--- title: "Conjugate beam theorems and derivation" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-beam-theorems" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-beam-theorems site: BCT Engineering --- # Conjugate beam theorems and derivation Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2073 Magh, 4 marks; 2076 Bhadra, 4 marks; 2072 Asoj, 4 marks; 2071 Magh (old course), 6 marks (asked 4 times) - Video: [Conjugate Beam Method (Concepts and Basics)](https://www.youtube.com/watch?v=57UiP6tqbqo) — DCBA online - 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/conjugate-beam-theorems - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question What are the conjugate beam theorems? State and prove them and explain their use with an example. ## Short answer The conjugate beam method gives slope as the shear and deflection as the moment in a conjugate beam loaded with the real beam's M/EI diagram. Proof: differential equations show θ = ∫(M/EI)dx = V' and y = M'. Supports are transformed accordingly. ## Answer The **conjugate beam method** converts the problem of finding the slope and deflection of a real beam into finding the shear and moment in a fictitious beam (the conjugate beam), which is loaded with the $M/EI$ diagram of the real beam. #### Theorems - **Theorem 1:** The slope at a point in the real beam is numerically equal to the shear in the conjugate beam at the corresponding point. - **Theorem 2:** The deflection at a point in the real beam is numerically equal to the bending moment in the conjugate beam at the corresponding point. #### Proof For a real beam, $\dfrac{d^2y}{dx^2} = \dfrac{M}{EI}$, with $\dfrac{d^2M}{dx^2} = w$, $\dfrac{dM}{dx} = V$ (load, shear, moment). For the conjugate beam, loaded by $w' = \dfrac{M}{EI}$: $$ \frac{dV'}{dx} = w' = \frac{M}{EI},\qquad \frac{d^2M'}{dx^2} = w' = \frac{M}{EI} $$ Comparing with the real beam relations: $$ \frac{d\theta}{dx} = \frac{M}{EI}\ \Rightarrow\ \theta = \int\frac{M}{EI}dx = V' $$ $$ \frac{d^2y}{dx^2} = \frac{M}{EI}\ \Rightarrow\ y = M' $$ The mathematical relations are the same, so $\theta = V'$ and $y = M'$ (provided the boundary conditions also match). #### Supports of the conjugate beam | Real beam | Conjugate beam | |---|---| | Fixed end ($\theta = 0$, $y = 0$) | Free end ($V' = 0$, $M' = 0$) | | Free end | Fixed end | | Simple end support ($y = 0$, $\theta \ne 0$) | Simple end support ($M' = 0$, $V' \ne 0$) | | Interior support ($y = 0$, $\theta$ continuous) | Internal hinge ($M' = 0$, $V'$ continuous) | | Internal hinge ($\theta$ discontinuous, $y \ne 0$) | Interior support | #### Procedure and example 1. Find $M$ of the real beam and draw the $M/EI$ diagram. 2. Load the conjugate beam with $M/EI$ (sagging $M/EI$ acts downward). 3. Find the reactions, then the shear (slope) and moment (deflection). **Example:** simply supported beam, span $L$, central load $P$. $M_{max} = PL/4$. Conjugate beam: simply supported, triangular load with peak $\dfrac{PL}{4EI}$. Total load $= \tfrac12\cdot L\cdot\dfrac{PL}{4EI} = \dfrac{PL^2}{8EI}$, so each reaction is $\dfrac{PL^2}{16EI}$. $$ \theta_A = V'_A = \frac{PL^2}{16EI} $$ At mid-span, the conjugate moment is $$ y_C = M'_C = \frac{PL^2}{16EI}\times\frac{L}{2} - \frac{PL^2}{16EI}\times\frac{L}{6}\ \ \Rightarrow\ \ y_C = \frac{PL^3}{32EI} - \frac{PL^3}{96EI} = \frac{PL^3}{48EI} $$ which is the known mid-span deflection. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What do the conjugate beam theorems state? A. Slope equals load; deflection equals shear B. Slope equals moment; deflection equals shear C. Slope of the real beam equals shear in the conjugate beam; deflection equals moment in the conjugate beam D. Deflection equals load Answer: C. The conjugate beam is loaded by M/EI; its shear equals the slope and its moment equals the deflection. (Key point: Theorems) 2. What is the conjugate of a fixed end of the real beam? A. Fixed end B. Free end C. Simple support D. Internal hinge Answer: B. A fixed end has zero slope and deflection, so V' = 0 and M' = 0, which is a free end. (Key point: Conjugate supports) 3. For a simply supported beam with a central load P, what is the conjugate end reaction, giving the end slope? A. PL^2/(8EI) B. PL^3/(48EI) C. PL^2/(16EI) D. PL/(4EI) Answer: C. Total triangular load is PL^2/(8EI), shared equally, so each reaction (the slope) is PL^2/(16EI). 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