Flexibility matrix: reactions at support D

Assumptions: A is fixed, D is hinged. AB = CD = 5 m high, BC = 4 m, EI constant. The 2 kN/m UDL acts horizontally (to the right) on AB and the 5 kN load acts to the right at C. Axial deformations are ignored. The hinge D gives two unknown reactions, so the frame is indeterminate to the 2nd degree. The redundants are the coordinates at D: 1 = vertical reaction (up +) and 2 = horizontal reaction (right +).

Step 1: Released structure

Remove the hinge at D. The frame A-B-C-D becomes a cantilever fixed at A with a free end D.

Step 2: Moment diagrams (positive = tension on the inside face)

For the loaded case, $M_L=-\left[(5-y)^2+5(5-y)\right]$ on AB (y up from A): the UDL gives $2(5-y)^2/2$ and the 5 kN load at C gives $5(5-y)$, both causing tension on the outside.

Step 3: Flexibility matrix ($f_{ij}=\sum\int m_im_j\,dx/EI$)

$$f_{11}=\frac{1}{EI}\left[4^2(5)+\frac{4^2(4)}{3}\right]=\frac{101.33}{EI},\quad f_{22}=\frac{1}{EI}\left[\frac{5^3}{3}+5^2(4)+\frac{5^3}{3}\right]=\frac{183.33}{EI}$$ $$f_{12}=f_{21}=\frac{1}{EI}\left[4\cdot\frac{5^2}{2}+5\cdot\frac{4\times4}{2}\right]=\frac{90.00}{EI}$$ $$[F]=\frac{1}{EI}\begin{bmatrix}101.33&90.00\\90.00&183.33\end{bmatrix}$$

Step 4: Displacements of D in the released structure due to the loads

$$\Delta_{1L}=\frac{1}{EI}\int m_1M_L\,dx=-416.67/EI,\qquad \Delta_{2L}=\frac{1}{EI}\int m_2M_L\,dx=-156.25/EI$$

(negative: D moves down and to the left in the released frame.)

Step 5: Compatibility (D cannot move)

$$[F]\begin{Bmatrix}R_1\\R_2\end{Bmatrix}=-\begin{Bmatrix}\Delta_{1L}\\\Delta_{2L}\end{Bmatrix} \;\Rightarrow\; \begin{Bmatrix}R_1\\R_2\end{Bmatrix}=\begin{Bmatrix}5.95\\-2.07\end{Bmatrix}\ \text{kN}$$

Reactions at A (equilibrium)

$H_A=-(2\times5+5)-(-2.07)=-12.93$ kN (to the left), $V_A=-5.95$ kN, $M_A=26.21$ kNm. Check $\sum F_x$: loads $15$ kN to the right, reactions -12.93 and -2.07 kN, total $=0$ ✓.

Final BM (inside tension +): $M_A=-26.21$, $M_B=13.45$, $M_C=-10.34$, $M_D=0$ kNm.

Answer: reactions at D: vertical $R_1=5.95$ kN upward and horizontal $R_2=2.07$ kN directed to the left.