Generate stiffness matrix of frame

Assumptions: the figure gives the beam BC (3I, 8 m), the two columns (I, 10 m, fixed at the base) and a short member CD (2I) at the right end of the girder ending on a roller. Its length is not legible, so $L_{CD}=4$ m is assumed (the general expression is also given). Coordinates: 1 = rotation of the left joint B, 2 = rotation of the right joint C, 3 = rotation of the far end D of the short member. Only rotational coordinates are required, so the stiffness matrix is $3\times3$.

Definition

$k_{ij}$ = moment needed at coordinate $i$ to produce a unit rotation at coordinate $j$ (all other coordinates held at zero). Use the member end stiffnesses: $4EI/L$ (rotation at the near end, far end fixed against rotation) and $2EI/L$ (moment carried to the far end).

Member stiffnesses

Column by column

• Unit rotation at 1 (θ₁ = 1): joint B needs $\frac{4EI}{10}+\frac{4(3EI)}{8}$; joint C receives the carry-over $\frac{2(3EI)}{8}$ from the beam; the roller end D is unaffected. $k_{11}=1.900EI$, $k_{21}=0.750EI$, $k_{31}=0$
• Unit rotation at 2 (θ₂ = 1): $k_{22}=\frac{4(3EI)}{8}+\frac{4EI}{10}+\frac{4(2EI)}{L_{CD}}$, $k_{12}=\frac{2(3EI)}{8}$, $k_{32}=\frac{2(2EI)}{L_{CD}}$ $k_{22}=3.900EI$, $k_{12}=0.750EI$, $k_{32}=1.000EI$
• Unit rotation at 3 (θ₃ = 1): $k_{33}=\frac{4(2EI)}{L_{CD}}=2.000EI$, $k_{23}=\frac{2(2EI)}{L_{CD}}=1.000EI$, $k_{13}=0$

Stiffness matrix

$$[K]=EI\begin{bmatrix}1.900&0.750&0\\0.750&3.900&1.000\\0&1.000&2.000\end{bmatrix}$$

In general, with $L_{CD}=L$: $K_{22}=1.9EI+\dfrac{8EI}{L}$, $K_{23}=\dfrac{4EI}{L}$, $K_{33}=\dfrac{8EI}{L}$.

The matrix is symmetric ($k_{ij}=k_{ji}$, Betti's law), positive definite and banded because joints 1 and 3 are not connected by a member. The equation of the structure is $\{P\}=[K]\{\theta\}$.