How are the overhangs EB and CF treated in moment distribution?

Their statically determinate moments (4x1 = 4 kNm and 2x1 = 2 kNm) act as fixed moments at B and C with DF = 0

As members with distribution factors

Their moments are carried over to the columns

Joints B and C have members of equal EI/L. What are the DFs?

0.5 for each of the column and beam, 0 for overhang

How is the final moment obtained?

M = Case I + k x Case II, with k = -R1/R2

2068 Baishakh · 20 marks

Moment distribution: frame with overhangs, AFD, SFD, BMD

Short answer

Moment distribution with sway correction: Case I with sway prevented and Case II with arbitrary sway give k=-R1/R2=-0.00429. Final end moments in kNm: AB -0.86, BA -2.14, BC -1.86, CB 0.14, CD 1.86, DC 1.14, overhang BE 4.00, CF -2.00. Draw AFD/SFD/BMD from these by statics.

Draw axial force, shear force and bending moment diagram for the given loaded frame. Use the moment distribution method. [Figure: portal frame, columns I of 4 m and beam I of 4 m span with 1 m overhangs on each side; 4000 N downward at the left tip and 2000 N downward at the right tip; fixed bases.]

Data: columns AB and CD are 4 m high, the beam BC has a span of 4 m with overhangs EB and CF of 1 m on each side; bases A and D are fixed; EI is constant. Loads: 4000 N $=4$ kN downward at E and 2000 N $=2$ kN downward at F. Work in kN and kNm.

The loading is not symmetric, so the frame sways (the antisymmetric part of the load). The moment distribution is carried out in two parts: sway prevented (Case I) and sway induced (Case II).

Step 1: Overhangs

$M_{BE}=4\times1=4$ kNm and $M_{CF}=2\times1=2$ kNm (both hogging). These are fixed moments acting on joints B and C.

Step 2: Distribution factors ( $k=EI/L=EI/4$ for all three members)

Joint B: relative stiffness $\Sigma k$ = 0.5000; BA: $k=EI/L$ = 0.2500, DF = 0.5000; BC: $k=EI/L$ = 0.2500, DF = 0.5000; BE: overhang, DF = 0 (its moment is a known fixed value)
Joint C: relative stiffness $\Sigma k$ = 0.5000; CB: $k=EI/L$ = 0.2500, DF = 0.5000; CD: $k=EI/L$ = 0.2500, DF = 0.5000; CF: overhang, DF = 0 (its moment is a known fixed value)

Case I: sway prevented by an imaginary support at B

	AB	BA	BC	CB	CD	DC	BE	CF
DF	-	0.500	0.500	0.500	0.500	-	0.000	0.000
FEM	0.00	0.00	0.00	0.00	0.00	0.00	4.00	-2.00
Balance	0.00	-2.00	-2.00	1.00	1.00	0.00	0.00	0.00
Carry-over	-1.00	0.00	0.50	-1.00	0.00	0.50	0.00	0.00
Balance	0.00	-0.25	-0.25	0.50	0.50	0.00	0.00	0.00
Carry-over	-0.12	0.00	0.25	-0.12	0.00	0.25	0.00	0.00
Balance	0.00	-0.13	-0.13	0.06	0.06	0.00	0.00	0.00
Carry-over	-0.06	0.00	0.03	-0.06	0.00	0.03	0.00	0.00
Balance	0.00	-0.02	-0.02	0.03	0.03	0.00	0.00	0.00
Carry-over	-0.01	0.00	0.02	-0.01	0.00	0.02	0.00	0.00
Further cycles (converged)	0.00	-0.01	-0.01	0.00	0.01	0.00	0.00	0.00
Final M	-1.20	-2.40	-1.60	0.40	1.60	0.80	4.00	-2.00

Horizontal reaction of the imaginary support: $R_1=0.300$ kN.

Case II: a sway of the girder with no external load

Take FEM $=-100$ kNm at the ends of both columns (same length, same EI, same $\Delta$ ).

	AB	BA	BC	CB	CD	DC	BE	CF
DF	-	0.500	0.500	0.500	0.500	-	0.000	0.000
FEM	-100.00	-100.00	0.00	0.00	-100.00	-100.00	0.00	0.00
Balance	0.00	50.00	50.00	50.00	50.00	0.00	0.00	0.00
Carry-over	25.00	0.00	25.00	25.00	0.00	25.00	0.00	0.00
Balance	0.00	-12.50	-12.50	-12.50	-12.50	0.00	0.00	0.00
Carry-over	-6.25	0.00	-6.25	-6.25	0.00	-6.25	0.00	0.00
Balance	0.00	3.12	3.12	3.12	3.12	0.00	0.00	0.00
Carry-over	1.56	0.00	1.56	1.56	0.00	1.56	0.00	0.00
Balance	0.00	-0.78	-0.78	-0.78	-0.78	0.00	0.00	0.00
Carry-over	-0.39	0.00	-0.39	-0.39	0.00	-0.39	0.00	0.00
Further cycles (converged)	0.08	0.16	0.23	0.23	0.16	0.08	0.00	0.00
Final M	-80.00	-60.00	60.00	60.00	-60.00	-80.00	0.00	0.00

Horizontal reaction of the imaginary support: $R_2=70.000$ kN.

Step 3: Sway correction

$k=-\dfrac{R_1}{R_2}=-\dfrac{0.300}{70.000}=-0.00429$

End	Case I	k × Case II	Final M (kNm)
AB	-1.20	0.34	-0.86
BA	-2.40	0.26	-2.14
BC	-1.60	-0.26	-1.86
BE	4.00	0.00	4.00
CB	0.40	-0.26	0.14
CD	1.60	0.26	1.86
CF	-2.00	0.00	-2.00
DC	0.80	0.34	1.14

Joint checks: B: $0.00$ , C: $0.00$ .

Bending moment diagram (kNm)

Member	BM at first end	BM at second end	Max within member
AB	-0.86	2.14	-
BC	-1.86	-0.14	-
CD	1.86	-1.14	-
EB	0.00	-4.00	-
CF	-2.00	0.00	-

Shear force and axial force

Member	Axial N (+ tension)	Shear at start	Shear at end
AB	-4.43	0.75	0.75
BC	0.75	0.43	0.43
CD	-1.57	-0.75	-0.75
EB	0.00	-4.00	-4.00
CF	0.00	2.00	2.00

Shear sign convention: positive when the force on the start end of the member acts along its local y axis (90° anticlockwise from the member direction A→B, B→C, C→D, E→B, C→F). Axial: negative = compression. The columns carry compression (4.43 kN in AB and 1.57 kN in CD) and the beam BC carries a small tension of 0.75 kN, equal to the horizontal reactions (0.75 kN each, equal and opposite). The shear in each overhang is constant along its length (4 kN in EB, 2 kN in CF); it jumps from zero to this value at the loaded free tip.

Reactions: A: $H=-0.75$ , $V=4.43$ , $M=0.86$ ; D: $H=0.75$ , $V=1.57$ , $M=-1.14$ .

Answer: $M_{AB}=-0.86$ , $M_{BA}=-2.14$ , $M_{BC}=-1.86$ , $M_{CB}=0.14$ , $M_{CD}=1.86$ , $M_{DC}=1.14$ kNm; $M_{BE}=4$ , $M_{CF}=2$ kNm (overhangs).

✓ Answer checked by Claude Opus 5.5 (high effort)

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