Moment distribution: frame with cantilever 15 kN

Assumptions: the beam AB = 5 m (2I) is fixed at A and carries 50 kN/m; it continues as a 1 m overhang BC (2I) with 15 kN at C; the column BD (I, 4 m) runs down from the junction B to its fixed base D. The fixed support A prevents sway, so only joint B rotates.

Step 1: Overhang BC

$M_{BC}=15\times1=15$ kNm (hogging), a known fixed moment on joint B.

Step 2: Distribution factors at B ($k=EI/L$)

• Joint B: relative stiffness $\Sigma k$ = 0.6500; BA: $k=EI/L$ = 0.4000, DF = 0.6154; BC: overhang, DF = 0 (its moment is a known fixed value); BD: $k=EI/L$ = 0.2500, DF = 0.3846

Step 3: Fixed-end moments (clockwise +)

• AB: $\mp\dfrac{wL^2}{12}=\mp\dfrac{50\times5^2}{12}=\mp104.17$ kNm
• BC (overhang): $-15$ kNm at B
• BD: 0

Step 4: Distribution

Unbalanced moment at B: $+104.17-15=89.17$ kNm. The balancing moment is $-89.17$ kNm, divided 0.615 : 0.385 between BA and BD, and half of each share is carried over to the fixed ends A and D.

Step 5: Results and BMD (kNm)

Joint B: $49.29+(-15.00)+(-34.29)=0.00$ ✓

BM sign: sagging positive for the horizontal members; for the column, positive = tension on the right-hand side when travelling from B down to D. The span AB has hogging at both ends and a sagging span moment; the overhang has a straight-line BMD from 0 at C to 15.00 kNm (hogging) at B.

Answer: $M_{AB}=-131.60$, $M_{BA}=49.29$, $M_{BD}=-34.29$, $M_{DB}=-17.15$ kNm, $M_{BC}=-15.00$ kNm (overhang, clockwise end moments +).