--- title: "Moment distribution: frame, BMD and SFD" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:moment-dist-frame-20knm-50kn-150kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-20knm-50kn-150kn site: BCT Engineering --- # Moment distribution: frame, BMD and SFD Theory of Structures II · Chapter 3: Displacement method - Asked in: 2066 Bhadra, 20 marks - Video: [Frame Bending Moment Diagram Prob. No. 1](https://www.youtube.com/watch?v=Lhbuvw9xeXY) — Dr. Rajeeth T J - 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-20knm-50kn-150kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Analyze the frame loaded as shown in figure given below. Use the moment distribution method. Draw BMD and SFD. [Figure: frame; beam A-B-C-D-E, A fixed, 20 kN/m UDL over B to D, 50 kN downward at the free end E; AB = 6 m (I), BC = 1.5I, CD = 2I, DE overhang; columns BE-type: column below B (2I) of 3 m + 3 m with 150 kN horizontal at B, fixed base E; column below C (I) fixed at F; horizontal spans 6 m, 6 m, 6 m, 3 m as marked.] ## Short answer Non-sway frame analyzed by moment distribution. Overhang DE yields fixed moment 150 kNm at D. Final end moments (kNm): A=-5.05, B(BA)=-10.11, B(BC)=-82.18, C(CB)=38.39, C(CD)=-29.03, B(BG)=92.28, G=-122.61, C(CF)=-9.35, F=-4.68. ## Answer #### Assumptions (figure is not fully clear) - Beam A-B-C-D-E is horizontal: AB = BC = CD = 6 m, overhang DE = 3 m. $I_{AB} = I$, $I_{BC} = 1.5I$, $I_{CD} = I_{DE} = 2I$. A is fixed. - Column BG ($2I$) and column CF ($I$) are 6 m high with fixed bases. The 150 kN horizontal load (taken as acting to the right) acts on column BG at mid-height (3 m + 3 m); the column below C is also taken as 6 m high. - D is a roller (vertical support), so the overhang DE can carry the 50 kN load. The beam is axially rigid and A is fixed, so the joints cannot sway: this is a **non-sway** frame. - Sign: end moments are clockwise (+) on the member end. #### Stiffness and distribution factors Relative stiffness $K = I/L$ (far end fixed), and $\tfrac34 I/L$ for CD because the far end D is a roller with a known moment. | Joint | Member | K | DF | |---|---|---|---| | B | BA | $I/6 = 0.1667I$ | 0.2222 | | B | BC | $1.5I/6 = 0.25I$ | 0.3333 | | B | BG | $2I/6 = 0.3333I$ | 0.4444 | | C | CB | $0.25I$ | 0.375 | | C | CD | $\tfrac34(2I/6) = 0.25I$ | 0.375 | | C | CF | $I/6 = 0.1667I$ | 0.25 | #### Fixed-end moments - BC and CD (20 kN/m): $wL^2/12 = 20\times 6^2/12 = 60$ kN·m. So $M_{BC}=-60$, $M_{CB}=+60$, and for CD $M_{CD}=-60$, $M_{DC}=+60$. - Overhang DE: $M_D = 50\times 3 = 150$ kN·m (hogging) is statically determinate, so the final $M_{DC} = +150$. Releasing D then gives the modified FEM $M_{CD} = -60 - \tfrac12(60-150) = -15$ kN·m. - BG: load at mid-height, $PL/8 = 150\times 6/8 = 112.5$ kN·m, so $M_{BG}=+112.5$, $M_{GB}=-112.5$. #### Moment distribution | Step | AB | BA | BC | CB | CD | BG | GB | CF | FC | |---|---|---|---|---|---|---|---|---|---| | DF | - | 0.2222 | 0.3333 | 0.3750 | 0.3750 | 0.4444 | - | 0.2500 | - | | FEM | 0 | 0 | -60.00 | 60.00 | -15.00 | 112.50 | -112.50 | 0 | 0 | | Dist 1 | 0 | -11.67 | -17.50 | -16.88 | -16.88 | -23.33 | 0 | -11.25 | 0 | | C.O. 1 | -5.83 | 0 | -8.44 | -8.75 | 0 | 0 | -11.67 | 0 | -5.62 | | Dist 2 | 0 | 1.88 | 2.81 | 3.28 | 3.28 | 3.75 | 0 | 2.19 | 0 | | C.O. 2 | 0.94 | 0 | 1.64 | 1.41 | 0 | 0 | 1.88 | 0 | 1.09 | | Dist 3 | 0 | -0.36 | -0.55 | -0.53 | -0.53 | -0.73 | 0 | -0.35 | 0 | | C.O. 3 | -0.18 | 0 | -0.26 | -0.27 | 0 | 0 | -0.36 | 0 | -0.18 | | **Final (converged)** | **-5.05** | **-10.11** | **-82.18** | **38.39** | **-29.03** | **92.28** | **-122.61** | **-9.35** | **-4.68** | Check at joint B: $-10.11 - 82.18 + 92.28 = 0$. At joint C: $38.39 - 29.03 - 9.35 \approx 0$. #### Final end moments (kN·m) - A: 5.05 hogging; B (in AB): 10.11 sagging (zero moment at about 2 m from A). - B (beam, right side): 82.18 hogging; C (left): 38.39 hogging; C (right): 29.03 hogging; D: 150.00 hogging. - Column BG: top 92.28, at the load point 117.55, base G 122.61. Column CF: top 9.35, base F 4.68. - Maximum sagging: in BC 31.05 kN·m at 3.36 m from B; in CD 10.65 kN·m at 1.99 m from C. #### Shear forces (kN) and reactions | Member | SF at start | SF at end | |---|---|---| | AB | +2.53 | +2.53 | | BC | +67.30 | -52.70 | | CD | +39.84 | -80.16 | | DE | +50.00 | +50.00 | | BG (B to load / load to G) | 69.95 | 80.05 | | CF | 2.34 | 2.34 | Reactions: A: 2.53 kN up, 67.61 kN left, M = 5.05 kN·m. D: 130.16 kN up. G: 64.77 kN up, 80.05 kN left, M = 122.61 kN·m. F: 92.54 kN up, 2.34 kN left, M = 4.68 kN·m. Check: $\Sigma V = 2.53+130.16+64.77+92.54 = 290.0 = 20\times12+50$; $\Sigma H = 67.61+80.05+2.34 = 150$. #### BMD and SFD shape - BMD (tension side): AB is a straight line from 5.05 (top tension at A) to 10.11 (bottom tension at B). BC is a parabola: hogging 82.18 at B, hogging 38.39 at C, sagging 31.05 near the middle. CD is a parabola from 29.03 hogging at C, through a small sagging 10.65, to 150 hogging at D. DE is a straight line falling from 150 at D to 0 at E (tension on top). Column BG: 92.28 at the top, 117.55 at the load point on the opposite face, 122.61 at the base; CF: a straight line from 9.35 to 4.68 (double curvature). - SFD: rectangular blocks in AB, DE (+50, jumping from -80.16 by the 130.16 kN reaction at D) and CF; straight sloping lines in BC and CD (slope 20 kN/m); BG steps from 69.95 to 80.05 at the 150 kN load. **Answer (magnitudes):** $M_{AB}=5.05$, $M_{BA}=10.11$, $M_{BC}=82.18$, $M_{CB}=38.39$, $M_{CD}=29.03$, $M_{DC}=150$, $M_{BG}=92.28$, $M_{GB}=122.61$, $M_{CF}=9.35$, $M_{FC}=4.68$ kN·m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the frame treated as non-sway? A. There are no horizontal loads B. The columns are hinged C. The beam is axially rigid and A is fixed, so joints cannot translate sideways D. The loads are symmetrical Answer: C. Fixed support A with an axially rigid beam prevents joint translation, so no sway correction is needed. (Key point: Non-sway frame) 2. Why is CD given stiffness (3/4)(I/L)? A. Its far end D is a roller/overhang with a known moment, so CD behaves as pinned at D B. Because CD carries a UDL C. Because CD is the longest D. Because C is fixed Answer: A. When the far end moment is known (released), the stiffness is 3EI/4L-type, i.e. 3/4 of the fixed-far-end value. (Key point: Modified stiffness) 3. How is the overhang DE with the 50 kN load handled? A. Its moment 50x3 = 150 kNm is applied at D, and the FEM_CD is modified to -15 B. It is ignored C. It is treated as another span with DF D. It is given a carry-over factor of 1 Answer: A. The overhang is statically determinate with M_D = 150, so D is released and FEM_CD = -60 - 1/2(60-150) = -15. (Key point: Overhang treatment) 4. FEM for column BG with 150 kN at mid-height (L = 6 m) is... A. wL^2/12 = 60 B. PL/8 = 112.5 kNm C. PL/4 = 225 D. Pab^2/L^2 = 56.25 Answer: B. Central point load on a fixed member gives PL/8 = 150x6/8 = 112.5. 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