Flexibility matrix: reactions at D, BMD

Reading of the figure (dimensions as assumed): column ABE with A fixed at the base, E 4 m above A (30 kN horizontal, towards the right) and B 8 m above A; beam BFC with $BF = 4$ m and $FC = 2$ m; column CD, 4 m high, so D is 4 m above the level of A and is hinged. EI is constant.

$r = 3 + 2 = 5$, so $D_s = 2$. Release the hinge at D and take $X_1$ = horizontal reaction ($\rightarrow$) and $X_2$ = vertical reaction ($\uparrow$). The primary structure is the cantilever fixed at A.

Step 1: Coordinates and unit-load moments

$M_0$ is due to the 30 kN load only: $M_0 = -30\times4 = -120$ kNm at A, zero above E. The unit moments $m_1$ and $m_2$ are those caused by a unit horizontal force and a unit vertical force at D (see the table below, + tension on the inner face).

Step 2: Flexibility matrix $[F]$ and load vector $\{\Delta_0\}$

$$\begin{aligned} f_{11} &= \frac{160.000}{EI} \\ f_{12} &= \frac{72.000}{EI} \\ f_{22} &= \frac{360.000}{EI} \\ \Delta_{10} &= \frac{640.0}{EI} \\ \Delta_{20} &= \frac{-1440.0}{EI} \end{aligned}$$ $$[F] = \frac{1}{EI}\begin{bmatrix} 160.00 & 72.00 \\ 72.00 & 360.00 \end{bmatrix},\qquad \{\Delta_0\} = \frac{1}{EI}\begin{Bmatrix} 640.0 \\ -1440.0 \end{Bmatrix}$$

Step 3: Compatibility, $[F]\{X\} = -\{\Delta_0\}$

$$\begin{aligned} 160.000\,X_1 + 72.000\,X_2 &= -640.0 \\ 72.000\,X_1 + 360.000\,X_2 &= 1440.0 \end{aligned}$$ $$X_1 = H_D = -6.37\ \text{kN},\qquad X_2 = V_D = 5.27\ \text{kN}$$

Step 4: Bending moments $M = M_0 + X_1m_1 + X_2m_2$ (kNm)

Step 5: Shear force (kN)

Reactions

• D: $H_D = 6.37$ kN (towards the left), $V_D = 5.27$ kN ($\uparrow$)
• A: $H_A = 23.63$ kN (towards the left), $V_A = 5.27$ kN ($\downarrow$), $M_A = 62.86$ kNm

Check: $\Sigma H$: $30 - 23.63 - 6.37 = 0$ and $\Sigma V$: $-5.27 + 5.27 = 0$.

Answer: reactions at D: $H_D = 6.37$ kN (towards the left) and $V_D = 5.27$ kN (upward).