Moment distribution: frame with 20 kN, 6 kN/m

Assumptions: the beam A-B-C is horizontal: A is a hinge, AB = 2 m + 3 m = 5 m with 20 kN at 2 m from A, BC = 9 m with 6 kN/m, C is a roller. The column BD is 4 m long (2 m + 2 m), fixed at D, with a 5 kN horizontal load (to the right) at its mid-height. EI is constant. The hinge at A prevents horizontal movement of the beam, so there is no sway; only joint B rotates.

Fixed-end moments (clockwise +)

• AB ($a=2$, $b=3$): $FEM_{AB}=-\dfrac{20\times2\times3^2}{5^2}=-14.40$, $FEM_{BA}=+\dfrac{20\times2^2\times3}{5^2}=9.60$ kNm; the far end A is hinged, so the modified $FEM_{BA}'=FEM_{BA}-\tfrac12FEM_{AB}=16.80$ kNm
• BC: $\mp\dfrac{6\times9^2}{12}=\mp40.5$ kNm; C is a roller, so $FEM_{BC}'=-40.5-\tfrac12(40.5)=-60.75$ kNm
• BD (5 kN at mid-height, $L=4$): $FEM_{BD}=2.50$, $FEM_{DB}=-2.50$ kNm ($\pm PL/8$ with the sign for a horizontal load to the right)

Distribution factors at B ($k=EI/L$; AB and BC use 3/4 because their far ends are hinged)

• Joint B: relative stiffness $\Sigma k$ = 0.4833; BA: $k=EI/L$ = 0.1500 (far end hinged: 3/4 factor), DF = 0.3103; BC: $k=EI/L$ = 0.0833 (far end hinged: 3/4 factor), DF = 0.1724; BD: $k=EI/L$ = 0.2500, DF = 0.5172

Moment distribution

Joint B: $29.66+(-53.60)+(23.94)=0.00$ ✓; $M_{AB}=M_{CB}=0$ (hinge and roller).

Bending moment diagram

Reactions: A: $H=-10.54$, $V=6.07$, $M=0.00$; C: $H=0.00$, $V=21.04$, $M=0.00$; D: $H=5.54$, $V=46.89$, $M=-8.22$.

Answer: $M_{BA}=29.66$, $M_{BC}=-53.60$, $M_{BD}=23.94$, $M_{DB}=8.22$ kNm; $M_{AB}=M_{CB}=0$.