--- title: "Principle of moment distribution method" subject: "Theory of Structures II" chapter: 3 key: "semester_5/theory_of_structures_ii:principle-moment-distribution" source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/principle-moment-distribution site: BCT Engineering --- # Principle of moment distribution method Theory of Structures II · Chapter 3: Displacement method - Asked in: 2079 Baishakh, 4 marks; 2075 Chaitra, 6 marks; 2074 Asoj, 6 marks (asked 3 times) - Video: [Explaining the Moment Distribution Method - Structural Analysis](https://www.youtube.com/watch?v=E3zLTnc_nlk) — structurefree - 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/principle-moment-distribution - Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-3 ## Question Explain the principle of the moment distribution method with a simple example, including the concept of distribution and carry over factors. ## Short answer Joints are first locked, fixed-end moments computed, then sequentially released. Unbalanced moment is distributed to members in proportion to their stiffness (DF = K/ΣK, K = 4EI/L for far end fixed). Half of each distributed moment is carried over to the far end (COF = 0.5). Iteration continues until moments converge. ## Answer #### Principle The moment distribution method (Hardy Cross, 1930) is an iterative method, based on the slope-deflection equations, for analysing continuous beams and frames without solving simultaneous equations. The idea is: 1. **Lock** every joint against rotation. Each loaded member then behaves as a fixed-ended beam and develops fixed-end moments (FEM). At a joint these are generally unequal, so there is an **unbalanced moment** = $\sum$FEM of the members meeting at the joint. 2. **Release** one joint at a time. It rotates until the joint is in equilibrium. The balancing moment (equal to the unbalanced moment with reversed sign) is **distributed** to the members in proportion to their stiffness. 3. A part of each distributed moment is **carried over** to the far end of the member (if the far end is fixed against rotation). 4. The carried-over moments disturb the neighbouring joints, so the balancing and carry-over are repeated until the carry-over moments are negligibly small. 5. The final end moment = FEM + all distributed moments + all carried-over moments. Equilibrium of every joint and slope compatibility of the members are therefore satisfied when the iteration converges. #### Distribution factor (DF) The fraction of the unbalanced moment taken by a member at a joint: $$ DF_{jk}=\frac{K_{jk}}{\sum K},\qquad K=\frac{4EI}{L}\ (\text{far end fixed}),\quad K=\frac{3EI}{L}\ (\text{far end hinged/roller}) $$ $\sum DF=1$ at every joint. For a fixed support $DF=0$ (it takes no rotation); for a simple end support with no member beyond it, $DF=1$. #### Carry-over factor (COF) A moment $M$ applied at the near end of a member whose far end is **fixed** produces $M/2$ at the far end, so $COF=\tfrac12$. It is 0 if the far end is hinged or roller (no moment can develop there). #### Example Beam ABC: AB = 6 m, BC = 4 m, EI constant; A is fixed, B and C are simple supports; AB carries 10 kN/m. Since the far end C is a simple (hinged) support, the stiffness of BC is reduced by the factor 3/4. **Distribution factors at B:** $K_{BA}=\dfrac{EI}{6}=0.1667EI$, $K_{BC}=\dfrac{3}{4}\cdot\dfrac{EI}{4}=0.1875EI$ (relative values). $DF_{BA}=0.471$, $DF_{BC}=0.529$. **Fixed-end moments:** $FEM_{AB}=-\dfrac{wL^2}{12}=-\dfrac{10\times6^2}{12}=-30$ kNm, $FEM_{BA}=+30$ kNm; BC has no load. (C is hinged: $FEM_{BC}=FEM_{CB}=0$.) **Distribution:** - Unbalanced moment at B = $+30$ kNm; balancing moment = $-30$ kNm. - $BA$ takes $-30\times0.471=-14.12$; $BC$ takes $-30\times0.529=-15.88$. - Carry-over to A: $\tfrac12\times(-14.12)=-7.06$. No carry-over to C (hinged). - Joint B is now balanced, and A (fixed) and C (hinged, already at zero moment) need no balancing, so the process ends in one cycle. | | AB | BA | BC | |---|---|---|---| | DF | - | 0.471 | 0.529 | | FEM | -30.00 | +30.00 | 0.00 | | Balance | | -14.12 | -15.88 | | Carry-over | -7.06 | | | | **Final M** | **-37.06** | **15.88** | **-15.88** | Joint check: $M_{BA}+M_{BC}=0.00$ ✓. The final moments equal the exact slope-deflection result, hence the method gives exact answers for this beam (several cycles are needed only when there are more joints to be balanced). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the distribution factor of a member at a joint defined, and how is a far end hinged treated? A. DF = K/sum K, with K = 4EI/L for a fixed far end and 3EI/L for a hinged far end B. DF = L/sum L, with no change for a hinged end C. DF = K/sum K, with K = 4EI/L for a hinged far end D. DF = 1/2 for every member Answer: A. Distribution is in proportion to stiffness. A hinged far end reduces the stiffness to 3EI/L (3/4 of the fixed value), and the DFs at a joint sum to 1. (Key point: Distribution factor) 2. What is the carry-over factor, and when is it zero? A. 1; never zero B. 1/2 when the far end is fixed; zero when it is hinged or a roller C. 1/2 when the far end is hinged; zero when it is fixed D. 1/4 for all members Answer: B. A moment M at the near end produces M/2 at a fixed far end. A hinge or roller cannot develop moment, so the carry-over is zero. (Key point: Carry-over factor) 3. At joint B of the example the unbalanced moment is +30 kNm. What is distributed to the members? A. +30 kNm in proportion to the stiffness B. -30 kNm shared equally C. -30 kNm shared in proportion to the DFs, giving -14.12 to BA and -15.88 to BC D. -30 kNm all to BA Answer: C. The balancing moment is the unbalanced moment with reversed sign, shared by DF 0.471 and 0.529. 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