--- title: "Conjugate beam: slopes at supports, deflection at E" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-slopes-e-udl-5knm-3kn" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-slopes-e-udl-5knm-3kn site: BCT Engineering --- # Conjugate beam: slopes at supports, deflection at E Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2070 Magh, 10 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-slopes-e-udl-5knm-3kn - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Find slopes at supports and deflection at E of the beam given in figure below. Use conjugate beam method for deflection and slope calculations. [Figure: beam ABC; A hinge, B roller, span AB = 20 m with 5 kN/m UDL on AB; E is 5 m from A; overhang BC = 3 m with 3 kN at C; EI constant.] ## Short answer Conjugate beam method with distributed M/EI load yields support slopes: 1636.67/EI rad clockwise at A, 1606.67/EI rad anticlockwise at B. Deflection at E (5 m from A) is 7281.25/EI kN·m³ downward. ## Answer #### Given data Beam ABC: $AB = 20$ m (hinge at A, roller at B) with UDL 5 kN/m; overhang $BC = 3$ m with 3 kN at C. E is 5 m from A. $EI$ constant. #### Step 1: Reactions $R_{A} = 49.55$ kN, $R_{B} = 53.45$ kN (positive = upward; check: $\sum F_y = 103.00$ kN equals the total load 103.00 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 20 | $- \frac{5 x^{2}}{2} + \frac{991 x}{20}$ | EI | 3243.33 | 9.908 | | 20 to 23 | $3 x - 69$ | EI | -13.50 | 21.000 | (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 real beam has a hinge at A, a roller at the second support and a free end at the overhang tip, so the conjugate beam has a hinge at A, an internal hinge at the support and a fixed end at the tip. 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 $x = 20$ m (conjugate moment is zero there): $$ R'_A\times20 = (3243.33)(10.092) = 32733.33 $$ $$ R'_A = \frac{1636.667}{EI}\ \text{(slope at A, clockwise)} $$ **At support A ($x = 0$ m):** $$ V' = R'_A - \sum A = 1636.667 - [0] = \frac{1636.667}{EI}\ \text{(slope, clockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 0.000 - [0] = \frac{0.000}{EI}\ \text{(deflection, zero)} $$ **At E (5 m from A) ($x = 5$ m):** $$ V' = R'_A - \sum A = 1636.667 - [(515.21)] = \frac{1121.458}{EI}\ \text{(slope, clockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 8183.333 - [(515.21)(1.751)] = \frac{7281.250}{EI}\ \text{(deflection, downward)} $$ **At support B ($x = 20$ m):** $$ V' = R'_A - \sum A = 1636.667 - [(3243.33)] = \frac{-1606.667}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 32733.333 - [(3243.33)(10.092)] = \frac{0.000}{EI}\ \text{(deflection, zero)} $$ **At C (free end) ($x = 23$ m):** $$ V' = R'_A - \sum A = 1636.667 - [(3243.33) + (-13.50)] = \frac{-1593.167}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 37643.333 - [(3243.33)(13.092) + (-13.50)(2.000)] = \frac{-4793.000}{EI}\ \text{(deflection, upward)} $$ **Answer:** slope at A $= \dfrac{1636.67}{EI}$ (clockwise); slope at B $= \dfrac{1606.67}{EI}$ (anticlockwise); deflection at E $= \dfrac{7281.25}{EI}$ downward (kNm$^2$/EI and kNm$^3$/EI). (At C the deflection is $4793/EI$ upward, which is not asked.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the real reactions at A and B? A. RA = 53.45 kN, RB = 49.55 kN B. RA = 49.55 kN, RB = 53.45 kN C. RA = 50 kN, RB = 53 kN D. RA = 51.5 kN, RB = 51.5 kN Answer: B. Moments about A give RB x 20 = 5x20x10 + 3x23 = 1069, so RB = 53.45 kN and RA = 49.55 kN. (Key point: Reactions) 2. How is the slope at A found and what is it? A. Conjugate reaction R'A, 1636.67/EI clockwise B. Conjugate moment at A, zero C. Conjugate shear at B, 1606.67/EI D. Conjugate reaction at B, 1606.67/EI clockwise Answer: A. Moments of the conjugate load about B give R'A = 32733.33/20 = 1636.67/EI. (Key point: Slope at A) 3. What is the deflection at E, 5 m from A? A. 7281.25/EI downward B. 1121.46/EI downward C. 4793/EI upward D. 8183.33/EI downward Answer: A. M'E = 8183.333 - 902.08 = 7281.25/EI, downward. (Key point: Deflection at E) 4. Why is the deflection at overhang tip C upward in this beam? A. The conjugate moment at C is negative (hogging) B. Because the load at C is upward C. Because slope at B is zero D. Because the UDL is on the overhang Answer: A. The conjugate moment at C is -4793/EI, negative, meaning upward deflection. 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