Stiffness matrix: BMD of frame

Assumptions: the girder A-B-C is horizontal with A fixed and C on a roller; AB = 4 m (2EI) with the 60 kN load at mid-span; BC = 4 m (EI) with 10 kN/m downward; the column BD (EI, 4 m) hangs below B and is fixed at D. The fixed support A prevents sway, so only $\theta_B$ and $\theta_C$ are unknown.

Fixed-end moments (clockwise +)

• AB: $\mp\dfrac{PL}{8}=\mp\dfrac{60\times4}{8}=\mp30$ kNm
• BC: $\mp\dfrac{wL^2}{12}=\mp\dfrac{10\times4^2}{12}=\mp13.33$ kNm
• BD: no load

Stiffness (slope-deflection) equations

• $M_{AB}=\frac{2(2EI)}{4}\left(\theta_{B}\right) - 30.00$
• $M_{BA}=\frac{2(2EI)}{4}\left(2\theta_{B}\right) + 30.00$
• $M_{BC}=\frac{2EI}{4}\left(2\theta_{B} + \theta_{C}\right) - 13.33$
• $M_{CB}=\frac{2EI}{4}\left(2\theta_{C} + \theta_{B}\right) + 13.33$
• $M_{BD}=\frac{2EI}{4}\left(2\theta_{B}\right)$
• $M_{DB}=\frac{2EI}{4}\left(\theta_{B}\right)$

Equilibrium

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

Writing these in terms of $\theta_B,\theta_C$ gives the stiffness matrix equation:

$$EI\begin{bmatrix}4.0000 & 0.5000 \\ 0.5000 & 1.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-16.667 \\ -13.333\end{Bmatrix}$$ $$EI\theta_{B}=-2.6667,\quad EI\theta_{C}=-12.0000$$

Final end moments (kNm, clockwise on the member end +)

Checks: Joint B: $24.67+(-22.00)+(-2.67)=0.00$ ✓; Roller C: $M_{CB}=0.00$ ✓.

Bending moment diagram ordinates (sagging + for the girder; column: tension on the right when going down, kNm)

Reactions: A: $H=1.00$, $V=32.00$, $M=32.67$; D: $H=-1.00$, $V=53.50$, $M=1.33$; C: $H=0.00$, $V=14.50$, $M=0.00$.

Answer: $M_{AB}=-32.67$, $M_{BA}=24.67$, $M_{BC}=-22.00$, $M_{CB}=0$, $M_{BD}=-2.67$, $M_{DB}=-1.33$ kNm.