--- 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-2knm-10kn-5kn-2068" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-udl-2knm-10kn-5kn-2068 site: BCT Engineering --- # Conjugate beam: slope and deflection at free end C Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2068 Bhadra, 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-2knm-10kn-5kn-2068 - 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 below. EI is constant. [Figure: beam ABC; A hinge, B roller 8 m from A; 2 kN/m UDL over AB; 10 kN at D, 4 m from A; overhang BC = 4 m with 5 kN at 2 m from B.] ## Short answer Conjugate beam loaded with the M/EI diagram gives at free end C: slope 46/EI rad anticlockwise and deflection 190.67/EI kN·m³ upward. ## Answer #### Given data Beam ABC: A hinge at 0, B roller at 8 m. UDL 2 kN/m over AB; 10 kN at D (4 m from A); overhang $BC = 4$ m with 5 kN at 2 m from B (at 10 m from A). C is the free end at 12 m. $EI$ constant. ```text 10 kN 2 kN/m 5 kN A ------v============ B ------v------- C ^ 4 m D 4 m ^ 2 m 2 m ``` #### Step 1: Reactions $R_{A} = 11.75$ kN, $R_{B} = 19.25$ kN (positive = upward; check: $\sum F_y = 31.00$ kN equals the total load 31.00 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 4 | $- x^{2} + \frac{47 x}{4}$ | EI | 72.67 | 2.569 | | 4 to 8 | $- x^{2} + \frac{7 x}{4} + 40$ | EI | 52.67 | 4.962 | | 8 to 10 | $5 x - 50$ | EI | -10.00 | 8.667 | | 10 to 12 | $0$ | EI | 0.00 | - | (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 $x = 8$ m (conjugate moment is zero there): $$ R'_A\times8 = (72.67)(5.431) + (52.67)(3.038) = 554.67 $$ $$ R'_A = \frac{69.333}{EI}\ \text{(slope at A, clockwise)} $$ **At D (10 kN) ($x = 4$ m):** $$ V' = R'_A - \sum A = 69.333 - [(72.67)] = \frac{-3.333}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 277.333 - [(72.67)(1.431)] = \frac{173.333}{EI}\ \text{(deflection, downward)} $$ **At C (free end) ($x = 12$ m):** $$ V' = R'_A - \sum A = 69.333 - [(72.67) + (52.67) + (-10.00)] = \frac{-46.000}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 832.000 - [(72.67)(9.431) + (52.67)(7.038) + (-10.00)(3.333)] = \frac{-190.667}{EI}\ \text{(deflection, upward)} $$ **Answer: slope at C $= \dfrac{46}{EI}$ (anticlockwise); deflection at C $= \dfrac{190.67}{EI}$ upward** (kNm$^2$/EI and kNm$^3$/EI). (The deflection under the 10 kN load at D is $173.33/EI$ downward.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the reactions at A and B? A. RA = 19.25 kN, RB = 11.75 kN B. RA = 11.75 kN, RB = 19.25 kN C. RA = 15.5 kN, RB = 15.5 kN D. RA = 13 kN, RB = 18 kN Answer: B. Moments about A: RB x 8 = 10x4 + 2x8x4 + 5x10 = 154, so RB = 19.25 kN and RA = 11.75 kN. (Key point: Reactions) 2. Why is the area for the portion 10 m to 12 m zero? A. The bending moment there is zero because no load acts beyond the 5 kN load B. The beam is stiffer there C. The UDL ends there D. Because it is a fixed support Answer: A. Beyond the last load on a free overhang the moment is zero, so M/EI loads nothing in CD. (Key point: Zero M/EI region) 3. What is the slope at the free end C? A. 46/EI anticlockwise B. 69.33/EI clockwise C. 3.33/EI anticlockwise D. 190.67/EI Answer: A. V' at C = 69.333 - 115.33 = -46/EI, an anticlockwise slope. (Key point: Slope at C) 4. What is the deflection at C and its direction? A. 173.33/EI downward B. 190.67/EI upward C. 190.67/EI downward D. 46/EI upward Answer: B. Conjugate moment at C is -190.667/EI (negative), so the tip deflects upward by 190.67/EI. 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