--- title: "Conjugate beam: deflection at B, slope at D" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-50kn-20knm-3-spans" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-50kn-20knm-3-spans site: BCT Engineering --- # Conjugate beam: deflection at B, slope at D Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2080 Chaitra, 8 marks - Video: [Beam Deflections - Conjugate Beam Method Example 1- Structural Analysis](https://www.youtube.com/watch?v=Vfuh4YGM57w) — structurefree - 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-50kn-20knm-3-spans - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question For the following beam find deflection at B and slope at D using the conjugate beam method. [Figure: beam ABCD; A hinge, C roller; AB = 4 m (EI) with 50 kN at B; BC = 4 m (EI) with 20 kN/m UDL; overhang CD = 3 m (2EI) with 30 kN at D.] ## Short answer Using the conjugate beam method, the conjugate beam has a hinge at A, an internal hinge at C, and a fixed end at D. The deflection at B is 706.67/EI downward, and the slope at D is 132.5/EI anticlockwise. ## Answer #### Given data Beam ABCD: A hinge at 0, B at 4 m (50 kN down), C roller at 8 m, D free end at 11 m. $AB$ and $BC$ have $EI$; $BC$ carries 20 kN/m; overhang $CD = 3$ m has $2EI$ with 30 kN at D. Required: deflection at B and slope at D. ```text 50 kN 20 kN/m 30 kN A ----v---------------------- C ---------v D ^ 4 m B 4 m (EI) ^ 3 m (2EI) ``` #### Step 1: Reactions $R_{A} = 33.75$ kN, $R_{C} = 126.25$ kN (positive = upward; check: $\sum F_y = 160.00$ kN equals the total load 160.00 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 | $\frac{135 x}{4}$ | EI | 270.00 | 2.667 | | 4 to 8 | $- 10 x^{2} + \frac{255 x}{4} + 40$ | EI | 196.67 | 4.475 | | 8 to 11 | $30 x - 330$ | 2EI | -67.50 | 9.000 | (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 real beam has a hinge at A, a roller at the second support C 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): $$ R'_A\times8 = (270.00)(5.333) + (196.67)(3.525) = 2133.33 $$ $$ R'_A = \frac{266.667}{EI}\ \text{(slope at A, clockwise)} $$ **At B ($x = 4$ m), deflection needed:** $$ V' = R'_A - \sum A = 266.667 - [(270.00)] = \frac{-3.333}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 1066.667 - [(270.00)(1.333)] = \frac{706.667}{EI}\ \text{(deflection, downward)} $$ **At D ($x = 11$ m), slope needed:** $$ V' = R'_A - \sum A = 266.667 - [(270.00) + (196.67) + (-67.50)] = \frac{-132.500}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 2933.333 - [(270.00)(8.333) + (196.67)(6.525) + (-67.50)(2.000)] = \frac{-465.000}{EI}\ \text{(deflection, upward)} $$ **Answer:** deflection at B $= \dfrac{706.67}{EI}$ downward; slope at D $= \dfrac{132.5}{EI}$ (anticlockwise, i.e. the free end tilts upward). (Values in kNm$^3$/EI and kNm$^2$/EI; with $EI$ in kNm$^2$ they are metres and radians.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the conjugate beam for this beam (A hinge, C roller, free tip D)? A. Hinge at A, hinge at C, free end at D B. Hinge at A, internal hinge at C, fixed end at D C. Fixed at A, roller at C, free at D D. Hinge at A, roller at C, fixed at D Answer: B. A stays a hinge, the interior support becomes an internal hinge (M' = 0), and the free tip becomes a fixed end. (Key point: Conjugate supports) 2. Why is the area of the M/EI piece for CD negative (-67.5)? A. Because 2EI is larger than EI B. Because the 30 kN load acts downward C. Because the overhang length is only 3 m D. Because the moment on the overhang is hogging (M = 30x - 330 is negative), so its conjugate load acts upward Answer: D. A downward load on an overhang gives hogging moment (-90 kNm at C), shown as negative M/EI and upward conjugate load, divided by 2EI. (Key point: Sign of M/EI on the overhang) 3. How is the deflection at B found from R'_A = 266.667/EI? A. Shear at B: 266.667 - 270 = -3.333 B. Conjugate moment at B: 266.667x4 - 270x1.333 = 706.67/EI, downward C. Moment at B equal to 270 x 4 D. R'_A divided by 4 Answer: B. Deflection is conjugate moment: R'_A times 4 m minus the triangle area (270) times its distance from B (1.333 m) = 706.67/EI. (Key point: Deflection at B) 4. The conjugate shear at the fixed end D is -132.5/EI. What is the slope at D? A. 132.5/EI anticlockwise (negative = anticlockwise) B. 132.5/EI clockwise C. 465/EI anticlockwise D. Zero, since D is free Answer: A. Slope = conjugate shear; positive is clockwise, so the negative value means an anticlockwise slope of 132.5/EI (the tip tilts upward). 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