Stiffness matrix: frame with 20 kN/m, 10 kN

Assumptions: A(0,0) and D are fixed; the girder BC (2I, 8 m) is horizontal at the top, AB = 6 m (1.5I) and CD = 4 m (I), so D is 2 m above A. The UDL of 20 kN/m acts downward on BC and the 10 kN load at C acts to the right. Axial deformations are ignored. Take $EI$ as the stiffness of I.

Unknown displacements

$\theta_B$, $\theta_C$ and the sway $\Delta$ of the girder. Chord rotations: $\psi_{AB}=\Delta/6$, $\psi_{CD}=\Delta/4$, $\psi_{BC}=0$.

Fixed-end moments

$FEM_{BC}=-\dfrac{20\times8^2}{12}=-106.67$ kNm, $FEM_{CB}=+106.67$ kNm. Columns carry no load.

Member stiffness equations

• $M_{AB}=\frac{2(1.5EI)}{6}\left(\theta_{B} - (0.5000)\Delta_1\right)$
• $M_{BA}=\frac{2(1.5EI)}{6}\left(2\theta_{B} - (0.5000)\Delta_1\right)$
• $M_{BC}=\frac{2(2EI)}{8}\left(2\theta_{B} + \theta_{C}\right) - 106.67$
• $M_{CB}=\frac{2(2EI)}{8}\left(2\theta_{C} + \theta_{B}\right) + 106.67$
• $M_{CD}=\frac{2EI}{4}\left(2\theta_{C} - (0.7500)\Delta_1\right)$
• $M_{DC}=\frac{2EI}{4}\left(\theta_{C} - (0.7500)\Delta_1\right)$

Equilibrium equations

• Joint B: $M_{BA}+M_{BC}=0$
• Joint C: $M_{CB}+M_{CD}=0$
• Storey shear: $\dfrac{M_{AB}+M_{BA}}{6}+\dfrac{M_{CD}+M_{DC}}{4}+10=0$

Combining gives the stiffness matrix equation:

$$EI\begin{bmatrix}2.0000 & 0.5000 & -0.2500 \\ 0.5000 & 2.0000 & -0.3750 \\ -0.2500 & -0.3750 & 0.2708\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}106.667 \\ -106.667 \\ 10.000\end{Bmatrix}$$

Typical terms: $K_{11}=\dfrac{4(1.5EI)}{6}+\dfrac{4(2EI)}{8}=2EI$, $K_{12}=\dfrac{2(2EI)}{8}=0.5EI$, $K_{13}=-\dfrac{6(1.5EI)}{6^2}=-0.25EI$ (sign convention of the rows above, positive sway to the right).

Solution

$$EI\theta_{B}=71.6049,\quad EI\theta_{C}=-70.1235,\quad EI\Delta_{1}=5.9259$$

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

Checks:

• Joint B: (70.12) + (-70.12) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (72.35) + (-72.35) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Storey shear: $\dfrac{104.44}{6}+\dfrac{-109.63}{4}+10=0.00$ ✓

Reactions: A: $H=17.41$, $V=79.72$, $M=-34.32$; D: $H=-27.41$, $V=80.28$, $M=37.28$.

Answer: $M_{AB}=34.32$, $M_{BA}=70.12$, $M_{BC}=-70.12$, $M_{CB}=72.35$, $M_{CD}=-72.35$, $M_{DC}=-37.28$ kNm.