Stiffness matrix: continuous beam BMD

Data: A is fixed; AB = 7 m with 10 kN at 2 m and 30 kN at 5 m from A; B is a roller support; BC = 8 m with 15 kN/m on the right-hand 4 m (from 4 m to 8 m from B); C is a roller (free to rotate). EI is constant.

Unknown displacements (coordinates)

1 = $\theta_B$, 2 = $\theta_C$ (clockwise +). $\theta_A=0$ and there are no support settlements.

Fixed-end moments (clockwise +, kNm)

• AB (point loads): load 10 kN ($a=2$, $b=5$): $\mp\dfrac{Pab^2}{L^2},\ \pm\dfrac{Pa^2b}{L^2}$ = $-10.204$, $+4.082$; load 30 kN ($a=5$, $b=2$): $-12.245$, $+30.612$. Total: $FEM_{AB}=-22.449$, $FEM_{BA}=34.694$.
• BC (partial UDL, 15 kN/m over 4 m, from the right end): $FEM_{BC}=-25.000$, $FEM_{CB}=55.000$ (integration of the point-load formulas over the loaded length).

Member stiffness equations

• $M_{AB}=\frac{2EI}{7}\left(\theta_{B}\right) - 22.45$
• $M_{BA}=\frac{2EI}{7}\left(2\theta_{B}\right) + 34.69$
• $M_{BC}=\frac{2EI}{8}\left(2\theta_{B} + \theta_{C}\right) - 25.00$
• $M_{CB}=\frac{2EI}{8}\left(2\theta_{C} + \theta_{B}\right) + 55.00$

Stiffness matrix equation ($[K]\{d\}=\{P\}$)

The roller C gives $M_{CB}=0$ and the joint B gives $M_{BA}+M_{BC}=0$:

$$EI\begin{bmatrix}1.0714 & 0.2500 \\ 0.2500 & 0.5000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-9.694 \\ -55.000\end{Bmatrix}$$

$K_{11}=\dfrac{4EI}{7}+\dfrac{4EI}{8}=1.0714EI$, $K_{12}=K_{21}=\dfrac{2EI}{8}=0.2500EI$, $K_{22}=\dfrac{4EI}{8}=0.5000EI$.

$$EI\theta_{B}=18.8140,\quad EI\theta_{C}=-119.4070$$

Final end moments (kNm)

Checks: Joint B: $45.44+(-45.44)=0.00$ ✓; roller C: $M_{CB}=0.00$ ✓.

Bending moment diagram (sagging +, kNm)

Reactions: A: $H=0.00$, $V=11.66$, $M=17.07$; B: $H=0.00$, $V=49.02$, $M=0.00$; C: $H=0.00$, $V=39.32$, $M=0.00$.

Answer: $M_{AB}=-17.07$, $M_{BA}=45.44$, $M_{BC}=-45.44$ kNm, $M_{CB}=0$; maximum sagging in BC = 51.53 kNm at 5.38 m from B.