Moment distribution: frame with 80 kN, 60 kN

Assumptions: the girder A-B-C-D-E has A fixed, a roller at D and rigid column connections at B (BG, 6 m, 2I) and C (CF, 6 m, I), both fixed at their bases. Spans AB = BC = CD = 6 m, overhang DE = 2 m with 60 kN at E; 10 kN/m UDL on BC and CD (B to D). Relative I: AB = I, BC = 1.5I, CD = 2I. The 80 kN horizontal load at B acts to the right.

Sway check

The girder is held horizontally by the fixed support at A, so it cannot sway. The 80 kN load is carried as axial force in the girder to support A and causes no bending moment. The analysis is therefore a non-sway moment distribution with joints B, C and D free to rotate.

Step 1: Overhang

$M_D=60\times2=120$ kNm (hogging), a fixed moment on joint D.

Step 2: Fixed-end moments (clockwise +)

For the UDL spans: $\mp wL^2/12=\mp10\times36/12=\mp30$ kNm on BC and CD.

• AB: $FEM_{AB}=0.00$, $FEM_{BA}=0.00$ kNm
• BC: $FEM_{BC}=-30.00$, $FEM_{CB}=30.00$ kNm
• CD: $FEM_{CD}=-30.00$, $FEM_{DC}=30.00$ kNm
• DE: $FEM_{DE}=-120.00$, $FEM_{ED}=0.00$ kNm
• BG: $FEM_{BG}=0.00$, $FEM_{GB}=0.00$ kNm
• CF: $FEM_{CF}=0.00$, $FEM_{FC}=0.00$ kNm

Step 3: Distribution factors ($k=EI/L$)

• Joint B: relative stiffness $\Sigma k$ = 0.7500; BA: $k=EI/L$ = 0.1667, DF = 0.2222; BC: $k=EI/L$ = 0.2500, DF = 0.3333; BG: $k=EI/L$ = 0.3333, DF = 0.4444
• Joint C: relative stiffness $\Sigma k$ = 0.7500; CB: $k=EI/L$ = 0.2500, DF = 0.3333; CD: $k=EI/L$ = 0.3333, DF = 0.4444; CF: $k=EI/L$ = 0.1667, DF = 0.2222
• Joint D: relative stiffness $\Sigma k$ = 0.3333; DC: $k=EI/L$ = 0.3333, DF = 1.0000; DE: overhang, DF = 0 (its moment is a known fixed value)

Step 4: Moment distribution (kNm)

Step 5: Final end moments and BMD

Joint B: $8.82+(-26.45)+(17.63)=0.00$ ✓; Joint C: $17.26+(-4.35)+(-12.90)=0.00$ ✓; Joint D: $120.00+(-120.00)=0.00$ ✓.

BM sign: positive = sagging for the girder, tension on the right-hand side when travelling along the member as listed.

Reactions: A: $H=-81.18$, $V=-2.20$, $M=-4.41$; G: $H=4.41$, $V=33.74$, $M=-8.82$; F: $H=-3.23$, $V=39.19$, $M=6.45$; D: $H=0.00$, $V=109.27$, $M=0.00$.

Answer: $M_{AB}=4.41$, $M_{BA}=8.82$, $M_{BC}=-26.45$, $M_{CB}=17.26$, $M_{CD}=-4.35$, $M_{DC}=120.00$, $M_{DE}=-120.00$ kNm; columns: $M_{BG}=17.63$, $M_{GB}=8.82$, $M_{CF}=-12.90$, $M_{FC}=-6.45$ kNm.