--- title: "Moment distribution: frame with 100 kN at D" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:moment-dist-frame-50knm-100kn-20kn" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-50knm-100kn-20kn site: BCT Engineering --- # Moment distribution: frame with 100 kN at D Theory of Structures II · Chapter 3: Displacement method - Asked in: 2070 Asar, 10 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-50knm-100kn-20kn - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Use the moment distribution method to analyse the frame loaded as shown in figure below. Also draw bending moment diagram. [Figure: frame; beam B-C-D with 50 kN/m UDL on BC (8 m), 100 kN at the point beyond C, D (2.5 m + 2.5 m) hinged; column AB of 4 m with 20 kN horizontal at B, A fixed; column CE of 6 m, E fixed.] ## Short answer The frame does not sway because hinge D prevents horizontal movement; the 20 kN load is axial. Moment distribution gives M_BC=-198.78, M_CB=253.36, M_CD=-169.35, M_BA=198.78, M_AB=99.39, M_CE=-84.00, M_EC=-42.00 kNm. ## Answer **Assumptions:** the girder B-C-D has BC = 8 m with 50 kN/m and CD = 5 m with 100 kN at its mid-point (2.5 m + 2.5 m); D is a hinge (the support prevents horizontal movement). Columns BA (4 m) and CE (6 m) are fixed at A and E. The 20 kN horizontal load acts at B (to the right). All members have the same EI. Because the hinge D prevents horizontal movement of the girder, the frame **does not sway**; the 20 kN load is carried by axial force in the girder and shared by the supports; it does not change the bending moments. #### Fixed-end moments (clockwise +) - BC: $\mp\dfrac{50\times8^2}{12}=\mp266.67$ kNm - CD: $\mp\dfrac{100\times5}{8}=\mp62.5$ kNm; D is hinged, so the modified FEM at C is $-62.5-\tfrac12(62.5)=-93.75$ kNm and the end moment at D is zero. #### Stiffness and distribution factors ($k=EI/L$; CD uses 3/4 because D is hinged) - Joint B: relative stiffness $\Sigma k$ = 0.3750; BC: $k=EI/L$ = 0.1250, DF = 0.3333; BA: $k=EI/L$ = 0.2500, DF = 0.6667 - Joint C: relative stiffness $\Sigma k$ = 0.4417; CB: $k=EI/L$ = 0.1250, DF = 0.2830; CD: $k=EI/L$ = 0.1500 (far end hinged: 3/4 factor), DF = 0.3396; CE: $k=EI/L$ = 0.1667, DF = 0.3774 #### Moment distribution (kNm) | | BC | CB | CD | BA | AB | CE | EC | |---|---|---|---|---|---|---|---| | DF | 0.333 | 0.283 | 0.340 | 0.667 | - | 0.377 | - | | FEM | -266.67 | 266.67 | -93.75 | 0.00 | 0.00 | 0.00 | 0.00 | | Balance | 88.89 | -48.94 | -58.73 | 177.78 | 0.00 | -65.25 | 0.00 | | Carry-over | -24.47 | 44.44 | 0.00 | 0.00 | 88.89 | 0.00 | -32.63 | | Balance | 8.16 | -12.58 | -15.09 | 16.31 | 0.00 | -16.77 | 0.00 | | Carry-over | -6.29 | 4.08 | 0.00 | 0.00 | 8.16 | 0.00 | -8.39 | | Balance | 2.10 | -1.15 | -1.39 | 4.19 | 0.00 | -1.54 | 0.00 | | Carry-over | -0.58 | 1.05 | 0.00 | 0.00 | 2.10 | 0.00 | -0.77 | | Balance | 0.19 | -0.30 | -0.36 | 0.38 | 0.00 | -0.40 | 0.00 | | Carry-over | -0.15 | 0.10 | 0.00 | 0.00 | 0.19 | 0.00 | -0.20 | | Balance | 0.05 | -0.03 | -0.03 | 0.10 | 0.00 | -0.04 | 0.00 | | Carry-over | -0.01 | 0.02 | 0.00 | 0.00 | 0.05 | 0.00 | -0.02 | | Further cycles (converged) | 0.00 | 0.00 | -0.01 | 0.01 | 0.01 | -0.01 | -0.01 | | **Final M** | **-198.78** | **253.36** | **-169.35** | **198.78** | **99.39** | **-84.00** | **-42.00** | Joint B: $-198.78+(198.78)=0.00$ ✓; Joint C: $253.36+(-169.35)+(-84.00)=0.00$ ✓; hinge D: $M_{DC}=0$. #### Bending moment diagram | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | BC | -198.78 | -253.36 | 174.40 at 3.86 m | | CD | -169.35 | 0.00 | 40.32 at 2.50 m | | BA | 198.78 | -99.39 | - | | CE | -84.00 | 42.00 | - | BM under the 100 kN load: 40.32 kNm. BM sign: sagging positive for the girder; for the columns positive = tension on the right-hand side when travelling from the girder end towards the base as listed. Reactions: A: $H=74.54$, $V=193.18$, $M=-99.39$; E: $H=-21.00$, $V=290.69$, $M=42.00$; D: $H=-73.54$, $V=16.13$, $M=0.00$. **Answer: $M_{BC}=-198.78$, $M_{CB}=253.36$, $M_{CD}=-169.35$, $M_{BA}=198.78$, $M_{AB}=99.39$, $M_{CE}=-84.00$, $M_{EC}=-42.00$ kNm (end moments in the order of the member names; clockwise +).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the 20 kN horizontal load at B not change the bending moments? A. Hinge D prevents horizontal movement, so there is no sway and the load goes into axial force and support reactions B. It is cancelled by the 100 kN load C. It acts at a fixed joint where it produces no moment D. It is applied to a column with infinite stiffness Answer: A. The girder cannot translate because D restrains it, so no sway occurs and the 20 kN is carried as axial force/reactions. (Key point: Why no sway) 2. For span CD with hinge D, what is the modified FEM at C? A. -62.5 - 1/2(62.5) = -93.75 kNm B. -62.5 kNm C. -31.25 kNm D. -125 kNm Answer: A. Releasing the hinge at D adds half of the FEM at D to the FEM at C, giving -93.75 kNm. (Key point: Modified FEM at hinged end) 3. How is the stiffness of CD treated in the DFs at C? A. 3/4 of EI/L since far end D is hinged B. Same as EI/L C. Twice EI/L D. Zero Answer: A. A hinged far end uses 3EI/L, i.e. 3/4 of EI/L, so k_CD = 0.15EI. (Key point: Stiffness of CD) 4. At joint B, BA (4 m column) and BC (8 m girder) have the same EI. What are the DFs? A. BA 0.6667 and BC 0.3333 B. BA 0.3333 and BC 0.6667 C. Both 0.5 D. BA 1 and BC 0 Answer: A. k = EI/L gives 0.25 for BA and 0.125 for BC, so DFs are 0.25/0.375 = 0.667 and 0.333. 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