--- title: "Conjugate beam: free end of overhang, UDL and point load" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-udl-10kn-4m-3m-2071b" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-udl-10kn-4m-3m-2071b site: BCT Engineering --- # Conjugate beam: free end of overhang, UDL and point load Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2071 Bhadra, 12 marks - Video: [Conjugate Beam Method | Example 6 | Simply Supported Beam with Overhang](https://www.youtube.com/watch?v=E1jCBl0r48w) — Being Civil - 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-10kn-4m-3m-2071b - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question A simply supported beam of span 4 m with an overhang of length 2 m on right side of the beam is loaded in the span with uniform distributed load of intensity 2 kN/m. The overhang is loaded with a concentrated force of magnitude 3 kN at the free end. Calculate the deflection of the free end of the overhang and slope at the support. Use conjugate beam method. ## Short answer Using conjugate beam, the M/EI diagram is split into parabolic and triangular areas. The conjugate beam reactions give the free end deflection as 13.33/EI kN·m³ downward, and the support slopes as 1.33/EI rad clockwise at A and 2.67/EI rad clockwise at B. ## Answer #### Given data Simply supported span $AB = 4$ m with an overhang $BC = 2$ m. UDL 2 kN/m over AB; 3 kN at the free end C. $EI$ constant. ```text 2 kN/m 3 kN A ============ B --------v C ^ 4 m ^ 2 m ``` #### Step 1: Reactions $R_{A} = 2.50$ kN, $R_{B} = 8.50$ kN (positive = upward; check: $\sum F_y = 11.00$ kN equals the total load 11.00 kN). #### Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces In AB, $M = 2.5x - x^2 = (4x - x^2) - 1.5x$: the free (simply supported) parabola of the UDL plus a hogging triangle caused by the overhang moment $M_B = -3\times2 = -6$ kNm. In BC, $M = 3x - 18$. | Piece | Area of $M/EI$ | $\bar x$ from A (m) | |---|---|---| | Parabola on AB (peak $wL^2/8 = 4$ kNm at mid-span) | $\tfrac23(4)(4) = +10.667$ | 2.000 | | Triangle on AB (0 at A to $-6$ at B) | $\tfrac12(4)(-6) = -12.000$ | 2.667 | | Triangle on BC ($-6$ at B to 0 at C) | $\tfrac12(2)(-6) = -6.000$ | 4.667 | (Areas are in units of $1/EI$; the net area on AB is $-1.333$.) #### Step 3: Conjugate beam The real beam has a hinge at A, a roller at B and a free end at C, so the conjugate beam has a simple support at A, an internal hinge at B and a fixed end at C. 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 span AB about the hinge B (conjugate moment is zero there): $$ R'_A\times4 = (10.667)(2.000) + (-12.000)(1.333) = 21.333 - 16.000 = 5.333 $$ $$ R'_A = \frac{1.333}{EI}\ \text{(slope at A, clockwise)} $$ **At support B ($x = 4$ m):** $$ V' = R'_A - \sum A = 1.333 - [(10.667) + (-12.000)] = \frac{2.667}{EI}\ \text{(slope, clockwise)} $$ (The conjugate moment at B is zero, so the deflection at B is zero, as it must be.) **At the free end C ($x = 6$ m):** $$ V' = R'_A - \sum A = 1.333 - [(10.667) + (-12.000) + (-6.000)] = \frac{8.667}{EI}\ \text{(slope, clockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 8.000 - [(10.667)(4.000) + (-12.000)(3.333) + (-6.000)(1.333)] = 8.000 + 5.333 = \frac{13.333}{EI}\ \text{(deflection, downward)} $$ (Check: $\theta_B\times2 + \dfrac{Pa^3}{3EI} = 2.667\times2 + \dfrac{3\times8}{3} = 13.33$.) **Answer:** deflection at the free end C $= \dfrac{13.33}{EI}$ downward (kNm$^3$/EI). Slope at support A $= \dfrac{1.333}{EI}$ (clockwise); slope at support B $= \dfrac{2.667}{EI}$ (clockwise) (kNm$^2$/EI). The slope at the free end C is $\dfrac{8.667}{EI}$ (clockwise). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the conjugate beam have at the free end C? A. A roller B. A fixed end C. A free end D. A hinge Answer: B. A free real end has nonzero slope and deflection, which requires a fixed end in the conjugate beam. (Key point: Conjugate supports) 2. How is the deflection of the free end C obtained and what is it? A. Conjugate shear at C, 8.667/EI B. Conjugate moment at C, 13.33/EI downward C. Conjugate moment at A, zero D. Conjugate shear at A, 1.333/EI Answer: B. Deflection is the conjugate moment at the fixed end C, which equals 13.33/EI downward. 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