Question B.1: Use the area-moment method to derive the flexibility matrix ......
Use the area-moment method to derive the flexibility matrix for the cantilever beam shown in Fig. B2 at (a), where EI is constant over the length of the beam.
The flexibility influence coefficient matrix is defined by:
\left \{ \begin{matrix} z_{1} \\ z_{2} \end{matrix} \right \} = \begin{bmatrix} \alpha_{11} & \alpha_{12} \\ \alpha_{21} & \alpha_{22} \end{bmatrix} \left \{ \begin{matrix} F_{1} \\ F_{2} \end{matrix} \right \} (A)where z_{1}, z_{2}, F_{1} \mathrm{and} F_{1} are defined by Fig. B2(a), and
\left[\alpha \right] = \left[K\right] ^{-1} = \begin{bmatrix} \alpha_{11} & \alpha_{12} \\ \alpha_{21} & \alpha_{22} \end{bmatrix} (B)is the flexibility influence coefficient matrix. The quantity \alpha_{12}, for example, is the displacement z_{1} \mathrm{when} F_{1} \mathrm{is zero and} F_{2} is a unit load.
Figure B2 at (b) shows the bending moment m_{1} due to a unit load at node 1, and Fig. B2 at (c) shows the bending moment m_{2} due to a unit load at node 2.
The individual terms of Eq. (A) are found as follows:
\alpha_{11} is the displacement z_{1}, at node 1, due to a unit load F_{1} also at node 1.
Then
but this is equal to the area moment of the triangle in Fig. B2 at (b) about node 1:
\alpha_{11} = \frac{1}{EI} \times \left[\frac{1}{2} \left(\frac{L}{2} \right)^{2} \right] \times \left[\frac{2}{3} \left(\frac{L}{2} \right) \right] = \frac{L^{3}}{24EI} (D)\alpha_{12} is the displacement z_{1} at node 1 due to a unit load F_{2} at node 2. This is given by the area moment of the shaded area in Fig. B2(c) about node 1. This can be found by splitting it into a triangle and a rectangle, and multiplying these areas by the distances of their centroids from node 1. It should be noted that \alpha_{12} is not given by the area moment of the whole triangle in Fig. B2(c) about node 1. Thus:
\alpha_{12} = \frac{1}{EI} \left\{ \left[\frac{1}{2} \left(\frac{L}{2} \right)^{2} \times \frac{L}{3} \right] + \left[\left(\frac{L}{2} \right)^{2} \times \frac{L}{4} \right] \right\} = \frac{5L^{3}}{48EI} (E)It can similarly be shown that:
\alpha_{21} = \alpha _{12} = \frac{5L^{3}}{48EI} (F)\\ \mathrm{and} \\ \alpha_{22} = \frac{L^{3}}{3EI} (G)The final flexibility matrix is
\left[\alpha \right] = \left[K\right]^{-1} = \begin{bmatrix} \alpha_{11} & \alpha_{12} \\ \alpha_{21} & \alpha_{22} \end{bmatrix} = \frac{L^{3}}{EI} \begin{bmatrix} \frac{1}{24} & \frac{5}{48} \\ \frac{5}{48} & \frac{1}{3} \end{bmatrix} (H)