Moment distribution: frame with 50 kNm

Assumptions: AB = 2 m + 4 m = 6 m (A fixed) carries a clockwise couple of 50 kNm at 2 m from A; BC = 5 m is a free-ended overhang with 30 kN/m; the column BD = 6 m is fixed at D. Relative stiffness: AB = 1.5I, BC = 2I, BD = I (E constant). No sway occurs because the fixed support at A prevents the horizontal movement of the girder.

Step 1: Overhang BC

The overhang is statically determinate: $M_{BC}=\dfrac{30\times5^2}{2}=375$ kNm (hogging). It is a fixed moment on joint B and takes no part in the distribution (DF = 0).

Step 2: Stiffness and distribution factors (only joint B is free)

• Joint B: relative stiffness $\Sigma k$ = 0.4167; BA: $k=EI/L$ = 0.2500, DF = 0.6000; BC: overhang, DF = 0 (its moment is a known fixed value); BD: $k=EI/L$ = 0.1667, DF = 0.4000

Step 3: Fixed-end moments (clockwise +)

For a couple $M$ at distance $a$ from A: $FEM_{AB}=\dfrac{Mb(2a-b)}{L^2}$, $FEM_{BA}=\dfrac{Ma(2b-a)}{L^2}$ with $a=2$, $b=4$, $L=6$: $FEM_{AB}=\dfrac{50\times4\times(4-4)}{36}=0$, $FEM_{BA}=\dfrac{50\times2\times(8-2)}{36}=16.67$ kNm. Overhang moment on B: $-375$ kNm. Column BD has no load.

Step 4: Moment distribution

Unbalanced moment at B = $16.67-375=-358.33$ kNm, so the balancing moment is $+358.33$ kNm, shared 0.6 : 0.4. Half of each share is carried over to the far fixed ends.

Step 5: Bending moment diagram (kNm)

• Joint B: $231.67+143.33-375=0$ ✓
• Column BD: top moment 143.33, base moment 71.67 (carry-over of one half)
• Overhang: 375 kNm at B (statics)

BM sign: positive = tension on the right-hand side when travelling A to B to C and B to D (sagging for the horizontal beam). In AB the shear is constant, $V_A=-(107.50+231.67+50)/6=-64.86$ kN, so the BMD is two parallel straight lines: from 107.50 at A to -22.22 just left of the couple, then a jump of +50 (the clockwise couple) to +27.78, then down to -231.67 at B. The overhang BC has a parabolic diagram from 0 at C to 375 kNm (hogging) at B.

Answer: $M_{AB}=107.50$, $M_{BA}=231.67$, $M_{BC}=-375.00$, $M_{BD}=143.33$, $M_{DB}=71.67$ kNm (clockwise end moments +).