Slope deflection: continuous beam, I = 4x10^7 mm^4

Data: $EI=200\ \text{kN/mm}^2\times4\times10^7\ \text{mm}^4=8\times10^{3}$ kNm². Relative stiffness: AB = EI, BC = 2EI, CD = EI. A and D fixed. AB = 4 m (no load), BC = 4 m with 10 kN/m, CD = 6 m with 60 kN at 2 m from C. No settlement, so the moments do not depend on the numerical value of EI.

Fixed-end moments (clockwise +)

• BC: $\mp\dfrac{10\times4^2}{12}=\mp13.33$ kNm
• CD ($a=2$, $b=4$): $FEM_{CD}=-\dfrac{60\times2\times4^2}{6^2}=-53.33$, $FEM_{DC}=+\dfrac{60\times2^2\times4}{6^2}=26.67$ kNm

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

• $M_{AB}=\frac{2EI}{4}\left(\theta_{B}\right)$
• $M_{BA}=\frac{2EI}{4}\left(2\theta_{B}\right)$
• $M_{BC}=\frac{2(2EI)}{4}\left(2\theta_{B} + \theta_{C}\right) - 13.33$
• $M_{CB}=\frac{2(2EI)}{4}\left(2\theta_{C} + \theta_{B}\right) + 13.33$
• $M_{CD}=\frac{2EI}{6}\left(2\theta_{C}\right) - 53.33$
• $M_{DC}=\frac{2EI}{6}\left(\theta_{C}\right) + 26.67$

Joint equilibrium

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

$$EI\begin{bmatrix}3.0000 & 1.0000 \\ 1.0000 & 2.6667\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}13.333 \\ 40.000\end{Bmatrix}$$ $$EI\theta_{B}=-0.6349,\quad EI\theta_{C}=15.2381$$

With $EI=8000$ kNm²: $\theta_B=-0.000079$ rad and $\theta_C=0.001905$ rad (clockwise positive).

Final end moments (kNm)

Checks:

• Joint B: (-0.63) + (0.63) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (43.17) + (-43.17) = 0.00 kNm (applied clockwise moment 0.00) ✓

Bending moment diagram ordinates (sagging +, kNm)

Reactions: A: $H=0.00$, $V=0.24$, $M=0.32$; B: $H=0.00$, $V=8.81$, $M=0.00$; C: $H=0.00$, $V=72.86$, $M=0.00$; D: $H=0.00$, $V=18.10$, $M=-31.75$.

Answer: $M_{AB}=-0.32$, $M_{BA}=-0.63$, $M_{BC}=0.63$, $M_{CB}=43.17$, $M_{CD}=-43.17$, $M_{DC}=31.75$ kNm (clockwise end moments +).