--- title: "Conjugate beam: slope and deflection at free end C" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-udl-15-3kn-2076b" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-udl-15-3kn-2076b site: BCT Engineering --- # Conjugate beam: slope and deflection at free end C Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2076 Baisakh, 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-udl-15-3kn-2076b - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Using conjugate beam method, calculate slope and deflection at point C, free end of the beam loaded as shown here. [Figure: beam ABC; A hinge, B roller; AB = 6 m (2EI) carries 15 kN/m UDL; overhang BC = 2 m (EI) with 3 kN at C.] ## Short answer By the conjugate beam method, the overhang is represented by a fixed end. The slope at free end C is 55.5/EI anticlockwise and the deflection is 115/EI upward. ## Answer #### Given data Beam ABC: A hinge, B roller, $AB = 6$ m ($2EI$) with UDL 15 kN/m, overhang $BC = 2$ m ($EI$) with 3 kN at C. ```text 15 kN/m (2EI) 3 kN A =============== B ----------v C ^ 6 m ^ 2 m (EI) ``` #### Step 1: Reactions $R_{A} = 44.00$ kN, $R_{B} = 49.00$ kN (positive = upward; check: $\sum F_y = 93.00$ kN equals the total load 93.00 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 6 | $- \frac{15 x^{2}}{2} + 44 x$ | 2EI | 126.00 | 2.929 | | 6 to 8 | $3 x - 24$ | EI | -6.00 | 6.667 | (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 $x = 6$ m (conjugate moment is zero there): $$ R'_A\times6 = (126.00)(3.071) = 387.00 $$ $$ R'_A = \frac{64.500}{EI}\ \text{(slope at A, clockwise)} $$ **At C (free end) ($x = 8$ m):** $$ V' = R'_A - \sum A = 64.500 - [(126.00) + (-6.00)] = \frac{-55.500}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 516.000 - [(126.00)(5.071) + (-6.00)(1.333)] = \frac{-115.000}{EI}\ \text{(deflection, upward)} $$ **Answer: slope at C $= \dfrac{55.5}{EI}$ (anticlockwise) and deflection at C $= \dfrac{115}{EI}$ upward** (the hogging moment from the UDL span lifts the overhang more than the 3 kN load pushes it down). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Moments about B for AB = 6 m with 15 kN/m UDL and 3 kN at 2 m beyond B give R_A = ? A. 45 kN B. 49 kN C. 46 kN D. 44 kN Answer: D. R_A x 6 = 15x6x3 - 3x2 = 270 - 6 = 264, so R_A = 44 kN (and R_B = 49 kN). (Key point: Reaction at A) 2. The M area of AB is 252 kNm^2. Why is the conjugate load 126/EI? A. AB has rigidity 2EI, so the load is M area / 2EI = 126/EI B. Only half of the span is loaded C. The parabola area is divided by 2 by its shape D. Because R_A is shared by two supports Answer: A. Integral of M over AB is 252, and with rigidity 2EI the M/EI area is 252/(2EI) = 126/EI. (Key point: Effect of 2EI on the area) 3. How is R'_A = 64.5/EI found? A. Sum of vertical conjugate loads equals R'_A B. Moments of the conjugate beam about the support B are zero (internal hinge): R'_A x 6 = 126 x 3.071 C. R'_A = 126/2 D. Moments about the fixed end C Answer: B. B becomes an internal hinge, so conjugate moment at B is zero: 6R'_A = 126(6 - 2.929) = 387, R'_A = 64.5/EI. (Key point: R'_A by internal hinge) 4. What do V' = -55.5/EI and M' = -115/EI at the fixed end C mean? A. Slope 55.5/EI clockwise, deflection 115/EI downward B. Slope 115/EI anticlockwise, deflection 55.5/EI upward C. Slope 55.5/EI anticlockwise, deflection 115/EI upward D. Both are zero at the free end Answer: C. Negative shear means anticlockwise slope and negative moment means upward deflection. The hogging moment from the UDL span lifts the overhang more than the 3 kN load pushes it down. 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