Stiffness matrix: frame with coordinates 1, 2, 3

Assumptions: A(0,0), B(0,10), C(10,10), D(10,5): AB = 10 m (1.5I), BC = 10 m (2I), CD = 5 m (I); A and D fixed; the 120 kN load acts to the right at B. Axial deformation is neglected. Coordinates: 1 = rotation of B, 2 = rotation of C, 3 = horizontal sway $\Delta$ (right +). EI denotes the stiffness of I.

Generation of the stiffness matrix (column by column)

Each $k_{ij}$ is the force needed at coordinate $i$ to give a unit displacement at coordinate $j$ with the other coordinates held at zero. Member stiffnesses: $4EI/L$, $2EI/L$, $6EI/L^2$, $12EI/L^3$.

• $k_{11}=\dfrac{4(1.5EI)}{10}+\dfrac{4(2EI)}{10}=1.400EI$, $k_{21}=k_{12}=\dfrac{2(2EI)}{10}=0.400EI$
• $k_{22}=\dfrac{4(2EI)}{10}+\dfrac{4EI}{5}=1.600EI$
• $k_{13}=-\dfrac{6(1.5EI)}{10^2}=-0.0900EI$, $k_{23}=-\dfrac{6EI}{5^2}=-0.2400EI$
• $k_{33}=\dfrac{12(1.5EI)}{10^3}+\dfrac{12EI}{5^3}=0.11400EI$

(The negative signs in the sway column follow from the sign convention: clockwise rotation positive, sway to the right causes clockwise chord rotation.)

$$[K]=EI\begin{bmatrix}1.4000 & 0.4000 & -0.0900 \\ 0.4000 & 1.6000 & -0.2400 \\ -0.0900 & -0.2400 & 0.1140\end{bmatrix}$$

Load vector and solution

Only the horizontal load at coordinate 3: $\{F\}=\{0,\;0,\;120\}^T$.

$$EI\begin{bmatrix}1.4000 & 0.4000 & -0.0900 \\ 0.4000 & 1.6000 & -0.2400 \\ -0.0900 & -0.2400 & 0.1140\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}0.000 \\ 0.000 \\ 120.000\end{Bmatrix}$$ $$EI\theta_{B}=35.8209,\quad EI\theta_{C}=223.8806,\quad EI\Delta_{1}=1552.2388$$

End moments from the member equations (kNm, clockwise +)

• $M_{AB}=\frac{2(1.5EI)}{10}\left(\theta_{B} - (0.3000)\Delta_1\right)$
• $M_{BA}=\frac{2(1.5EI)}{10}\left(2\theta_{B} - (0.3000)\Delta_1\right)$
• $M_{BC}=\frac{2(2EI)}{10}\left(2\theta_{B} + \theta_{C}\right)$
• $M_{CB}=\frac{2(2EI)}{10}\left(2\theta_{C} + \theta_{B}\right)$
• $M_{CD}=\frac{2EI}{5}\left(2\theta_{C} - (0.6000)\Delta_1\right)$
• $M_{DC}=\frac{2EI}{5}\left(\theta_{C} - (0.6000)\Delta_1\right)$

Horizontal reactions

Column shear = (sum of end moments)/height: $H_A=\dfrac{M_{AB}+M_{BA}}{10}=-24.72$ kN, $H_D=\dfrac{M_{CD}+M_{DC}}{5}=-95.28$ kN (both resist the load, so they act to the left); check: $120.00\approx120$ kN ✓.

Full reactions: A: $H=-24.72$, $V=-31.16$, $M=128.96$; D: $H=-95.28$, $V=31.16$, $M=282.99$.

Answer: $M_{AB}=-128.96$, $M_{BA}=-118.21$, $M_{BC}=118.21$, $M_{CB}=193.43$, $M_{CD}=-193.43$, $M_{DC}=-282.99$ kNm; $H_A=24.72$ kN and $H_D=95.28$ kN (to the left).