Stiffness matrix: BMD of frame, 60 kN

Assumptions: the beam ABC is a horizontal member of 4 m (2 m + 2 m) with the 60 kN load downward at B, its mid-point; A is fixed. Column CD is 5 m high, rigidly connected at C and hinged at D. EI is constant. Because A is fixed, the frame cannot sway. The unknown rotations are $\theta_C$ and $\theta_D$ (clockwise +).

Treat AC as one member (the load at B is a member load).

Fixed-end moments (clockwise +)

$FEM_{AC}=-\dfrac{PL}{8}=-\dfrac{60\times4}{8}=-30$ kNm, $FEM_{CA}=+30$ kNm. CD is unloaded.

Stiffness equations of the members

• $M_{AC}=\frac{2EI}{4}\left(\theta_{C}\right) - 30.00$
• $M_{CA}=\frac{2EI}{4}\left(2\theta_{C}\right) + 30.00$
• $M_{CD}=\frac{2EI}{5}\left(2\theta_{C} + \theta_{D}\right)$
• $M_{DC}=\frac{2EI}{5}\left(2\theta_{D} + \theta_{C}\right)$

Equilibrium

• Joint C: $M_{CA}+M_{CD}=0$
• Hinge D: $M_{DC}=0$

Stiffness matrix equation

$$EI\begin{bmatrix}1.8000 & 0.4000 \\ 0.4000 & 0.8000\end{bmatrix}\begin{Bmatrix}\theta_{C} \\ \theta_{D}\end{Bmatrix}=\begin{Bmatrix}-30.000 \\ 0.000\end{Bmatrix}$$

Here $K_{11}=\dfrac{4EI}{4}+\dfrac{4EI}{5}=1.8EI$, $K_{12}=K_{21}=\dfrac{2EI}{5}=0.4EI$, $K_{22}=\dfrac{4EI}{5}=0.8EI$.

$$EI\theta_{C}=-18.7500,\quad EI\theta_{D}=9.3750$$

Final end moments (kNm)

Checks: Joint C: $11.25+(-11.25)=0.00$ ✓; hinge $M_{DC}=0.00$ ✓.

Bending moment diagram

Under the 60 kN load the sagging moment is 34.69 kNm. BM sign: sagging for the beam; for the column, tension on the right when travelling from C to D.

Reactions: A: $H=2.25$, $V=37.03$, $M=39.38$; D: $H=-2.25$, $V=22.97$, $M=0.00$.

Answer: $M_{AC}=-39.38$ kNm, $M_{CA}=11.25$ kNm, $M_{CD}=-11.25$ kNm, $M_{DC}=0$; BM under the load = 34.69 kNm.