--- title: "Moment distribution: frame, 150 kN, 60 kN/m" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:moment-dist-frame-150kn-50kn-60knm" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-150kn-50kn-60knm site: BCT Engineering --- # Moment distribution: frame, 150 kN, 60 kN/m Theory of Structures II · Chapter 3: Displacement method - Asked in: 2072 Kartik, 15 marks - Video: [Moment Distribution Method easy and Quick Excel Example 12-2 Hibbeler](https://www.youtube.com/watch?v=arlAdtg4P8g) — Noman Khan - 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-150kn-50kn-60knm - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Analyse the frame shown in figure below by using the moment distribution method. [Figure: frame; column AB (2I) of 10 m (8 m + 2 m marks), A fixed; 150 kN horizontal at B; beam B-C-D with 50 kN loads at 2 m from B and at 4 m, spans 2 m + 4 m + 2 m; BC (2I) and C-D (I) of 5 m with 60 kN/m UDL, D roller; column CE (I) of 6 m, E fixed.] ## Answer **Assumptions (figure partly illegible):** column AB (2I) is 10 m high, fixed at A; the girder is B-C-D with BC = 4 m (2I) carrying a 50 kN load at mid-span (2 m from B), CD = 5 m (I) carrying 60 kN/m, and D on a roller; column CE (I, 6 m) runs down from C to its fixed base E. A second 50 kN load acts vertically at C (it passes through the axis of column CE and produces no bending moment). The 150 kN load at B acts to the right. The roller at D allows horizontal movement, so the frame sways; the moment distribution is done for the non-sway case and the sway case. #### Distribution factors ($k=EI/L$; the hinged-end member CD gets the factor 3/4) - Joint B: relative stiffness $\Sigma k$ = 0.7000; BA: $k=EI/L$ = 0.2000, DF = 0.2857; BC: $k=EI/L$ = 0.5000, DF = 0.7143 - Joint C: relative stiffness $\Sigma k$ = 0.8167; CB: $k=EI/L$ = 0.5000, DF = 0.6122; CD: $k=EI/L$ = 0.1500 (far end hinged: 3/4 factor), DF = 0.1837; CE: $k=EI/L$ = 0.1667, DF = 0.2041 #### Case I: sway prevented by an imaginary support at B Fixed-end moments: BC $\mp\dfrac{50\times4}{8}=\mp25$ kNm; CD $\mp\dfrac{60\times5^2}{12}=\mp125$ kNm (modified at C for the hinge at D: -187.50 kNm). | | AB | BA | BC | CB | CD | CE | EC | |---|---|---|---|---|---|---|---| | DF | - | 0.286 | 0.714 | 0.612 | 0.184 | 0.204 | - | | FEM | 0.00 | 0.00 | -25.00 | 25.00 | -187.50 | 0.00 | 0.00 | | Balance | 0.00 | 7.14 | 17.86 | 99.49 | 29.85 | 33.16 | 0.00 | | Carry-over | 3.57 | 0.00 | 49.74 | 8.93 | 0.00 | 0.00 | 16.58 | | Balance | 0.00 | -14.21 | -35.53 | -5.47 | -1.64 | -1.82 | 0.00 | | Carry-over | -7.11 | 0.00 | -2.73 | -17.77 | 0.00 | 0.00 | -0.91 | | Balance | 0.00 | 0.78 | 1.95 | 10.88 | 3.26 | 3.63 | 0.00 | | Carry-over | 0.39 | 0.00 | 5.44 | 0.98 | 0.00 | 0.00 | 1.81 | | Balance | 0.00 | -1.55 | -3.88 | -0.60 | -0.18 | -0.20 | 0.00 | | Carry-over | -0.78 | 0.00 | -0.30 | -1.94 | 0.00 | 0.00 | -0.10 | | Further cycles (converged) | -0.05 | -0.09 | 0.39 | 1.14 | 0.38 | 0.42 | 0.21 | | **Final M** | **-3.97** | **-7.94** | **7.94** | **120.64** | **-155.83** | **35.19** | **17.59** | The horizontal reaction of the imaginary support (positive to the right) is $R_1=-157.61$ kN; this is the force that the imaginary support must supply to prevent sway and that the sway correction has to remove. #### Case II: sway $\Delta$ with no loads Give the girder a sway to the right such that the fixed-end moment in AB is $-100$ kNm at each end. Since $FEM\propto EI/L^2$ and both columns move equally, column CE (EI, 6 m) gets $-100\times\dfrac{(1/36)}{(2/100)}=-138.89$ kNm. | | AB | BA | BC | CB | CD | CE | EC | |---|---|---|---|---|---|---|---| | DF | - | 0.286 | 0.714 | 0.612 | 0.184 | 0.204 | - | | FEM | -100.00 | -100.00 | 0.00 | 0.00 | 0.00 | -138.89 | -138.89 | | Balance | 0.00 | 28.57 | 71.43 | 85.03 | 25.51 | 28.34 | 0.00 | | Carry-over | 14.29 | 0.00 | 42.52 | 35.71 | 0.00 | 0.00 | 14.17 | | Balance | 0.00 | -12.15 | -30.37 | -21.87 | -6.56 | -7.29 | 0.00 | | Carry-over | -6.07 | 0.00 | -10.93 | -15.18 | 0.00 | 0.00 | -3.64 | | Balance | 0.00 | 3.12 | 7.81 | 9.30 | 2.79 | 3.10 | 0.00 | | Carry-over | 1.56 | 0.00 | 4.65 | 3.90 | 0.00 | 0.00 | 1.55 | | Balance | 0.00 | -1.33 | -3.32 | -2.39 | -0.72 | -0.80 | 0.00 | | Carry-over | -0.66 | 0.00 | -1.20 | -1.66 | 0.00 | 0.00 | -0.40 | | Further cycles (converged) | 0.11 | 0.22 | 0.97 | 1.12 | 0.25 | 0.28 | 0.14 | | **Final M** | **-90.78** | **-81.56** | **81.56** | **93.97** | **21.28** | **-115.25** | **-127.07** | The horizontal reaction of the imaginary support is $R_2=57.62$ kN. #### Combination $R_1+kR_2=0\Rightarrow k=-\dfrac{-157.61}{57.62}=2.7353$ | End | Case I | k × Case II | Final M (kNm) | |---|---|---|---| | AB | -3.97 | -248.31 | -252.28 | | BA | -7.94 | -223.09 | -231.03 | | BC | 7.94 | 223.09 | 231.03 | | CB | 120.64 | 257.04 | 377.68 | | CD | -155.83 | 58.20 | -97.63 | | CE | 35.19 | -315.23 | -280.05 | | EC | 17.59 | -347.57 | -329.97 | Checks: Joint B: $0.00$; Joint C: $0.00$; hinge D: $M_{DC}=0$. | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | AB | -252.28 | 231.03 | - | | BC | 231.03 | -377.68 | - | | CD | -97.63 | 0.00 | 141.86 at 2.83 m | | CE | -280.05 | 329.97 | - | **Answer: $M_{AB}=-252.28$, $M_{BA}=-231.03$, $M_{BC}=231.03$, $M_{CB}=377.68$, $M_{CD}=-97.63$, $M_{CE}=-280.05$, $M_{EC}=-329.97$ kNm (clockwise end moments +).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is CD given a factor 3/4 in the distribution factors? A. CD is shorter B. CD has no load C. Its far end D is a roller (hinged), so stiffness is 3EI/L D. D is fixed Answer: C. With a hinged far end, the stiffness is 3EI/L = 3/4 of 4EI/L. (Key point: Modified stiffness) 2. Why are two cases (I and II) analysed? A. Two loads B. Two supports C. To check symmetry D. The frame sways: non-sway case with imaginary support, then sway case Answer: D. Sway is first prevented by an imaginary support, then its reaction is removed by a sway analysis. (Key point: Two cases) 3. How is the sway factor k found? A. R1 + k R2 = 0 B. R1 = R2 C. k = R1 x R2 D. k from DF Answer: A. The imaginary support force must vanish in the final result, giving k = 2.7353. (Key point: Combination) 4. Why is FEM in CE -138.89 when AB has -100? A. Random B. FEM is proportional to EI/L^2 with same sway C. Equal stiffness D. CE is longer so smaller Answer: B. -100 x (1/36)/(2/100) = -138.89, same sway Delta for both columns. 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