Flexibility matrix: frame with 48 kNm moment

Assumptions: A(0,0) is hinged at the base of the 6 m left column AB; the 4 m beam BC is horizontal at the top of AB; the 4 m right column CD is fixed at D, whose base is 2 m above A. The 48 kNm moment acts clockwise at joint B. EI is the same for all members. Axial deformation is ignored.

Degree of indeterminacy

Reactions: hinge A (2) + fixed D (3) = 5; equations = 3. So the frame is indeterminate to the 2nd degree. Redundants: the reactions at the hinge A, coordinate 1 = horizontal $X_1$ (right +) and coordinate 2 = vertical $X_2$ (up +). Released structure: cantilever fixed at D with A free.

Moment diagrams of the released structure (positive = inside tension)

(In case 1 the moment equals the height of the section above A; in case 2 it equals the horizontal distance from A.)

Flexibility coefficients ($\int m_im_j\,dx/EI$)

$$f_{11}=\frac{1}{EI}\left[\frac{6^3}{3}+6^2(4)+\frac{6^3-2^3}{3}\right]=\frac{285.33}{EI},\quad f_{22}=\frac{1}{EI}\left[\frac{4^3}{3}+4^2(4)\right]=\frac{85.33}{EI},\quad f_{12}=f_{21}=\frac{-112.00}{EI}$$ $$\Delta_{1L}=\frac{1}{EI}[(-6)(48)(4)+(-1)(48)(16)]=\frac{-1920}{EI},\qquad \Delta_{2L}=\frac{1}{EI}\left[\frac{4}{2}(48)(4)+4(48)(4)\right]=\frac{1152}{EI}$$

Compatibility (A cannot move)

$$\begin{bmatrix}285.33&-112.00\\-112.00&85.33\end{bmatrix}\begin{Bmatrix}X_1\\X_2\end{Bmatrix}=-\begin{Bmatrix}-1920\\1152\end{Bmatrix} \;\Rightarrow\; X_1=2.95\ \text{kN},\quad X_2=-9.63\ \text{kN}$$

So the hinge A develops a horizontal reaction of 2.95 kN to the right and a vertical reaction of 9.63 kN downward.

Reactions at D

$H_D=-2.95$ kN (left), $V_D=9.63$ kN (up), $M_D=3.59$ kNm (anticlockwise). Check, moments about A: $V_D(4)+|H_D|(2)+M_D=48.00$ kNm, equal to the applied 48 kNm ✓.

Final bending moments ($M=M_L+X_1m_1+X_2m_2$, inside tension +)

Joint B: 30.30 - (-17.70) = 48 kNm, equal to the applied moment ✓. Negative values mean tension on the outside face. The BM varies linearly in every member (no loads between joints).

Answer: $X_1=H_A=2.95$ kN and $X_2=V_A=9.63$ kN (downward); $M_B=-17.70$ kNm in AB and $30.30$ kNm in BC, $M_C=-8.21$ kNm, $M_D=3.59$ kNm.