Moment distribution: frame with SFD, AFD, BMD

Assumptions (the figure is only described): the beam A-B-C is horizontal, 9 m long: A is a hinge (x = 0), a 50 kN load acts at 1 m, the column BD (5 m, fixed at D) is attached at B (x = 4 m), an 80 kN load acts at 5 m, the beam is fixed at C (x = 9 m) and a UDL of 30 kN/m acts on the last 4 m (x = 5 m to 9 m). The 15 kN horizontal load acts on the column at 3 m above its base D (2 m below B), pointing to the right. EI is the same for all members. The fixed support at C prevents sway, so the frame is solved without a sway case.

Members

AB = 4 m (50 kN at 1 m from A), BC = 5 m (80 kN at 1 m from B; 30 kN/m from 1 m to 5 m), BD = 5 m (15 kN at 2 m from B). A is a hinge: the member AB is stiffened by 3/4.

Fixed-end moments (clockwise +, kNm)

• AB: $FEM_{AB}=-28.13$, $FEM_{BA}=9.37$ kNm
• BC: $FEM_{BC}=-102.40$, $FEM_{CB}=73.60$ kNm
• BD: $FEM_{BD}=10.80$, $FEM_{DB}=-7.20$ kNm The FEM of AB at B is modified for the hinge at A: $FEM'_{BA}=FEM_{BA}-\tfrac12FEM_{AB}=23.44$ kNm.

Distribution factors at B ($k=EI/L$)

• Joint B: relative stiffness $\Sigma k$ = 0.5875; BA: $k=EI/L$ = 0.1875 (far end hinged: 3/4 factor), DF = 0.3191; BC: $k=EI/L$ = 0.2000, DF = 0.3404; BD: $k=EI/L$ = 0.2000, DF = 0.3404

Moment distribution

Joint check at B: $45.19+(-79.20)+(34.00)=0.00$ ✓. Hinge A: $M_{AB}=0$.

Bending moment diagram (sagging/inside tension +, kNm)

Shear force and axial force (end values from the member equilibrium)

Shear sign: positive when the force on the start end of the member acts along its local y-axis (90° anticlockwise from the direction A to B, B to C, B to D). Axial force: negative = compression. The column pushes joint B to the right with 16.68 kN. With the girder axially rigid, the split of this force between A and C depends on the axial stiffnesses; for equal $EA$ it is in inverse proportion to the lengths, so AB takes $16.68\times5/9=9.27$ kN (tension) and BC takes $16.68\times4/9=7.41$ kN (compression).

Shear in the girder (kN): AB start 26.20, before the 50 kN load 26.20, after the load -23.80, at B -23.80; BC starts with 110.80, falls to 30.80 after the 80 kN load and decreases linearly under the UDL to -89.20 at C.

Reactions: A: $H=-9.27$, $V=26.20$, $M=0.00$; C: $H=-7.41$, $V=89.20$, $M=-85.20$; D: $H=1.68$, $V=134.60$, $M=-4.40$.

Answer: $M_{BA}=45.19$, $M_{BC}=-79.20$, $M_{CB}=85.20$, $M_{BD}=34.00$, $M_{DB}=4.40$ kNm, $M_{AB}=0$; the SFD, AFD and BMD ordinates are tabulated above.