--- title: "Conjugate beam: slope and deflection at free end" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-beam-24kn-10knm" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-beam-24kn-10knm site: BCT Engineering --- # Conjugate beam: slope and deflection at free end Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2081 Chaitra, 8 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-beam-24kn-10knm - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Determine the slope and deflections at the free end of the beam shown in figure below using conjugate beam method. [Figure: beam ABC; A hinge; D at 4 m from A carries a 24 kN downward load; B is a support 2 m from D; the portion A to B has EI; overhang BC of 4 m has 2EI and carries a 10 kN/m UDL.] ## Short answer By conjugate beam method, real free end C becomes fixed, interior support B becomes internal hinge. M/EI areas with variable EI (EI for AB, 2EI for BC) give shear and moment. Slope at C = 160/EI clockwise; deflection at C = 586.7/EI downward (rad and m for EI in kNm^2). ## Answer #### Given data Beam ABC: A hinge at the left end, D at 4 m from A (24 kN down), B roller support at 6 m from A, overhang BC = 4 m (stiffness $2EI$) with UDL 10 kN/m. Portion AB has $EI$. Required: slope and deflection at the free end C. ```text 24 kN 10 kN/m A ------v------- B ======================= C ^ 4 m D 2 m ^ 4 m (2EI) (EI) ``` #### Step 1: Reactions and bending moments Moments about B: $R_A\times6 = 24\times2 - 40\times2$, so $R_A = -5.333$ kN (downward); $R_B = 24 + 40 + 5.333 = 69.33$ kN. $$ M_D = -5.333\times4 = -21.33\ \text{kNm},\qquad M_B = -5.333\times6 - 24\times2 = -80\ \text{kNm (hogging)} $$ (Check from the overhang: $M_B = -10\times4^2/2 = -80$ kNm.) #### Step 2: $M/EI$ diagram (hogging, so conjugate loads act upward) | Portion | $M/EI$ | Area (units $1/EI$) | Centroid | |---|---|---|---| | A-D triangle | 0 to 21.33 | $\tfrac12\times4\times21.33 = 42.67$ | 2.667 m from A | | D-B: rectangle | 21.33 | $2\times21.33 = 42.67$ | 5.0 m from A | | D-B: triangle | 0 to 58.67 | $\tfrac12\times2\times58.67 = 58.67$ | 5.333 m from A | | B-C (parabola, $2EI$) | $\dfrac{5s^2}{2EI}$, $s$ from C; 40 at B | $\tfrac13\times4\times40 = 53.33$ | 3 m from C | #### Step 3: Conjugate beam Real hinge A stays a hinge; the interior support B becomes an **internal hinge**; the free end C becomes a **fixed end**. Moment about the hinge B of the part AB (moment of the conjugate beam at B = 0): $$ R'_A\times6 = 42.67(6-2.667) + 42.67(6-5) + 58.67(6-5.333) = 142.2 + 42.67 + 39.11 = 224 $$ $$ R'_A = \theta_A = \frac{37.33}{EI}\ \text{(anticlockwise)} $$ #### Step 4: Slope at C (shear at fixed end C) Total upward load on AB $= 42.67 + 42.67 + 58.67 = 144$; shear at B (continuous through the internal hinge): $$ \theta_B = 37.33 - 144 = -\frac{106.67}{EI} $$ Shear at C: $\theta_C = -106.67 - 53.33 = -\dfrac{160}{EI}$ #### Step 5: Deflection at C (moment at fixed end C) $$ y_C = \theta_B\times4 + (-53.33)(3) = -426.67 - 160 = -\frac{586.67}{EI} $$ The negative sign means downward. **Answer: slope at C $= 160/EI$ (clockwise); deflection at C $= 586.67/EI$ downward** (kNm$^2$ and kNm$^3$ per $EI$ respectively; with $EI$ in kNm$^2$ the answers are in rad and m). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For this beam (A hinge, B roller interior support, C free end), what is the conjugate beam arrangement? A. A hinge, B internal hinge, C fixed end B. A fixed, B roller, C free C. A hinge, B fixed, C free D. A free, B internal hinge, C hinge Answer: A. The simple end A stays a hinge, the interior support B becomes an internal hinge, and the free end C becomes a fixed end. (Key point: Conjugate supports) 2. What is the bending moment at B, and what does it do to the conjugate loading? A. +80 kNm sagging; conjugate load downward B. -80 kNm hogging; conjugate load acts upward C. -40 kNm hogging; conjugate load upward D. 0, since B is a roller Answer: B. From the overhang M_B = -10x4^2/2 = -80 kNm (hogging). Hogging M/EI loads the conjugate beam upward. (Key point: Bending moment at B) 3. How is the conjugate reaction R'_A (slope at A) found? A. Sum of vertical forces in AB equals 144 B. Take the conjugate moment about the fixed end C C. Divide the total load by 2 D. Take moments about the internal hinge B for part AB, setting conjugate moment at B equal to zero Answer: D. The conjugate moment (deflection) at B is zero, so moments of R'_A and the AB loads about B balance: R'_A x 6 = 224, giving 37.33/EI. (Key point: Finding R'_A) 4. How are the slope and deflection of the free end C obtained? A. Slope = moment at C, deflection = shear at C B. Both from the reaction at A C. Slope = shear and deflection = moment at the fixed end C of the conjugate beam D. Slope = deflection / 4 m Answer: C. At the conjugate fixed end the shear gives the slope (160/EI) and the moment gives the deflection (586.67/EI, downward since negative). 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