--- title: "Conjugate beam: rotation at A, deflection at C" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-a-rotation-c-defl-60knm" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-a-rotation-c-defl-60knm site: BCT Engineering --- # Conjugate beam: rotation at A, deflection at C Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2065 Chaitra, 8 marks - Video: [Conjugate Beam Method Slope and Deflection - Introduction and Step by Step Example](https://www.youtube.com/watch?v=QsNvwRQVkBE) — AF Math & Engineering - 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-a-rotation-c-defl-60knm - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Determine the rotation at A and deflection at C in the overhanging beam shown in fig-4 by using conjugate beam method. [Figure: beam ABC; A hinge, B roller; AB = 6 m (2I) and BC = 2 m (I); 60 kN/m UDL over the whole length.] ## Short answer Using conjugate beam, rotation at A is 210/EI rad clockwise; deflection at C is 180/EI upward. The conjugate beam had A simply supported, B internal hinge, C fixed, with M/(2EI) on AB and M/EI on BC. ## Answer **Data:** $AB = 6$ m ($2I$), $BC = 2$ m ($I$) overhang, $w = 60$ kN/m on the whole beam. Units: kN, m, $EI$ taken for the section $I$. #### Real beam: reactions and BM $$ \begin{aligned} R_B\times 6 &= 60\times 8\times 4 \Rightarrow R_B = 320\ \text{kN}, \quad R_A = 480-320 = 160\ \text{kN}\\ M_x &= 160x - 30x^2 \ (0\le x\le 6), \quad M_B = -60\times\tfrac{2^2}{2} = -120\ \text{kN m} \end{aligned} $$ The loading on the conjugate beam is $M/EI$, so in $AB$ it is $M/(2EI)$ and in $BC$ it is $M/EI$. #### Conjugate beam Real hinge $A$ gives a simple support. Real roller $B$ (with overhang) gives an internal hinge. Real free end $C$ gives a fixed end. ```text Real: A o=======o B========= C (free) Conj.: A o=======o hinge====== C (fixed) load = M/EI (sagging down, hogging up) ``` #### Span AB (simply supported at A, hinge at B) Load $q' = \dfrac{160x-30x^2}{2EI} = \dfrac{80x-15x^2}{EI}$; total load $=\int_0^6 q'\,dx = \dfrac{360}{EI}$. Taking moments about the hinge $B$ (moment of conjugate beam is zero there): $$ R'_A = \frac{1}{6}\int_0^6 (80x-15x^2)(6-x)\,dx = \frac{1260}{6EI} = \frac{210}{EI} $$ $$ R'_B = \frac{360-210}{EI} = \frac{150}{EI} $$ #### Span BC (cantilever, fixed at C) $M/EI$ in $BC$ is negative (hogging), $M = -30(2-x')^2$ with $x'$ from $B$. Its load is upward on the conjugate beam: total $= \int_0^2 30(2-x')^2dx' = 80/EI$, with moment about $C$ equal to $\int_0^2 30(2-x')^3dx' = 120/EI$. The hinge $B$ passes the force $150/EI$ downward onto the cantilever. Moment at the fixed end $C$: $$ M'_C = 150\times 2 - 120 = \frac{180}{EI}\ (\text{hogging}) $$ #### Results - Rotation at $A$ = shear in conjugate beam at $A$: $\theta_A = \dfrac{210}{EI}$ (clockwise). - Deflection at $C$ = moment in conjugate beam at $C$: $y_C = \dfrac{180}{EI}$, and since the conjugate moment is hogging it is **upward**. **Answer: $\theta_A = 210/EI$ kN m² (clockwise), $y_C = 180/EI$ kN m³ (upward).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the reactions of the real beam? A. RA = 320 kN, RB = 160 kN B. RA = 160 kN, RB = 320 kN C. RA = 240 kN, RB = 240 kN D. RA = 480 kN, RB = 0 Answer: B. RB x 6 = 60x8x4 gives RB = 320 kN; RA = 480 - 320 = 160 kN. (Key point: Real reactions) 2. What loading is applied on span AB of the conjugate beam? A. M/EI = (160x - 30x^2)/EI B. M/(2EI) = (80x - 15x^2)/EI C. M x 2EI D. Constant 120/EI Answer: B. AB has rigidity 2EI, so the load is half of M/EI. (Key point: Conjugate in span AB) 3. How is the rotation at A found and what is it? A. Conjugate shear at A = R'A = 210/EI, clockwise B. Conjugate moment at C = 180/EI C. R'B = 150/EI D. Total load 360/EI Answer: A. Moments about hinge B give R'A = 1260/6 = 210/EI, equal to the slope. (Key point: Rotation at A) 4. The conjugate moment at C is hogging; what does this mean for the real deflection at C? A. It is 180/EI upward B. It is 180/EI downward C. It is zero D. It is 120/EI upward Answer: A. M'C = 150x2 - 120 = 180/EI hogging, and hogging conjugate moment means upward deflection. 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