Stiffness matrix: continuous beam reactions and BMD

Data: AB = 3 m (1.5 m + 1.5 m) with 10 kN at mid-span, A is a hinge; BC = 4 m with 6 kN/m, C is fixed; B is the middle support. EI is constant.

Unknown displacements

1 = $\theta_A$ (rotation of the hinged end), 2 = $\theta_B$ (clockwise +). $\theta_C=0$; no settlements.

Fixed-end moments (clockwise +)

• AB: $\mp\dfrac{PL}{8}=\mp\dfrac{10\times3}{8}=\mp3.75$ kNm
• BC: $\mp\dfrac{wL^2}{12}=\mp\dfrac{6\times4^2}{12}=\mp8$ kNm

Member stiffness equations

• $M_{AB}=\frac{2EI}{3}\left(2\theta_{A} + \theta_{B}\right) - 3.75$
• $M_{BA}=\frac{2EI}{3}\left(2\theta_{B} + \theta_{A}\right) + 3.75$
• $M_{BC}=\frac{2EI}{4}\left(2\theta_{B}\right) - 8.00$
• $M_{CB}=\frac{2EI}{4}\left(\theta_{B}\right) + 8.00$

Stiffness matrix equation

Equilibrium: $M_{AB}=0$ (hinge) and $M_{BA}+M_{BC}=0$ (joint B).

$$EI\begin{bmatrix}1.3333 & 0.6667 \\ 0.6667 & 2.3333\end{bmatrix}\begin{Bmatrix}\theta_{A} \\ \theta_{B}\end{Bmatrix}=\begin{Bmatrix}3.750 \\ 4.250\end{Bmatrix}$$

with $K_{11}=\dfrac{4EI}{3}$, $K_{12}=K_{21}=\dfrac{2EI}{3}$, $K_{22}=\dfrac{4EI}{3}+\dfrac{4EI}{4}$.

$$EI\theta_{A}=2.2188,\quad EI\theta_{B}=1.1875$$

End moments (kNm)

Checks: $M_{AB}=0.00$ ✓; $M_{BA}+M_{BC}=0.00$ ✓.

Support reactions

Using the end moments and the equilibrium of each span:

• Span AB (taking moments about B): $R_A=\dfrac{10\times1.5-6.81}{3}=2.73$ kN (the hogging moment at B reduces the reaction at A)
• $R_B=18.83$ kN (sum of the shears of AB and BC at B)
• $R_C=12.45$ kN, $M_C=-8.59$ kNm (anticlockwise +), i.e. fixed-end moment 8.59 kNm (hogging)

Check: $R_A+R_B+R_C=34.00$ kN $=10+6\times4=34$ kN ✓.

Bending moment diagram (sagging +, kNm)

Answer: $M_{BA}=6.81$, $M_{BC}=-6.81$, $M_{CB}=8.59$ kNm, $M_{AB}=0$; $R_A=2.73$ kN, $R_B=18.83$ kN, $R_C=12.45$ kN, $M_C=8.59$ kNm.