Stiffness method: frame, flexural deformations only

Assumptions (figure partly described): the frame is A-B-C: column AB is 5 m high with a hinge at A; the beam BC is 3 m long, fixed at C. At joint B a 10 kN horizontal force (to the right) and a 40 kNm clockwise moment act. EI is constant and only flexural deformation is considered (axial and shear deformations neglected). Since C is fixed, B cannot translate horizontally; the vertical displacement of B is prevented by the column (axially rigid), so no sway occurs and the unknowns are the rotations $\theta_A$ (hinge) and $\theta_B$ (clockwise +).

Member stiffness equations (no member loads)

• $M_{AB}=\frac{2EI}{5}\left(2\theta_{A} + \theta_{B}\right)$
• $M_{BA}=\frac{2EI}{5}\left(2\theta_{B} + \theta_{A}\right)$
• $M_{BC}=\frac{2EI}{3}\left(2\theta_{B}\right)$
• $M_{CB}=\frac{2EI}{3}\left(\theta_{B}\right)$ The 10 kN force acts at the joint B and is carried directly to the supports by axial force in BC (it produces no bending moment).

Equilibrium

• Hinge A: $M_{AB}=0$
• Joint B: $M_{BA}+M_{BC}=40$ (applied clockwise moment)

Stiffness matrix equation

$$EI\begin{bmatrix}0.8000 & 0.4000 \\ 0.4000 & 2.1333\end{bmatrix}\begin{Bmatrix}\theta_{A} \\ \theta_{B}\end{Bmatrix}=\begin{Bmatrix}0.000 \\ 40.000\end{Bmatrix}$$

$K_{11}=\dfrac{4EI}{5}=0.8EI$, $K_{12}=\dfrac{2EI}{5}=0.4EI$, $K_{22}=\dfrac{4EI}{5}+\dfrac{4EI}{3}=2.1333EI$.

$$EI\theta_{A}=-10.3448,\quad EI\theta_{B}=20.6897$$

End moments (kNm, clockwise +)

Checks: $M_{AB}=0.00$ ✓; Joint B: $12.41+27.59=40.00$ kNm = applied 40 kNm ✓.

Reactions

A: $H=2.48$, $V=-13.79$, $M=0.00$; C: $H=-12.48$, $V=13.79$, $M=-13.79$ (M anticlockwise +). Horizontally the supports give 2.48 kN at A (equal to the column shear caused by the end moments) and -12.48 kN at C; their sum balances the 10 kN load. The column shear is $(M_{AB}+M_{BA})/5=2.48$ kN.

Bending moment diagram

Answer: $M_{AB}=0$, $M_{BA}=12.41$ kNm, $M_{BC}=27.59$ kNm, $M_{CB}=13.79$ kNm; $\theta_B=20.690/EI$ (clockwise).