Slope deflection: BMD and SFD of beam

Assumptions (the figure is only described): the continuous beam has a free left end A with a 30 kNm clockwise couple, a 2 m overhang AB (EI), then BC = 6 m (2I) with 20 kN/m, CD = 10 m (3I) unloaded, DE = 6 m (3I) with 100 kN at mid-span (3 m from D), and EF = 6 m (3I) unloaded with the fixed end at F. B, C, D and E are simple supports. No settlement.

Step 1: Overhang AB

The couple at the free end A is transmitted unchanged along the overhang, so the moment at B is 30 kNm (sagging). For the right-hand end B of member AB this is an anticlockwise end moment, i.e. $M_{BA}=-30.00$ kNm in the clockwise-positive convention. It is a known external moment on joint B.

Step 2: Fixed-end moments (clockwise +)

• BC: $\mp\dfrac{20\times6^2}{12}=\mp60$ kNm
• DE: $\mp\dfrac{PL}{8}=\mp\dfrac{100\times6}{8}=\mp75$ kNm
• CD, EF: 0

Step 3: Slope-deflection equations

• $M_{BC}=\frac{2(2EI)}{6}\left(2\theta_{B} + \theta_{C}\right) - 60.00$
• $M_{CB}=\frac{2(2EI)}{6}\left(2\theta_{C} + \theta_{B}\right) + 60.00$
• $M_{CD}=\frac{2(3EI)}{10}\left(2\theta_{C} + \theta_{D}\right)$
• $M_{DC}=\frac{2(3EI)}{10}\left(2\theta_{D} + \theta_{C}\right)$
• $M_{DE}=\frac{2(3EI)}{6}\left(2\theta_{D} + \theta_{E}\right) - 75.00$
• $M_{ED}=\frac{2(3EI)}{6}\left(2\theta_{E} + \theta_{D}\right) + 75.00$
• $M_{EF}=\frac{2(3EI)}{6}\left(2\theta_{E}\right)$
• $M_{FE}=\frac{2(3EI)}{6}\left(\theta_{E}\right)$

Step 4: Joint equilibrium (B, C, D, E)

Joint B: $M_{BC}+M_{BA}=0$ with $M_{BA}=-30.00$. Joints C, D, E: $\sum M=0$.

$$EI\begin{bmatrix}1.3333 & 0.6667 & 0.0000 & 0.0000 \\ 0.6667 & 2.5333 & 0.6000 & 0.0000 \\ 0.0000 & 0.6000 & 3.2000 & 1.0000 \\ 0.0000 & 0.0000 & 1.0000 & 4.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \theta_{D} \\ \theta_{E}\end{Bmatrix}=\begin{Bmatrix}90.000 \\ -60.000 \\ 75.000 \\ -75.000\end{Bmatrix}$$ $$EI\theta_{B}=97.3532,\quad EI\theta_{C}=-59.7064,\quad EI\theta_{D}=43.9233,\quad EI\theta_{E}=-29.7308$$

Step 5: Final end moments (kNm)

Checks:

• Joint B: (-30.00) + (30.00) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (45.29) + (-45.29) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint D: (16.88) + (-16.88) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint E: (59.46) + (-59.46) = 0.00 kNm (applied clockwise moment 0.00) ✓

Bending moment diagram (sagging +, kNm)

Shear force diagram (kN, positive when the left part is pushed upward)

The shear in BC decreases linearly by 20 kN/m, and in DE it changes by 100 kN at mid-span. Reactions: B: $H=0.00$, $V=47.45$, $M=0.00$; C: $H=0.00$, $V=75.39$, $M=0.00$; D: $H=0.00$, $V=40.06$, $M=0.00$; E: $H=0.00$, $V=71.96$, $M=0.00$; F: $H=0.00$, $V=-14.87$, $M=29.73$.

Answer: $M_{BC}=30.00$, $M_{CB}=45.29$, $M_{CD}=-45.29$, $M_{DC}=16.88$, $M_{DE}=-16.88$, $M_{ED}=59.46$, $M_{EF}=-59.46$, $M_{FE}=-29.73$ kNm (clockwise end moments +); the BMD and SFD ordinates are tabulated above.