--- title: "Conjugate beam: slope and deflection at free end" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-slope-defl-free-end-2073" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-slope-defl-free-end-2073 site: BCT Engineering --- # Conjugate beam: slope and deflection at free end Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2073 Bhadra, 12 marks - Video: [Conjugate Beam Method (Ex 2 - Overhanging Beam)](https://www.youtube.com/watch?v=ekiKJFUpGdU) — Engr. Clidez - 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-free-end-2073 - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Determine slope and deflection at free end and 2 m from left support. [Figure: beam ABC; A hinge, B roller; AB = 6 m (2EI) carries 40 kN/m UDL; overhang BC = 2 m (EI) with 30 kN at C.] ## Short answer Using conjugate beam method: at 2 m from left support, slope = 66.67/EI clockwise, deflection = 240/EI downward; at free end C, slope = 60/EI anticlockwise, deflection = 160/EI upward. ## Answer #### Given data Beam ABC: A hinge, B roller, $AB = 6$ m ($2EI$) with UDL 40 kN/m, overhang $BC = 2$ m ($EI$) with 30 kN at C. ```text 40 kN/m (2EI) 30 kN A =============== B ---------v C ^ 6 m ^ 2 m (EI) ``` #### Step 1: Reactions $R_{A} = 110.00$ kN, $R_{B} = 160.00$ kN (positive = upward; check: $\sum F_y = 270.00$ kN equals the total load 270.00 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 | $- 20 x^{2} + 110 x$ | 2EI | 270.00 | 2.667 | | 6 to 8 | $30 x - 240$ | EI | -60.00 | 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): $$ R'_A\times6 = (270.00)(3.333) = 900.00 $$ $$ R'_A = \frac{150.000}{EI}\ \text{(slope at A, clockwise)} $$ **At the point 2 m from A ($x = 2$ m):** $$ V' = R'_A - \sum A = 150.000 - [(83.33)] = \frac{66.667}{EI}\ \text{(slope, clockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 300.000 - [(83.33)(0.720)] = \frac{240.000}{EI}\ \text{(deflection, downward)} $$ **At C (free end) ($x = 8$ m):** $$ V' = R'_A - \sum A = 150.000 - [(270.00) + (-60.00)] = \frac{-60.000}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 1200.000 - [(270.00)(5.333) + (-60.00)(1.333)] = \frac{-160.000}{EI}\ \text{(deflection, upward)} $$ **Answer:** at 2 m from A: slope $= \dfrac{66.67}{EI}$ (clockwise), deflection $= \dfrac{240}{EI}$ downward. At the free end C: slope $= \dfrac{60}{EI}$ (anticlockwise), deflection $= \dfrac{160}{EI}$ **upward** (kNm$^2$/EI and kNm$^3$/EI). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For this overhanging beam, what does the conjugate beam have at A, B and C? A. Hinge at A, internal hinge at B, fixed end at C B. Fixed at A, roller at B, free at C C. Hinge at A, roller at B, free at C D. Fixed at all three Answer: A. A real hinge end maps to a hinge, the intermediate support to an internal hinge, and the free end to a fixed end. (Key point: Conjugate supports) 2. What is the sign of M/EI in the overhang BC and how does it load the conjugate beam? A. Sagging, downward load B. Hogging, upward load C. Zero D. Sagging, upward load Answer: B. The 30 kN load at C gives M = 30x - 240 < 0 in BC, so it is hogging and acts upward on the conjugate beam. (Key point: Sign of overhang M/EI) 3. What is R'A, the slope at A? A. 110/EI B. 150/EI clockwise C. 270/EI D. 60/EI Answer: B. R'A x 6 = 270 x 3.333 = 900, so R'A = 150/EI. (Key point: Slope at A) 4. What is the deflection at the free end C? A. 160/EI downward B. 160/EI upward C. 240/EI upward D. 60/EI downward Answer: B. The conjugate moment at C is -160/EI, negative, meaning upward deflection of 160/EI. 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