--- title: "Conjugate beam: overhanging beam free end" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:conjugate-overhang-300-150" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/conjugate-overhang-300-150 site: BCT Engineering --- # Conjugate beam: overhanging beam free end Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2079 Chaitra, 10 marks - Video: [Conjugate Beam Method (Ex 2 - Overhanging Beam)](https://www.youtube.com/watch?v=ekiKJFUpGdU) — Engr. Clidez - 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-overhang-300-150 - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Determine deflection and rotation at free end of the overhanging beam shown in figure below. Use conjugate beam method. [Figure: beam ABCD; A hinge, C roller; AB = 6 m and BC = 6 m with flexural rigidity 2EI and 300 kN at B; overhang CD = 2 m (EI) with 150 kN at D.] ## Short answer By the conjugate beam method, the conjugate beam has hinges at A and C and a fixed end at the overhang tip. The free end D deflects upward by 1100/EI and rotates 450/EI anticlockwise. ## Answer #### Given data Beam ABCD: A hinge at 0, B at 6 m (300 kN), C roller at 12 m, D free end at 14 m with 150 kN. $AC$ ($= 12$ m) has $2EI$; overhang $CD = 2$ m has $EI$. ```text 300 kN 150 kN A ------v------------------ C --------v D ^ 6 m B 6 m (2EI) ^ 2 m (EI) ``` #### Step 1: Reactions $R_{A} = 125.00$ kN, $R_{C} = 325.00$ kN (positive = upward; check: $\sum F_y = 450.00$ kN equals the total load 450.00 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 6 | $125 x$ | 2EI | 1125.00 | 4.000 | | 6 to 12 | $1800 - 175 x$ | 2EI | 675.00 | 6.667 | | 12 to 14 | $150 x - 2100$ | EI | -300.00 | 12.667 | (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 $x = 12$ m (conjugate moment is zero there): $$ R'_A\times12 = (1125.00)(8.000) + (675.00)(5.333) = 12600.00 $$ $$ R'_A = \frac{1050.000}{EI}\ \text{(slope at A, clockwise)} $$ **At B (300 kN) ($x = 6$ m):** $$ V' = R'_A - \sum A = 1050.000 - [(1125.00)] = \frac{-75.000}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 6300.000 - [(1125.00)(2.000)] = \frac{4050.000}{EI}\ \text{(deflection, downward)} $$ **At D (free end) ($x = 14$ m):** $$ V' = R'_A - \sum A = 1050.000 - [(1125.00) + (675.00) + (-300.00)] = \frac{-450.000}{EI}\ \text{(slope, anticlockwise)} $$ $$ M' = R'_A\,x - \sum A(x-\bar x) = 14700.000 - [(1125.00)(10.000) + (675.00)(7.333) + (-300.00)(1.333)] = \frac{-1100.000}{EI}\ \text{(deflection, upward)} $$ **Answer:** at the free end D the deflection is $\dfrac{1100}{EI}$ **upward** (the 150 kN load is small compared with the span load, so the overhang rises), and the rotation is $\dfrac{450}{EI}$ anticlockwise (kNm$^3$/EI and kNm$^2$/EI). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For beam ABCD with hinge A, roller C and free end D, which conjugate beam is used? A. Hinge at A, internal hinge at C, fixed end at D B. Hinge at A, roller at C, free end at D C. Fixed at A, hinge at C, free at D D. Same as the real beam Answer: A. Interior support C becomes an internal hinge and the free end D becomes a fixed end. (Key point: Conjugate supports) 2. How is the area 1125 of the portion A to B obtained? A. 1/2 x 6 x 750 (M not divided by stiffness) B. 6 x 375 C. 1/2 x 6 x 375, since M at B = 125x6 = 750 kNm divided by 2EI D. 1/2 x 6 x 125 Answer: C. M_B = 750 kNm; with rigidity 2EI the M/EI ordinate is 375/EI, and the triangle area is 0.5 x 6 x 375 = 1125/EI. (Key point: M/EI area with 2EI) 3. R'_A = 1050/EI. How is the slope at D found? A. Take total conjugate load only: 1125 + 675 + 300 B. V' at D = 1050 - (1125 + 675 - 300) = -450/EI, i.e. 450/EI anticlockwise C. V' at D = 1050 - 1125 - 675 - 300 D. Take the moment at D, 1100/EI Answer: B. Slope is the conjugate shear at the fixed end. The overhang area (-300) is hogging and acts upward, so V' = 1050 - 1500 = -450/EI. (Key point: Slope at free end) 4. The conjugate moment at D is -1100/EI. What does this mean for the real beam? A. D deflects 1100/EI downward B. D deflects 1100/EI upward, because a negative conjugate moment means upward deflection C. D has no deflection D. D rotates but does not deflect Answer: B. Positive conjugate moment is downward deflection, so the negative value means D rises by 1100/EI. The large span moment lifts the overhang more than the 150 kN load pushes it down. (Key point: Direction of end deflection) ## Related in Theory of Structures I - [Conjugate beam: slope and deflection at free end C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-udl-15-3kn-2076b) — Theory of Structures I, chapter 4, asked 2× - [Moment area: overhang beam free-end deflection](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-overhang-10kn-m-2078) — Theory of Structures I, chapter 4 - [Conjugate beam: slope and deflection at free end C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-udl-2knm-10kn-5kn-2068) — Theory of Structures I, chapter 4, asked 2× - [Conjugate beam: slope and deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-slope-defl-free-end-2073) — Theory of Structures I, chapter 4 - [Conjugate beam: slope and deflection at free end](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-beam-24kn-10knm) — Theory of Structures I, chapter 4 - [Conjugate beam: vertical deflection and rotation at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aconjugate-udl-5knm-overhang-2071m) — Theory of Structures I, chapter 4 ## Related in other subjects - [Flexural rigidity; T-beam bending stresses (60 kN loads)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aflexural-rigidity-t-beam-60kn) — Strength of Materials, chapter 7 - [Rotation and deflection at free end, BMD](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Abeam-fem-rotation-deflection-free-end) — Computational Techniques in Civil Engineering, chapter 4 - [Slope deflection: beam with overhang](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-40kn-30kn) — Theory of Structures II, chapter 3 - [Slope deflection: continuous beam, 90 kN and 30 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-15knm-90kn-30kn) — Theory of Structures II, chapter 3