--- title: "Moment distribution: frame with 80 kN, 60 kN" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:moment-dist-frame-80kn-10knm-60kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-80kn-10knm-60kn site: BCT Engineering --- # Moment distribution: frame with 80 kN, 60 kN Theory of Structures II · Chapter 3: Displacement method - Asked in: 2078 Bhadra, 10 marks - Video: [Moment Distribution Method - Analysis of Frames - Problem No 10](https://www.youtube.com/watch?v=wzQPEz-5Kcw) — Stan Academy - 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-80kn-10knm-60kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Use the moment distribution method to analyze the frame loaded as shown below. Also draw BMD. [Figure: beam A-B-C-D-E, A fixed, supports at C and D; spans 6 m, 6 m, 6 m and overhang 2 m with 60 kN at E; 10 kN/m UDL on BD; AB = I, BC = 1.5I, CD = 2I; columns BG (2I) and CF (I) each with 3 m + 3 m height, fixed bases; 80 kN horizontal at B.] ## Short answer Fixed support A prevents sway; the 80 kN horizontal load causes no bending. Overhang DE gives fixed moment -120 kNm at D. Moment distribution at B, C, D (DFs given) yields final moments (kNm): AB 4.41, BA 8.82, BC -26.45, CB 17.26, CD -4.35, DC 120.00, DE -120.00, BG 17.63, GB 8.82, CF -12.90, FC -6.45. ## Answer **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) | | AB | BA | BC | CB | CD | DC | DE | BG | GB | CF | FC | |---|---|---|---|---|---|---|---|---|---|---|---| | DF | - | 0.222 | 0.333 | 0.333 | 0.444 | 1.000 | 0.000 | 0.444 | - | 0.222 | - | | FEM | 0.00 | 0.00 | -30.00 | 30.00 | -30.00 | 30.00 | -120.00 | 0.00 | 0.00 | 0.00 | 0.00 | | Balance | 0.00 | 6.67 | 10.00 | 0.00 | 0.00 | 90.00 | 0.00 | 13.33 | 0.00 | 0.00 | 0.00 | | Carry-over | 3.33 | 0.00 | 0.00 | 5.00 | 45.00 | 0.00 | 0.00 | 0.00 | 6.67 | 0.00 | 0.00 | | Balance | 0.00 | 0.00 | 0.00 | -16.67 | -22.22 | 0.00 | 0.00 | 0.00 | 0.00 | -11.11 | 0.00 | | Carry-over | 0.00 | 0.00 | -8.33 | 0.00 | 0.00 | -11.11 | 0.00 | 0.00 | 0.00 | 0.00 | -5.56 | | Balance | 0.00 | 1.85 | 2.78 | 0.00 | 0.00 | 11.11 | 0.00 | 3.70 | 0.00 | 0.00 | 0.00 | | Carry-over | 0.93 | 0.00 | 0.00 | 1.39 | 5.56 | 0.00 | 0.00 | 0.00 | 1.85 | 0.00 | 0.00 | | Balance | 0.00 | 0.00 | 0.00 | -2.31 | -3.09 | 0.00 | 0.00 | 0.00 | 0.00 | -1.54 | 0.00 | | Carry-over | 0.00 | 0.00 | -1.16 | 0.00 | 0.00 | -1.54 | 0.00 | 0.00 | 0.00 | 0.00 | -0.77 | | Balance | 0.00 | 0.26 | 0.39 | 0.00 | 0.00 | 1.54 | 0.00 | 0.51 | 0.00 | 0.00 | 0.00 | | Carry-over | 0.13 | 0.00 | 0.00 | 0.19 | 0.77 | 0.00 | 0.00 | 0.00 | 0.26 | 0.00 | 0.00 | | Further cycles (converged) | 0.02 | 0.04 | -0.12 | -0.34 | -0.37 | 0.00 | 0.00 | 0.08 | 0.04 | -0.25 | -0.12 | | **Final M** | **4.41** | **8.82** | **-26.45** | **17.26** | **-4.35** | **120.00** | **-120.00** | **17.63** | **8.82** | **-12.90** | **-6.45** | #### 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: $120.00+(-120.00)=0.00$ ✓. | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | AB | 4.41 | -8.82 | - | | BC | -26.45 | -17.26 | 23.26 at 3.15 m | | CD | -4.35 | -120.00 | 1.40 at 1.07 m | | DE | -120.00 | 0.00 | - | | BG | 17.63 | -8.82 | - | | CF | -12.90 | 6.45 | - | 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the 80 kN horizontal load at B cause no bending moment? A. Because it passes through the centroid of the column B. Because it acts at the fixed support C. Because the columns are infinitely stiff D. Because the girder is held by fixed support A and the load goes to A as axial force in the girder Answer: D. A prevents horizontal movement of the girder, so there is no sway and the 80 kN is carried as axial force. Only a non-sway distribution is needed. (Key point: No sway) 2. What is the moment at D from the overhang DE (2 m, 60 kN at E), and what is the DF of DE? A. 30 kNm; DF = 0.5 B. 120 kNm hogging; DF = 0, so DF of DC = 1 C. 60 kNm; DF = 0 D. 120 kNm; DF = 1 Answer: B. M_D = 60 x 2 = 120 kNm, a known fixed moment. It takes no part in the distribution, so DC takes the whole balancing moment (DF = 1). (Key point: Overhang moment) 3. At B, k_BA = I/6, k_BC = 1.5I/6 and k_BG = 2I/6. What is DF_BG? A. 0.4444 B. 0.3333 C. 0.2222 D. 0.5 Answer: A. The total is 0.1667 + 0.25 + 0.3333 = 0.75, so DF_BG = 0.3333/0.75 = 0.4444. (Key point: Distribution factor at B) 4. FEM_DC = +30 and FEM_DE = -120 at D. What balancing moment is applied to DC? A. -90 kNm B. +150 kNm C. +120 kNm D. +90 kNm Answer: D. The unbalanced moment is 30 - 120 = -90 kNm. The balancing moment is +90 kNm, all taken by DC since its DF is 1. (Key point: First balance at D) ## Related in Theory of Structures II - [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 - [Moment distribution: frame, 150 kN, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-150kn-50kn-60knm) — Theory of Structures II, chapter 3 - [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 - [Force method: frame with 60 kN/m, BMD](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aforce-method-frame-60knm-80kn-2i) — Theory of Structures II, chapter 2 - [Moment distribution: frame with 100 kN at D](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-50knm-100kn-20kn) — Theory of Structures II, chapter 3 - [Moment distribution: frame with cantilever 15 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Amoment-dist-frame-50knm-15kn-cantilever) — Theory of Structures II, chapter 3 ## Related in other subjects - [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 60 kN/m UDL](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aframe-afd-sfd-bmd-60knm-hinge) — Applied Mechanics, chapter 6 - [Frame AFD, SFD, BMD (10 kN/m, 60 kN, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-10knm-60kn) — Strength of Materials, chapter 1 - [Frame AFD, SFD, BMD (80 kN, 30 kNm, hinge)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aframe-afd-sfd-bmd-80kn-30knm) — Strength of Materials, chapter 1