--- title: "Moment area vs conjugate beam method" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:moment-area-vs-conjugate" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-vs-conjugate site: BCT Engineering --- # Moment area vs conjugate beam method Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2069 Bhadra, 4 marks - Video: [Moment Area Method and Conjugate Beam method](https://www.youtube.com/watch?v=2PBIHYGxtWA) — Sumit Paudel - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-vs-conjugate - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Explain the difference between moment area method and conjugate beam method with suitable examples. ## Short answer Moment area method applies two theorems directly on the real beam using tangents, best for cantilevers. Conjugate beam method replaces the real beam with a fictitious beam having modified supports, then uses statics to find shear (slope) and moment (deflection), convenient for simply supported and multiple-point problems. Both yield identical results, e.g., cantilever end deflection PL^3/(3EI) or simply supported central load deflection PL^3/(48EI). ## Answer Both methods use the $M/EI$ diagram of the beam to find slope and deflection, but they apply it in different ways. | Point | Moment-area method | Conjugate beam method | |---|---|---| | Basis | Two theorems (area and moment of $M/EI$ area) | Analogy between $M/EI$ loading and shear and moment | | Slope | Change in slope = area of $M/EI$ | Slope = shear in conjugate beam | | Deflection | Tangent deviation = moment of $M/EI$ area | Deflection = moment in conjugate beam | | Beam used | Real beam with tangents | Fictitious (conjugate) beam | | Supports | No change | Changed (fixed becomes free, interior support becomes hinge, etc.) | | Procedure | Needs reference tangent (horizontal at fixed end or found from deviations) | Routine statics: reactions, shear, moment | | Best for | Cantilevers and quick single-point answers | Simply supported and overhanging beams, many points | | Variable EI | Possible, but areas get complex | Convenient: loads become $M/EI$ with different EI per segment | #### Example 1: moment-area (cantilever) Cantilever with end load $P$: $\Delta_B = \dfrac{PL^2}{2EI}\times\dfrac{2L}{3} = \dfrac{PL^3}{3EI}$ (tangent horizontal at the fixed end). #### Example 2: conjugate beam (simply supported beam, central load) The conjugate beam is simply supported with a triangular load of peak $\dfrac{PL}{4EI}$. The conjugate reaction (the slope at the support) is $\dfrac{PL^2}{16EI}$ and the conjugate moment at mid-span is $\dfrac{PL^3}{48EI}$ (the deflection). Both methods give the same results. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is slope obtained in the conjugate beam method compared with the moment-area method? A. Conjugate: moment of the M/EI area; moment-area: shear B. Both use the moment of the M/EI area C. Conjugate: shear in the conjugate beam; moment-area: change in slope = area of M/EI D. Conjugate: area of the shear diagram; moment-area: tangent deviation Answer: C. In the conjugate method slope = shear in the conjugate beam. In the moment-area method the change in slope between two points equals the area of the M/EI diagram. (Key point: How slope is obtained) 2. Which statement about supports is correct? A. Moment-area changes supports; conjugate keeps the real supports B. Neither method changes the supports C. Both methods convert fixed ends to free ends D. Conjugate changes supports (fixed becomes free, interior support becomes hinge); moment-area keeps the real beam Answer: D. The conjugate beam is a fictitious beam with modified supports, while the moment-area method works on the real beam using tangents. (Key point: Support changes) 3. For a cantilever with end load P, why can the moment-area method give the tip deflection directly? A. The tangent at the fixed end is horizontal, so the deviation equals the deflection: PL^3/(3EI) B. The M/EI diagram is rectangular C. The shear is constant along the beam D. The conjugate beam is simply supported Answer: A. At the fixed end the tangent stays horizontal, so the tangent deviation (area PL^2/(2EI) times 2L/3) is the deflection PL^3/(3EI). (Key point: Moment-area cantilever example) 4. For a simply supported beam with a central load P, what loads the conjugate beam and what gives the mid-span deflection? A. A uniform load PL/4EI; deflection = shear at mid-span B. A triangular load of peak PL/(4EI); deflection = conjugate moment at mid-span = PL^3/(48EI) C. A point load P/EI; deflection = reaction D. A triangular load of peak PL/(4EI); deflection = conjugate shear at mid-span Answer: B. The M/EI diagram is a triangle of peak PL/4EI. Deflection is the conjugate moment at mid-span, PL^3/(48EI), while the support reaction PL^2/(16EI) is the slope. 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