Flexibility matrix for coordinates 1 and 2

Reading of the figure: a beam of two spans AB and BC, each $L = 10$ m, with constant $EI$; A is a fixed end, B an interior support and C a roller. Coordinate 1 is the rotation at the fixed end A, and coordinate 2 is the rotation at the interior support B. To define flexibility coefficients, the restraints against these rotations are released: A becomes a hinge and the beam is cut into two simply supported spans by a hinge over B, so the primary structure is made of the two simply supported spans AB and BC. A unit moment at coordinate 1 acts at A and a unit moment at coordinate 2 acts at B (on both spans).

 (1) A o------------o B (2)------------o C
        |<-- 10 -->|     |<-- 10 -->|

Flexibility coefficients

The flexibility coefficient $f_{ij}$ is the rotation at coordinate $i$ due to a unit moment at coordinate $j$.

Unit moment at coordinate 1 (A): it acts on span AB only. For a simply supported span of length $L$ with a unit moment at one end, the rotation at that end is $\dfrac{L}{3EI}$ and the rotation at the far end is $\dfrac{L}{6EI}$:

$$f_{11} = \frac{L}{3EI} = \frac{10}{3EI} = \frac{3.333}{EI},\qquad f_{21} = \frac{L}{6EI} = \frac{10}{6EI} = \frac{1.667}{EI}$$

Unit moment at coordinate 2 (B): it acts on both spans, so each span turns by $\dfrac{L}{3EI}$ at B, and span AB rotates $\dfrac{L}{6EI}$ at A:

$$f_{22} = \frac{L}{3EI} + \frac{L}{3EI} = \frac{20}{3EI} = \frac{6.667}{EI},\qquad f_{12} = \frac{L}{6EI} = \frac{1.667}{EI}$$

($f_{12} = f_{21}$, as required by Maxwell's theorem.)

Flexibility matrix

$$[F] = \frac{1}{EI}\begin{bmatrix} 3.333 & 1.667 \\ 1.667 & 6.667 \end{bmatrix} = \frac{L}{6EI}\begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix}$$

The matrix is symmetric with positive diagonal terms. Its inverse is the stiffness matrix for the same coordinates:

$$[K] = [F]^{-1} = \frac{6EI}{7L}\begin{bmatrix} 4 & -1 \\ -1 & 2 \end{bmatrix}$$