--- title: "Conjugate beam: stepped symmetric beam" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-sym-beam-3portions" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-sym-beam-3portions site: BCT Engineering --- # Conjugate beam: stepped symmetric beam Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2075 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-sym-beam-3portions - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question A symmetrical beam ABCD is simply supported at its ends A and D over a span of 6 m. It is made up of three portions with different values of I, the length of the middle portion with the value of 2I is 3 m and the portion with the value of I is 1.5 m each. The beam carries two point loads of 20 kN at B and C. Find the slope and deflection at A, B, C and D using the conjugate beam method. Take $E = 200\ \text{kN/mm}^2$ and $I = 2\times10^{10}\ \text{mm}^4$. [Figure: AB = 2 m? (I), BC (2I), CD (I); loads 20 kN at B and C; AB = BC = CD = 2 m as dimensioned.] ## Short answer With EI = 4×10⁶ kN·m², the conjugate beam method gives slopes: θ_A = 1.125×10⁻⁵ rad clockwise, θ_B = 5.625×10⁻⁶ rad clockwise, θ_C = 5.625×10⁻⁶ rad anticlockwise, θ_D = 1.125×10⁻⁵ rad anticlockwise. Deflections: y_B = y_C = 0.0141 mm downward; y_A = y_D = 0. ## Answer #### Given data Symmetrical beam ABCD, simply supported at A and D, span 6 m, in three portions: AB (1.5 m, $I$), BC (3 m, $2I$), CD (1.5 m, $I$) as stated in the question (so B and C are 1.5 m from the ends). Loads 20 kN at B and at C. $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $I = 2\times10^{10}$ mm$^4 = 0.02$ m$^4$, so $EI = 4,000,000$ kNm$^2$. ```text 20 kN 20 kN A -----v------------v----- D ^ 1.5 m B 3 m (2I) C 1.5 m ^ (I) (I) ``` #### Step 1: Reactions $R_{A} = 20.00$ kN, $R_{D} = 20.00$ kN (positive = upward; check: $\sum F_y = 40.00$ kN equals the total load 40.00 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 1.5 | $20 x$ | EI | 22.50 | 1.000 | | 1.5 to 4.5 | $30$ | 2EI | 45.00 | 3.000 | | 4.5 to 6 | $120 - 20 x$ | EI | 22.50 | 5.000 | (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 conjugate of a simply supported beam is a simply supported beam of the same span. 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): $$ R'_A\times6 = (22.50)(5.000) + (45.00)(3.000) + (22.50)(1.000) = 270.00 $$ $$ R'_A = \frac{45.000}{EI}\ \text{(slope at A, clockwise)} $$ **At A ($x = 0$ m):** $$ V' = R'_A - \sum A = 45.000 - [0] = \frac{45.000}{EI}\ \text{(slope, clockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 0.000 - [0] = \frac{0.000}{EI}\ \text{(deflection, zero)} $$ **At B ($x = 1.5$ m):** $$ V' = R'_A - \sum A = 45.000 - [(22.50)] = \frac{22.500}{EI}\ \text{(slope, clockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 67.500 - [(22.50)(0.500)] = \frac{56.250}{EI}\ \text{(deflection, downward)} $$ **At C ($x = 4.5$ m):** $$ V' = R'_A - \sum A = 45.000 - [(22.50) + (45.00)] = \frac{-22.500}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 202.500 - [(22.50)(3.500) + (45.00)(1.500)] = \frac{56.250}{EI}\ \text{(deflection, downward)} $$ **At D ($x = 6$ m):** $$ V' = R'_A - \sum A = 45.000 - [(22.50) + (45.00) + (22.50)] = \frac{-45.000}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 270.000 - [(22.50)(5.000) + (45.00)(3.000) + (22.50)(1.000)] = \frac{0.000}{EI}\ \text{(deflection, zero)} $$ #### Results (symmetry check) By symmetry $\theta_D = -\theta_A$ and $y_C = y_B$. With $EI = 4,000,000$ kNm$^2$: | Point | Slope (rad) | Deflection (mm, downward) | |---|---|---| | A | $1.125\times10^{-5}$ | 0.0000 | | B | $5.625\times10^{-6}$ | 0.0141 | | C | $5.625\times10^{-6}$ | 0.0141 | | D | $1.125\times10^{-5}$ | 0.0000 | Slopes: A is clockwise, B is clockwise, C is anticlockwise, D is anticlockwise. **Answer:** $\theta_A = \theta_D = \dfrac{45}{EI} = 1.125\times10^{-5}$ rad, $\theta_B = \theta_C = \dfrac{22.5}{EI} = 5.625\times10^{-6}$ rad (opposite senses), and $y_B = y_C = \dfrac{56.25}{EI} = 0.0141$ mm downward; $y_A = y_D = 0$. **If the figure's dimensions are used instead** (AB = BC = CD = 2 m, with $I$, $2I$, $I$): $\theta_A = \dfrac{60}{EI} = 1.500\times10^{-5}$ rad, $\theta_B = \dfrac{20}{EI} = 5.000\times10^{-6}$ rad, $y_B = y_C = \dfrac{93.33}{EI} = 0.0233$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the middle portion BC (stiffness 2EI), how is the conjugate beam load obtained from the bending moment? A. M/EI is used unchanged B. M/(2EI) is used, so the loading is halved there C. M x 2EI is used D. The middle portion is ignored Answer: B. The conjugate load is M divided by the local flexural rigidity, so in the 2EI portion the ordinate is M/(2EI). (Key point: Effect of 2I on M/EI) 2. What is the conjugate beam of a simply supported real beam? A. A cantilever of the same length B. A simply supported beam of the same span loaded with M/EI C. A fixed-fixed beam D. A beam with internal hinges at A and D Answer: B. End supports with zero deflection and nonzero slope map to simple supports in the conjugate beam, loaded with M/EI. (Key point: Conjugate of simple supports) 3. In the conjugate beam, the slope at A equals which quantity, and what is its value here? A. Conjugate moment at A, 45/EI B. Conjugate shear (reaction) at A, 45/EI C. Conjugate shear at A, 22.5/EI D. Conjugate moment at A, zero Answer: B. Slope = conjugate shear; R'A = 270/6 = 45/EI by moments of the three M/EI areas about D. (Key point: Slope at A) 4. With EI = 4,000,000 kNm^2, what is the deflection at B (and C) for the 1.5 m / 3 m / 1.5 m reading? A. 0.0141 mm downward B. 0.0141 mm upward C. 0.0563 mm downward D. Zero by symmetry Answer: A. y_B = M'_B = 56.25/EI = 56.25/4,000,000 m = 0.0141 mm, downward; symmetry gives y_C = y_B, not zero. 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