Generate stiffness matrix for portal frame

Assumptions: portal frame A-B-C-D with fixed bases A and D, columns AB and CD of height 4 m and rigidity $EI$, beam BC of span 5 m and rigidity $3EI$. Axial deformations are ignored. Coordinates (positive directions): 1 = rotation of joint B, 2 = rotation of joint C (both clockwise), 3 = horizontal sway of the beam (to the right).

Member stiffness coefficients

Column-by-column generation ($k_{ij}$ = force at $i$ for a unit displacement at $j$, others zero)

• Unit rotation at 1 (B turns clockwise by 1): $k_{11}=\dfrac{4EI}{4}+\dfrac{4(3EI)}{5}=3.4000EI$; $k_{21}=\dfrac{2(3EI)}{5}=1.2000EI$; $k_{31}$: the column AB gives a shear $\dfrac{6EI}{4^2}$ which must be resisted by the sway restraint, $k_{31}=-0.3750EI$ (negative: on its own, a clockwise rotation of B would make the top of column AB move to the right, so the restraint must push the girder to the left).
• Unit rotation at 2: $k_{22}=\dfrac{4(3EI)}{5}+\dfrac{4EI}{4}=3.4000EI$; $k_{12}=1.2000EI$; $k_{32}=-0.3750EI$.
• Unit sway at 3: $k_{33}=2\times\dfrac{12EI}{4^3}=0.3750EI$; $k_{13}=k_{23}=-\dfrac{6EI}{4^2}=-0.3750EI$.

Stiffness matrix

$$[K]=EI\begin{bmatrix}3.4000 & 1.2000 & -0.3750 \\ 1.2000 & 3.4000 & -0.3750 \\ -0.3750 & -0.3750 & 0.3750\end{bmatrix}$$

The matrix is symmetric and, because the frame is symmetrical, $k_{11}=k_{22}$ and $k_{13}=k_{23}$. The equilibrium equation for any loading is $\{P\}=[K]\{d\}$ with $\{d\}=\{\theta_B,\theta_C,\Delta\}^T$ and $P$ the equivalent joint loads (applied loads minus fixed-end effects).