Flexibility matrix: frame reactions at D, SFD, BMD

Reading of the figure: A (0, 0) fixed; column AB is 5 m high with 50 kN horizontal at B (towards the right); beam BC is 5 m long with a couple of 30 kNm (taken clockwise) applied at F, 2 m from B; column CD is 3 m high (D is 2 m above the level of A) and D is hinged. EI is constant.

$D_s = 3 + 2 - 3 = 2$. Release the hinge at D: $X_1$ = horizontal reaction ($\rightarrow$), $X_2$ = vertical reaction ($\uparrow$).

Flexibility matrix and load vector (flexibility method)

The primary structure is the cantilever fixed at A. The loads (50 kN at B and the couple at F) give $M_0$, and unit forces at D give $m_1$, $m_2$ (the table of moments below lists their values; + means tension on the inner face).

$$\begin{aligned} f_{11} &= \frac{65.667}{EI} \\ f_{12} &= \frac{50.000}{EI} \\ f_{22} &= \frac{166.667}{EI} \\ \Delta_{10} &= \frac{-46.7}{EI} \\ \Delta_{20} &= \frac{-4115.0}{EI} \end{aligned}$$

$[F]\{X\} = -\{\Delta_0\}$:

$$\begin{aligned} 65.667\,X_1 + 50.000\,X_2 &= 46.7 \\ 50.000\,X_1 + 166.667\,X_2 &= 4115.0 \end{aligned}$$ $$X_1 = H_D = -23.44\ \text{kN},\qquad X_2 = V_D = 31.72\ \text{kN}$$

Final bending moments $M = M_0 + X_1m_1 + X_2m_2$ (kNm)

Shear force (kN)

Reactions

• D (hinge): $H_D = 23.44$ kN (towards the left), $V_D = 31.72$ kN ($\uparrow$)
• A (fixed): $H_A = 26.56$ kN (towards the left), $V_A = 31.72$ kN (downward), $M_A = 74.50$ kNm

Check: $\Sigma H$: $50 - 26.56 - 23.44 = 0$; $\Sigma V$: $-31.72 + 31.72 = 0$ (no vertical load).

The BMD is linear in each member, with a jump of 30 kNm at F from the couple.

Answer: at D, $H_D = 23.44$ kN (towards the left) and $V_D = 31.72$ kN (upward).