Slope deflection: continuous beam, SFD and BMD

Data: A is fixed; AB = 5 m + 3 m = 8 m (2.5EI) with 50 kN at 5 m from A; BC = 6 m (3EI) with 10 kN/m; the overhang CD = 2 m (EI) carries 20 kN at D; B and C are simple supports. No settlement.

Step 1: Overhang CD

$M_C=20\times2=40$ kNm (hogging), a known moment at C.

Step 2: Fixed-end moments (clockwise +)

• AB ($a=5$, $b=3$): $FEM_{AB}=-\dfrac{50\times5\times3^2}{8^2}=-35.16$, $FEM_{BA}=+\dfrac{50\times5^2\times3}{8^2}=58.59$ kNm
• BC: $\mp\dfrac{10\times6^2}{12}=\mp30$ kNm

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

• $M_{AB}=\frac{2(2.5EI)}{8}\left(\theta_{B}\right) - 35.16$
• $M_{BA}=\frac{2(2.5EI)}{8}\left(2\theta_{B}\right) + 58.59$
• $M_{BC}=\frac{2(3EI)}{6}\left(2\theta_{B} + \theta_{C}\right) - 30.00$
• $M_{CB}=\frac{2(3EI)}{6}\left(2\theta_{C} + \theta_{B}\right) + 30.00$

Step 4: Equilibrium

Joint B: $M_{BA}+M_{BC}=0$. Joint C: $M_{CB}+M_{CD}=0$ with $M_{CD}=-40$ kNm.

$$EI\begin{bmatrix}3.2500 & 1.0000 \\ 1.0000 & 2.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-28.594 \\ 10.000\end{Bmatrix}$$ $$EI\theta_{B}=-12.2159,\quad EI\theta_{C}=11.1080$$

Step 5: Final end moments (kNm)

Step 6: Shear force and bending moment

Shear forces follow from the equilibrium of each span with its end moments (positive when the left-hand part is pushed upward):

SFD: AB: 18.68 kN from A to the 50 kN load, then -31.32 kN to B; BC: 30.55 kN at B falling uniformly ($-10$ kN/m) to -29.45 kN at C; CD: $+20$ kN constant (dropping to zero at the free end after the 20 kN load). Reactions: A: $H=0.00$, $V=18.68$, $M=42.79$; B: $H=0.00$, $V=61.87$, $M=0.00$; C: $H=0.00$, $V=49.45$, $M=0.00$.

BMD (sagging +, kNm):

Answer: $M_{AB}=-42.79$, $M_{BA}=43.32$, $M_{BC}=-43.32$, $M_{CB}=40.00$ kNm, $M_{CD}=-40$ kNm; $R_B=61.87$ kN, $R_C=49.45$ kN, $V_A=18.68$ kN.