--- title: "Conjugate beam: deflection at B and D" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-udl-8knm-20kn-bd" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-udl-8knm-20kn-bd site: BCT Engineering --- # Conjugate beam: deflection at B and D Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2073 Magh, 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-8knm-20kn-bd - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Calculate deflection at point B and D using conjugate beam method. [Figure: beam ABCD; A hinge, C roller; AC with 2EI, AD = 5 m with a 20 kN load at D, DC = 3 m; overhang CB = 3 m with EI carrying 8 kN/m UDL.] ## Short answer Reactions: 3 kN at A, 41 kN at C. The conjugate beam with stiffness 2EI on AC and EI on CB gives deflection at D (under 20 kN) = 20.625/EI downward and at free end B = 103.125/EI downward, where EI is the reference flexural rigidity. ## Answer #### Given data Beam ABCD as labelled: A hinge at 0, D at 5 m (20 kN down), C roller at 8 m ($DC = 3$ m), overhang $CB = 3$ m with UDL 8 kN/m and free end B at 11 m. $AC$ has $2EI$ and the overhang $CB$ has $EI$. Required: deflection at B (free end) and D. ```text 20 kN 8 kN/m (EI) A -------v---------- C =============== B ^ 5 m D 3 m (2EI) ^ 3 m ``` #### Step 1: Reactions $R_{A} = 3.00$ kN, $R_{C} = 41.00$ kN (positive = upward; check: $\sum F_y = 44.00$ kN equals the total load 44.00 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 5 | $3 x$ | 2EI | 18.75 | 3.333 | | 5 to 8 | $100 - 17 x$ | 2EI | -15.75 | 7.714 | | 8 to 11 | $- 4 x^{2} + 88 x - 484$ | EI | -36.00 | 8.750 | (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): $$ R'_A\times8 = (18.75)(4.667) + (-15.75)(0.286) = 83.00 $$ $$ R'_A = \frac{10.375}{EI}\ \text{(slope at A, clockwise)} $$ **At D (20 kN) ($x = 5$ m):** $$ V' = R'_A - \sum A = 10.375 - [(18.75)] = \frac{-8.375}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 51.875 - [(18.75)(1.667)] = \frac{20.625}{EI}\ \text{(deflection, downward)} $$ **At B (free end) ($x = 11$ m):** $$ V' = R'_A - \sum A = 10.375 - [(18.75) + (-15.75) + (-36.00)] = \frac{43.375}{EI}\ \text{(slope, clockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 114.125 - [(18.75)(7.667) + (-15.75)(3.286) + (-36.00)(2.250)] = \frac{103.125}{EI}\ \text{(deflection, downward)} $$ **Answer: deflection at D $= \dfrac{20.625}{EI}$ downward; deflection at the free end B $= \dfrac{103.125}{EI}$ downward** (kNm$^3$/EI). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a real beam with a hinge at A, a roller at C and a free overhang end B, what does the conjugate beam have at the support C and at B? A. Roller at C and free end at B B. Internal hinge at C and fixed end at B C. Fixed end at C and hinge at B D. Hinge at C and roller at B Answer: B. Free end (slope and deflection nonzero) becomes a fixed end; an interior support (zero deflection) becomes an internal hinge. (Key point: Conjugate of free end) 2. How does the M/EI loading treat the different rigidities of AC (2EI) and CB (EI)? A. M is divided by 2EI on AC and by EI on CB B. M is divided by EI everywhere C. M is divided by 2EI on CB D. Only the overhang is loaded Answer: A. Each segment's moment is divided by its own rigidity before being applied as conjugate load. (Key point: Stiffness change) 3. What are the real reactions at A and C? A. RA = 41 kN, RC = 3 kN B. RA = 22 kN, RC = 22 kN C. RA = 3 kN, RC = 41 kN D. RA = 20 kN, RC = 24 kN Answer: C. Taking moments about C gives RA = 3 kN, and RC = 44 - 3 = 41 kN. (Key point: Reactions of real beam) 4. What deflections are obtained at D and at the free end B? A. D = 103.125/EI, B = 20.625/EI B. D = 20.625/EI, B = 103.125/EI, both downward C. D = 20.625/EI downward, B = 103.125/EI upward D. Both equal 43.375/EI Answer: B. The conjugate moments give 20.625/EI at D and 103.125/EI at B, both downward. 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