--- title: "Moment distribution: frame with cantilever" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:moment-dist-frame-3kn-15knm-3ei" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/moment-dist-frame-3kn-15knm-3ei site: BCT Engineering --- # Moment distribution: frame with cantilever Theory of Structures II · Chapter 3: Displacement method - Asked in: 2076 Chaitra, 12 marks - Video: [SA40: Moment Distribution Method: Frames (No Sidesway)](https://www.youtube.com/watch?v=BjbVlb6HvMc) — Dr. Structure - 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-3kn-15knm-3ei - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Draw BMD of the given frame using the moment distribution method. [Figure: horizontal member F-E-D-C with 3 kN at the end F, 15 kN/m UDL on ED, spans 1.5 m, 4 m and 1.5 m, members 3EI, C roller; column EA (2EI) fixed at A, 2.5 m + 1.5 m heights; column DB (EI) fixed at B.] ## Short answer Sway frame analyzed by moment distribution in two cases: non-sway plus trial sway. Final member-end moments: EF=4.50, ED=-10.26, DE=19.22, DC=-15.08, EA=5.76, AE=1.26, DB=-4.14, BD=-2.88 kNm; C end moment is zero. ## Answer **Assumptions:** the girder F-E-D-C has a 1.5 m cantilever FE (3EI) with 3 kN at F, a 4 m span ED (3EI) with 15 kN/m, and a 1.5 m span DC (3EI) ending on a roller at C. Columns EA (2EI) and DB (EI) are each 4 m high and fixed at their bases. The roller at C allows horizontal movement, so the frame can **sway**; the moment distribution is done in two cases (sway prevented, then sway induced). #### Step 1: Overhang FE $M_{EF}=3\times1.5=4.5$ kNm (hogging), fixed moment on joint E. #### Step 2: Distribution factors - Joint E: relative stiffness $\Sigma k$ = 1.2500; EF: overhang, DF = 0 (its moment is a known fixed value); ED: $k=EI/L$ = 0.7500, DF = 0.6000; EA: $k=EI/L$ = 0.5000, DF = 0.4000 - Joint D: relative stiffness $\Sigma k$ = 2.5000; DE: $k=EI/L$ = 0.7500, DF = 0.3000; DC: $k=EI/L$ = 1.5000 (far end hinged: 3/4 factor), DF = 0.6000; DB: $k=EI/L$ = 0.2500, DF = 0.1000 (Roller end C: member DC is stiffened by 3/4 because its far end is hinged and the end moment at C is zero.) #### Step 3: Case I, sway prevented by an imaginary horizontal support at E Fixed-end moments from the loads: $\mp15\times4^2/12=\mp20$ kNm on ED. | | EF | ED | DE | DC | EA | AE | DB | BD | |---|---|---|---|---|---|---|---|---| | DF | 0.000 | 0.600 | 0.300 | 0.600 | 0.400 | - | 0.100 | - | | FEM | 4.50 | -20.00 | 20.00 | 0.00 | 0.00 | 0.00 | 0.00 | 0.00 | | Balance | 0.00 | 9.30 | -6.00 | -12.00 | 6.20 | 0.00 | -2.00 | 0.00 | | Carry-over | 0.00 | -3.00 | 4.65 | 0.00 | 0.00 | 3.10 | 0.00 | -1.00 | | Balance | 0.00 | 1.80 | -1.40 | -2.79 | 1.20 | 0.00 | -0.47 | 0.00 | | Carry-over | 0.00 | -0.70 | 0.90 | 0.00 | 0.00 | 0.60 | 0.00 | -0.23 | | Balance | 0.00 | 0.42 | -0.27 | -0.54 | 0.28 | 0.00 | -0.09 | 0.00 | | Carry-over | 0.00 | -0.14 | 0.21 | 0.00 | 0.00 | 0.14 | 0.00 | -0.05 | | Balance | 0.00 | 0.08 | -0.06 | -0.13 | 0.05 | 0.00 | -0.02 | 0.00 | | Carry-over | 0.00 | -0.03 | 0.04 | 0.00 | 0.00 | 0.03 | 0.00 | -0.01 | | Further cycles (converged) | 0.00 | 0.02 | 0.00 | -0.03 | 0.02 | 0.01 | -0.01 | 0.00 | | **Final M** | **4.50** | **-12.25** | **18.07** | **-15.49** | **7.75** | **3.87** | **-2.58** | **-1.29** | Reaction of the imaginary support (from the column shears): $R_1=-1.94$ kN (acting to the left). #### Step 4: Case II, sway $\Delta_0$ without external loads Give the girder a trial sway to the right and choose its size so that the fixed-end moment in column EA is $-100$ kNm at both ends. Since $FEM=-6EI\Delta/L^2$ is proportional to $EI/L^2$ and both columns have the same height and the same $\Delta$, the column DB (EI) gets half of this: $-50$ kNm at both ends. All other members have zero FEM. | | EF | ED | DE | DC | EA | AE | DB | BD | |---|---|---|---|---|---|---|---|---| | DF | 0.000 | 0.600 | 0.300 | 0.600 | 0.400 | - | 0.100 | - | | FEM | 0.00 | 0.00 | 0.00 | 0.00 | -100.00 | -100.00 | -50.00 | -50.00 | | Balance | 0.00 | 60.00 | 15.00 | 30.00 | 40.00 | 0.00 | 5.00 | 0.00 | | Carry-over | 0.00 | 7.50 | 30.00 | 0.00 | 0.00 | 20.00 | 0.00 | 2.50 | | Balance | 0.00 | -4.50 | -9.00 | -18.00 | -3.00 | 0.00 | -3.00 | 0.00 | | Carry-over | 0.00 | -4.50 | -2.25 | 0.00 | 0.00 | -1.50 | 0.00 | -1.50 | | Balance | 0.00 | 2.70 | 0.67 | 1.35 | 1.80 | 0.00 | 0.22 | 0.00 | | Carry-over | 0.00 | 0.34 | 1.35 | 0.00 | 0.00 | 0.90 | 0.00 | 0.11 | | Balance | 0.00 | -0.20 | -0.40 | -0.81 | -0.13 | 0.00 | -0.13 | 0.00 | | Carry-over | 0.00 | -0.20 | -0.10 | 0.00 | 0.00 | -0.07 | 0.00 | -0.07 | | Further cycles (converged) | 0.00 | 0.12 | 0.07 | 0.03 | 0.08 | 0.04 | 0.00 | 0.00 | | **Final M** | **0.00** | **61.26** | **35.34** | **12.57** | **-61.26** | **-80.63** | **-47.91** | **-48.95** | Reaction of the imaginary support: $R_2=59.69$ kN. #### Step 5: Correction for sway The imaginary support does not exist, so the restraint force must vanish: $R_1+k\,R_2=0$ $$ k=-\frac{R_1}{R_2}=-\frac{-1.94}{59.69}=0.0325 $$ Final moments $M=M_I+kM_{II}$ (kNm): | End | Case I | k × Case II | Final | |---|---|---|---| | EF | 4.50 | 0.00 | 4.50 | | EA | 7.75 | -1.99 | 5.76 | | ED | -12.25 | 1.99 | -10.26 | | DE | 18.07 | 1.15 | 19.22 | | DB | -2.58 | -1.56 | -4.14 | | DC | -15.49 | 0.41 | -15.08 | Support moments: $M_{AE}=1.26$, $M_{BD}=-2.88$ kNm. Joint E check: $M_{ED}+M_{EA}+M_{EF}=0.00$ ✓. Joint D check: $M_{DE}+M_{DB}+M_{DC}=0.00$ ✓. $M_{CD}=0$. #### Bending moment ordinates (kNm) | Member | BM at first end | BM at second end | Max within member | |---|---|---|---| | FE | 0.00 | -4.50 | - | | ED | -10.26 | -19.22 | 15.43 at 1.85 m | | DC | -15.08 | 0.00 | - | | EA | 5.76 | -1.26 | - | | DB | -4.14 | 2.88 | - | **Answer: $M_{EA}=5.76$, $M_{AE}=1.26$, $M_{ED}=-10.26$, $M_{DE}=19.22$, $M_{DB}=-4.14$, $M_{BD}=-2.88$, $M_{DC}=-15.08$ kNm (clockwise end moments +); $M_{EF}=4.50$ kNm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the moment distribution done in two cases (sway prevented, then sway induced)? A. The UDL is unsymmetrical B. The roller at C lets the frame sway C. Column stiffnesses differ D. The overhang is statically indeterminate Answer: B. The roller allows horizontal movement, so a sway correction is needed on top of the no-sway analysis. (Key point: Sway case) 2. How is the overhang FE treated at joint E? A. It is distributed with DF = 0.5 B. It is ignored C. Its moment 3 x 1.5 = 4.5 kNm is a known fixed value with distribution factor 0 D. It is carried over to F Answer: C. A cantilever moment is statically determinate, so it is applied to joint E as known and takes no distribution. (Key point: Overhang) 3. Why is DC's stiffness multiplied by 3/4? A. DC has the largest EI B. C is fixed C. DC carries the UDL D. Its far end C is a roller/hinge with zero moment Answer: D. With the far end hinged, the member is stiffer by 3EI/L instead of 4EI/L and no carry-over occurs. (Key point: Hinged far end) 4. In case II, why does column DB get -50 when EA gets -100 for the same Δ? A. FEM = -6EIΔ/L^2 is proportional to EI, and DB's EI is half of EA's B. DB is shorter C. DB is fixed D. DB is hinged Answer: A. The columns are of equal height and have the same Δ, so FEM scales with EI: EI against 2EI. 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