Slope deflection: beam with 50 kN and 30 kN

Data: A fixed; AB = 2.5 m + 1.5 m = 4 m (3EI) with 50 kN at 2.5 m from A; BC = 4 m (2EI) with 10 kN/m; CD = 2 m + 2 m = 4 m (1.5EI) with 30 kN at mid-span; D is a roller; B and C simple supports. No settlement.

Fixed-end moments (clockwise +)

• AB ($a=2.5$, $b=1.5$): $FEM_{AB}=-\dfrac{50\times2.5\times1.5^2}{4^2}=-17.58$, $FEM_{BA}=+\dfrac{50\times2.5^2\times1.5}{4^2}=29.30$ kNm
• BC: $\mp\dfrac{10\times4^2}{12}=\mp13.33$ kNm
• CD: $\mp\dfrac{30\times4}{8}=\mp15$ kNm

Slope-deflection equations

• $M_{AB}=\frac{2(3EI)}{4}\left(\theta_{B}\right) - 17.58$
• $M_{BA}=\frac{2(3EI)}{4}\left(2\theta_{B}\right) + 29.30$
• $M_{BC}=\frac{2(2EI)}{4}\left(2\theta_{B} + \theta_{C}\right) - 13.33$
• $M_{CB}=\frac{2(2EI)}{4}\left(2\theta_{C} + \theta_{B}\right) + 13.33$
• $M_{CD}=\frac{2(1.5EI)}{4}\left(2\theta_{C} + \theta_{D}\right) - 15.00$
• $M_{DC}=\frac{2(1.5EI)}{4}\left(2\theta_{D} + \theta_{C}\right) + 15.00$

Equilibrium

• Joint B: $M_{BA}+M_{BC}=0$ Joint C: $M_{CB}+M_{CD}=0$ Roller D: $M_{DC}=0$

$$EI\begin{bmatrix}5.0000 & 1.0000 & 0.0000 \\ 1.0000 & 3.5000 & 0.7500 \\ 0.0000 & 0.7500 & 1.5000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \theta_{D}\end{Bmatrix}=\begin{Bmatrix}-15.964 \\ 1.667 \\ -15.000\end{Bmatrix}$$ $$EI\theta_{B}=-4.0378,\quad EI\theta_{C}=4.2254,\quad EI\theta_{D}=-12.1127$$

Final end moments (kNm)

Checks:

• Joint B: (17.18) + (-17.18) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (17.75) + (-17.75) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint D: (0.00) = 0.00 kNm (applied clockwise moment 0.00) ✓

Bending moment diagram ordinates (sagging +, kNm)

Reactions: A: $H=0.00$, $V=20.36$, $M=23.63$; B: $H=0.00$, $V=49.50$, $M=0.00$; C: $H=0.00$, $V=39.58$, $M=0.00$; D: $H=0.00$, $V=10.56$, $M=0.00$.

Answer: $M_{AB}=-23.63$, $M_{BA}=17.18$, $M_{BC}=-17.18$, $M_{CB}=17.75$, $M_{CD}=-17.75$, $M_{DC}=0$ kNm; BM at the load points: 27.27 kNm under 50 kN, 21.13 kNm under 30 kN.