--- title: "Moment distribution: frame with overhangs, AFD, SFD, BMD" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:moment-dist-frame-4000n-2000n-afd" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-4000n-2000n-afd site: BCT Engineering --- # Moment distribution: frame with overhangs, AFD, SFD, BMD Theory of Structures II · Chapter 3: Displacement method - Asked in: 2068 Baishakh, 20 marks - Video: [MDM Sway frame Analysis_ Lect09 Part-II](https://www.youtube.com/watch?v=15eNbNpAUNY) — Shrinivas R. Suryawanshi - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-4000n-2000n-afd - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question 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.] ## 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. ## Answer **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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the overhangs EB and CF treated in moment distribution? A. As members with distribution factors B. Their statically determinate moments (4x1 = 4 kNm and 2x1 = 2 kNm) act as fixed moments at B and C with DF = 0 C. They are removed D. Their moments are carried over to the columns Answer: B. Overhang moments are known; they are applied as fixed moments on the joints. (Key point: Overhang moments) 2. Why must the sway case be considered? A. The loading is not symmetric, so the frame sways B. The bases are hinged C. The columns are unequal D. The beam is flexible Answer: A. Unsymmetric loads (4 kN and 2 kN) cause sway; it is handled by Case I (prevented) and Case II (induced). (Key point: Why sway) 3. Joints B and C have members of equal EI/L. What are the DFs? A. 0.5 for each of the column and beam, 0 for overhang B. 0.33 each C. 1 for beam D. 0.25 each Answer: A. Equal stiffness of two members gives 0.5 each; the overhang has DF = 0. (Key point: Distribution factors) 4. How is the final moment obtained? A. M = Case I + k x Case II, with k = -R1/R2 B. M = Case I - Case II C. M = Case II only D. M = k x Case I Answer: A. The sway case is scaled so the net imaginary-support reaction becomes zero: k = -R1/R2. (Key point: Sway correction) ## Related in Theory of Structures II - [Moment distribution: portal frame, AFD, SFD, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-60kn-15knm-afd) — Theory of Structures II, chapter 3 - [Moment distribution: frame with SFD, AFD, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-sfd-afd-bmd) — Theory of Structures II, chapter 3 - [Moment distribution: multi-bay frame, AFD, SFD, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-multi-column-frame-afd) — Theory of Structures II, chapter 3 - [Moment distribution: frame, BMD and SFD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-20knm-50kn-150kn) — Theory of Structures II, chapter 3 - [Force method: frame, SFD and BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-5knm-20kn-sfd) — Theory of Structures II, chapter 2 - [Consistent deformation: frame, BMD, SFD, normal thrust](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aconsistent-deformation-frame-15kn-10knm-sfd) — Theory of Structures II, chapter 2 ## Related in other subjects - [AFD, SFD, BMD for frame with UDL and couple](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-30kn-4knm-8kn) — Applied Mechanics, chapter 6 - [AFD, SFD, BMD of frame with 60 kN/m UDL](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-60knm-20knm-12kn) — Applied Mechanics, chapter 6 - [AFD, SFD, BMD of frame with hinge at D](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-8knm-hinge-50kn) — Applied Mechanics, chapter 6 - [Frame AFD, SFD, BMD (20 kNm, 15 kN)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-20knm-15kn) — Strength of Materials, chapter 1