Slope deflection: beam with overhang

Assumptions: the figure shows AB = 2 m + 4 m = 6 m with the 40 kN load 2 m from A, BC = 10 m, CD = 2 m overhang with 30 kN at D. A is taken as a fixed end; B and C are simple supports. Stiffness: AB = 1.5EI, BC = 2EI, CD = EI.

Step 1: Overhang CD

Moment at C due to the overhang: $M_C=30\times2=60$ kNm (hogging). This is a known moment acting on joint C.

Step 2: Fixed-end moments (clockwise +)

• AB: $FEM_{AB}=-\dfrac{40\times2\times4^2}{6^2}=-35.56$, $FEM_{BA}=+\dfrac{40\times2^2\times4}{6^2}=17.78$ kNm
• BC: $FEM_{BC}=-\dfrac{10\times10^2}{12}=-83.33$, $FEM_{CB}=+83.33$ kNm

No support settles ($\psi=0$) and $\theta_A=0$.

Step 3: Slope-deflection equations

• $M_{AB}=\frac{2(1.5EI)}{6}\left(\theta_{B}\right) - 35.56$
• $M_{BA}=\frac{2(1.5EI)}{6}\left(2\theta_{B}\right) + 17.78$
• $M_{BC}=\frac{2(2EI)}{10}\left(2\theta_{B} + \theta_{C}\right) - 83.33$
• $M_{CB}=\frac{2(2EI)}{10}\left(2\theta_{C} + \theta_{B}\right) + 83.33$

Step 4: Joint equilibrium

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

$$EI\begin{bmatrix}1.8000 & 0.4000 \\ 0.4000 & 0.8000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}65.556 \\ -23.333\end{Bmatrix}$$ $$EI\theta_{B}=48.2639,\quad EI\theta_{C}=-53.2986$$

Step 5: Final end moments (kNm)

Bending moment diagram

Reactions: A: $H=0.00$, $V=17.56$, $M=11.42$; B: $H=0.00$, $V=73.04$, $M=0.00$; C: $H=0.00$, $V=79.40$, $M=0.00$.

Answer: $M_{AB}=-11.42$, $M_{BA}=66.04$, $M_{BC}=-66.04$, $M_{CB}=60.00$ kNm (A fixed); $M_C=60$ kNm hogging; span BC sagging maximum about 62.00 kNm.