Slope deflection: three-span beam with settlement

Data: three spans AB = 6 m ($EI$, 2 kN/m), BC = 5 m ($2EI$, 5 kN at 2 m from C, i.e. 3 m from B), CD = 5 m ($EI$, 8 kN at mid-span); A and D are fixed. Support B settles 2 mm and support C is lifted 3 mm. $EI=8\times10^{11}$ N mm² $=800$ kNm². Units: kN, m.

Chord rotations (clockwise +)

• $\psi_{AB}=\dfrac{0.002}{6}=0.000333$ (B goes down)
• $\psi_{BC}=\dfrac{-(0.003+0.002)}{5}=-0.001000$ (C is 5 mm above B)
• $\psi_{CD}=\dfrac{0.003}{5}=0.000600$ (D is 3 mm below C)

Fixed-end moments (kNm, clockwise +)

• AB: $\mp\dfrac{2\times6^2}{12}=\mp6.000$
• BC ($a=3$, $b=2$): $FEM_{BC}=-\dfrac{5\times3\times2^2}{5^2}=-2.400$, $FEM_{CB}=+\dfrac{5\times3^2\times2}{5^2}=3.600$
• CD: $\mp\dfrac{8\times5}{8}=\mp5.000$

Slope-deflection equations ($\theta_A=\theta_D=0$)

• $M_{AB}=266.67\left(\theta_{B}\right) - 6.27$
• $M_{BA}=266.67\left(2\theta_{B}\right) + 5.73$
• $M_{BC}=640.00\left(2\theta_{B} + \theta_{C}\right) - 0.48$
• $M_{CB}=640.00\left(2\theta_{C} + \theta_{B}\right) + 5.52$
• $M_{CD}=320.00\left(2\theta_{C}\right) - 5.58$
• $M_{DC}=320.00\left(\theta_{C}\right) + 4.42$

Equilibrium and solution

Joint B: $M_{BA}+M_{BC}=0$; joint C: $M_{CB}+M_{CD}=0$.

$$\begin{bmatrix}1813.3333 & 640.0000 \\ 640.0000 & 1920.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-5.253 \\ 0.056\end{Bmatrix}$$ $$\theta_{B}=-0.003295,\quad \theta_{C}=0.0011275$$

Final end moments (kNm)

Checks:

• Joint B: (3.98) + (-3.98) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (4.85) + (-4.85) = 0.00 kNm (applied clockwise moment 0.00) ✓

Bending moment diagram (sagging +, kNm)

Reactions (kN, kNm; support moment M anticlockwise +): A: $H=0.00$, $V=6.53$, $M=7.15$; B: $H=0.00$, $V=7.30$, $M=0.00$; C: $H=0.00$, $V=7.19$, $M=0.00$; D: $H=0.00$, $V=3.99$, $M=-4.78$.

Answer: $M_{AB}=-7.15$, $M_{BA}=3.98$, $M_{BC}=-3.98$, $M_{CB}=4.85$, $M_{CD}=-4.85$, $M_{DC}=4.78$ kNm.