--- title: "Conjugate beam: slope and deflection at C" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-slope-defl-c-120kn-3ei" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-slope-defl-c-120kn-3ei site: BCT Engineering --- # Conjugate beam: slope and deflection at C Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2069 Bhadra, 12 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-slope-defl-c-120kn-3ei - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Using conjugate beam method, find slope and deflection at point C of the following loaded beam. [Figure: beam AB, A hinge, B roller, span 12 m; portion AD of 7 m has flexural rigidity 3EI with a 120 kN load near the left end; portion DB of 5 m has EI and carries 10 kN/m UDL over its last 3 m, C being the end of the UDL region. Figure partly illegible.] ## Short answer At point C, by conjugate beam method, slope = 311.42/EI anticlockwise and deflection = 1339.27/EI downward. ## Answer #### Given data and assumptions Simply supported beam AB, span 12 m. The portion from $0$ to $7$ m (A to D) has rigidity $3EI$ and the portion from $7$ to $12$ m (D to B) has $EI$. A 120 kN load acts 3 m from A; a UDL of 10 kN/m acts over the last 3 m (from 9 m to 12 m). The point C is taken at the start of the UDL, $x = 9$ m (the figure is partly illegible; this is the reading used). ```text 120 kN 10 kN/m A -----v--------- 3EI -------D--- EI ---C=========B ^<-3 m->|<------- 4 m ------>|<- 2 m ->|<- 3 m ->^ ``` #### Step 1: Reactions $R_{A} = 93.75$ kN, $R_{B} = 56.25$ kN (positive = upward; check: $\sum F_y = 150.00$ kN equals the total load 150.00 kN). #### Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces | Portion (m) | $M$ (kNm) | stiffness | Area of $M/EI$ | $\bar x$ from A (m) | |---|---|---|---|---| | 0 to 3 | $\frac{375 x}{4}$ | 3EI | 140.62 | 2.000 | | 3 to 7 | $360 - \frac{105 x}{4}$ | 3EI | 305.00 | 4.847 | | 7 to 9 | $360 - \frac{105 x}{4}$ | EI | 300.00 | 7.942 | | 9 to 12 | $- 5 x^{2} + \frac{255 x}{4} - 45$ | EI | 208.12 | 10.054 | (Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.) #### Step 3: Conjugate beam The conjugate of a simply supported beam is a simply supported beam of the same span. It carries the $M/EI$ diagram of Step 2 as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward). #### Step 4: Conjugate reaction at A (slope at A) Taking moments of the conjugate beam about the support at $x = 12$ m (conjugate moment is zero there): $$ R'_A\times12 = (140.62)(10.000) + (305.00)(7.153) + (300.00)(4.058) + (208.12)(1.946) = 5210.42 $$ $$ R'_A = \frac{434.201}{EI}\ \text{(slope at A, clockwise)} $$ **At the 120 kN load ($x = 3$ m):** $$ V' = R'_A - \sum A = 434.201 - [(140.62)] = \frac{293.576}{EI}\ \text{(slope, clockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 1302.604 - [(140.62)(1.000)] = \frac{1161.979}{EI}\ \text{(deflection, downward)} $$ **At C ($x = 9$ m):** $$ V' = R'_A - \sum A = 434.201 - [(140.62) + (305.00) + (300.00)] = \frac{-311.424}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 3907.812 - [(140.62)(7.000) + (305.00)(4.153) + (300.00)(1.058)] = \frac{1339.271}{EI}\ \text{(deflection, downward)} $$ **Answer: at C, slope $= \dfrac{311.42}{EI}$ (anticlockwise) and deflection $= \dfrac{1339.27}{EI}$ downward** (kNm$^2$/EI and kNm$^3$/EI). Slope at A $= 434.20/EI$ (clockwise). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this solution where is C taken, and what is the nature of the loading beyond it? A. At the end of the UDL, x = 12 m B. At the start of the UDL, x = 9 m C. At the 120 kN load, x = 3 m D. At the junction of 3EI and EI, x = 7 m Answer: B. The figure is partly illegible, so C is taken at the start of the UDL region, x = 9 m. (Key point: Reading of point C) 2. Which portion has its M/EI ordinates reduced by dividing by 3EI? A. Portion 0 to 7 m B. Portion 7 to 12 m C. Only the UDL region D. The whole beam Answer: A. The first 7 m has rigidity 3EI, so M is divided by 3EI there and by EI in 7 to 12 m. (Key point: Stiffness in conjugate loading) 3. What is the slope at C and its sense? A. 434.20/EI clockwise B. 311.42/EI anticlockwise C. 1339.27/EI anticlockwise D. 311.42/EI clockwise Answer: B. V' at C = 434.201 - (140.62 + 305 + 300) = -311.42/EI, a negative (anticlockwise) slope. (Key point: Slope at C) 4. How is the deflection at C obtained, and what is it? A. Conjugate shear at C, 311.42/EI B. Conjugate moment at C, 1339.27/EI downward C. Conjugate moment at A, zero D. Conjugate shear at A, 434.20/EI Answer: B. Deflection equals the conjugate moment: 3907.812 - 2568.541 = 1339.27/EI downward. 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