ kNm and $M_{CD}=-60.00$ kNm (overhang moment, hogging 60 kNm).