--- title: "Moment distribution: frame with SFD, AFD, BMD" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:moment-dist-frame-sfd-afd-bmd" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-sfd-afd-bmd site: BCT Engineering --- # Moment distribution: frame with SFD, AFD, BMD Theory of Structures II · Chapter 3: Displacement method - Asked in: 2070 Chaitra (old course), 16 marks - Video: [Frame Analysis || Shear Force & Bending Moment Diagram](https://www.youtube.com/watch?v=0pG1to_L44M) — Civil Engineering - 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-sfd-afd-bmd - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Analyse the frame shown in figure below using the moment distribution method. Also draw shear force, axial force and bending moment diagram. [Figure: beam (A hinged at the left end) with 50 kN and 80 kN loads, 30 kN/m UDL on the right portion, fixed right end; column of 5 m with fixed base carrying 15 kN horizontal at the 3 m level; horizontal dimensions 1 m, 3 m, 1 m, 4 m as marked.] ## Short answer By moment distribution, the final end moments are M_AB = 0, M_BA = 45.19 kNm, M_BC = -79.20 kNm, M_CB = 85.20 kNm, M_BD = 34.00 kNm, M_DB = 4.40 kNm. Shear, axial force, and bending moment diagrams are drawn from these moments and member loads. ## Answer **Assumptions (the figure is only described):** the beam A-B-C is horizontal, 9 m long: A is a hinge (x = 0), a 50 kN load acts at 1 m, the column BD (5 m, fixed at D) is attached at B (x = 4 m), an 80 kN load acts at 5 m, the beam is fixed at C (x = 9 m) and a UDL of 30 kN/m acts on the last 4 m (x = 5 m to 9 m). The 15 kN horizontal load acts on the column at 3 m above its base D (2 m below B), pointing to the right. EI is the same for all members. The fixed support at C prevents sway, so the frame is solved without a sway case. #### Members AB = 4 m (50 kN at 1 m from A), BC = 5 m (80 kN at 1 m from B; 30 kN/m from 1 m to 5 m), BD = 5 m (15 kN at 2 m from B). A is a hinge: the member AB is stiffened by 3/4. #### Fixed-end moments (clockwise +, kNm) - AB: $FEM_{AB}=-28.13$, $FEM_{BA}=9.37$ kNm - BC: $FEM_{BC}=-102.40$, $FEM_{CB}=73.60$ kNm - BD: $FEM_{BD}=10.80$, $FEM_{DB}=-7.20$ kNm The FEM of AB at B is modified for the hinge at A: $FEM'_{BA}=FEM_{BA}-\tfrac12FEM_{AB}=23.44$ kNm. #### Distribution factors at B ($k=EI/L$) - Joint B: relative stiffness $\Sigma k$ = 0.5875; BA: $k=EI/L$ = 0.1875 (far end hinged: 3/4 factor), DF = 0.3191; BC: $k=EI/L$ = 0.2000, DF = 0.3404; BD: $k=EI/L$ = 0.2000, DF = 0.3404 #### Moment distribution | | BA | BC | CB | BD | DB | |---|---|---|---|---|---| | DF | 0.319 | 0.340 | - | 0.340 | - | | FEM | 23.44 | -102.40 | 73.60 | 10.80 | -7.20 | | Balance | 21.75 | 23.20 | 0.00 | 23.20 | 0.00 | | Carry-over | 0.00 | 0.00 | 11.60 | 0.00 | 11.60 | | **Final M** | **45.19** | **-79.20** | **85.20** | **34.00** | **4.40** | Joint check at B: $45.19+(-79.20)+(34.00)=0.00$ ✓. Hinge A: $M_{AB}=0$. #### Bending moment diagram (sagging/inside tension +, kNm) | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | AB | 0.00 | -45.19 | 26.20 at 1.00 m | | BC | -79.20 | -85.20 | 47.41 at 2.02 m | | BD | 34.00 | -4.40 | - | Under the 50 kN load: 26.20 kNm; under the 80 kN load: 31.60 kNm. #### Shear force and axial force (end values from the member equilibrium) | Member | Axial N (+ tension) | Shear at start | Shear at end | |---|---|---|---| | AB | 9.27 | 26.20 | -23.80 | | BC | -7.41 | 110.80 | -89.20 | | BD | -134.60 | -16.68 | -1.68 | Shear sign: positive when the force on the start end of the member acts along its local y-axis (90° anticlockwise from the direction A to B, B to C, B to D). Axial force: negative = compression. The column pushes joint B to the right with 16.68 kN. With the girder axially rigid, the split of this force between A and C depends on the axial stiffnesses; for equal $EA$ it is in inverse proportion to the lengths, so AB takes $16.68\times5/9=9.27$ kN (tension) and BC takes $16.68\times4/9=7.41$ kN (compression). Shear in the girder (kN): AB start 26.20, before the 50 kN load 26.20, after the load -23.80, at B -23.80; BC starts with 110.80, falls to 30.80 after the 80 kN load and decreases linearly under the UDL to -89.20 at C. Reactions: A: $H=-9.27$, $V=26.20$, $M=0.00$; C: $H=-7.41$, $V=89.20$, $M=-85.20$; D: $H=1.68$, $V=134.60$, $M=-4.40$. **Answer: $M_{BA}=45.19$, $M_{BC}=-79.20$, $M_{CB}=85.20$, $M_{BD}=34.00$, $M_{DB}=4.40$ kNm, $M_{AB}=0$; the SFD, AFD and BMD ordinates are tabulated above.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the frame solved without a sway case? A. The 15 kN horizontal load is too small to cause sway B. The fixed support at C prevents sway, so only joint B rotates C. All members have the same EI D. The hinge at A allows free horizontal movement Answer: B. The fixed support at C restrains horizontal movement of the girder, so no sway occurs and only rotation of B is distributed. (Key point: Why no sway case) 2. For member AB with a hinge at A, what is the modified FEM at B? A. FEM_BA + 1/2 FEM_AB B. FEM_BA - 1/2 FEM_AB = 9.37 + 14.06 = 23.44 kNm C. FEM_BA only = 9.37 kNm D. 2 x FEM_BA Answer: B. Releasing the hinge at A removes FEM_AB, and half of it would have been carried to B, so FEM'_BA = FEM_BA - 1/2 FEM_AB = 23.44 kNm. (Key point: Hinged end modification of FEM) 3. How is the stiffness of AB (hinged at A) entered in the DF calculation at B? A. EI/L multiplied by 3/4 B. EI/L multiplied by 4/3 C. EI/L unchanged D. Zero, since A is hinged Answer: A. With the far end hinged the stiffness is 3EI/L = 3/4 of 4EI/L, so k is multiplied by 3/4 (0.1875 instead of 0.25). (Key point: Stiffness of hinged member) 4. Which check confirms the final moments at joint B? A. The three moments at B (45.19, -79.20, 34.00) sum to zero B. The moments of BA and BC are equal C. M_BD equals M_DB D. The moments at B sum to the 15 kN load Answer: A. With no external moment at B, the member end moments at the joint must sum to zero: 45.19 - 79.20 + 34.00 = 0. 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