--- title: "Moment area: overhang beam free-end deflection" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:moment-area-overhang-10kn-m-2078" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-overhang-10kn-m-2078 site: BCT Engineering --- # Moment area: overhang beam free-end deflection Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2078 Chaitra, 6 marks - Video: [Area Moment Method (Overhanging Beam - Deflection at the Free End) - PART 1](https://www.youtube.com/watch?v=t04b_1uwpAo) — 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/moment-area-overhang-10kn-m-2078 - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Calculate vertical deflection at free end of the given overhanging beam using moment area method. Take EI to be constant. [Figure: beam ABC; A hinge and B roller 4 m apart; AB (2EI) carries 10 kN/m UDL; overhang BC = 2 m (EI) carries 100 kN at C.] ## Short answer Applying the moment-area method, the tangent at A yields θA = 53.33/EI anticlockwise. The vertical deflection at the free end C is 506.67/EI downward. ## Answer #### Given data Overhanging beam ABC: A hinge, B roller, $AB = 4$ m ($2EI$, UDL 10 kN/m), overhang $BC = 2$ m ($EI$) with 100 kN at C. ```text 10 kN/m (2EI) 100 kN A =============== B ------------v C ^ 4 m ^ 2 m (EI) ``` #### Step 1: Reactions $R_{A} = -30.00$ kN, $R_{B} = 170.00$ kN (positive = upward; check: $\sum F_y = 140.00$ kN equals the total load 140.00 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 | $- 5 x^{2} - 30 x$ | 2EI | -173.33 | 2.769 | | 4 to 6 | $100 x - 600$ | EI | -200.00 | 4.667 | (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: Tangent at A Deviation of the support at $x = 4$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support): $$ t = (-173.33)(1.231) = \frac{-213.33}{EI} $$ That support does not deflect, so $\theta_A = -t/4$: $$ \theta_A = \frac{53.333}{EI}\ \text{(anticlockwise)} $$ **At C (free end) ($x = 6$ m):** $$ \theta = \theta_A + \sum A = 53.333 + [(-173.33) + (-200.00)] = \frac{-320.000}{EI}\ \text{(clockwise)} $$ $$ y = \theta_A\,x + t_{P/A} = 320.000 + [(-173.33)(3.231) + (-200.00)(1.333)] = \frac{-506.667}{EI}\ \text{(downward)} $$ **Answer: vertical deflection at the free end C $= \dfrac{506.67}{EI}$ (kNm$^3$/EI), downward.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Moments about B give R_A = -30 kN. What does this tell you and what is M at B? A. R_A acts upward 30 kN; M_B = 0 B. R_A acts downward 30 kN (the overhang load lifts A); M_B = -200 kNm hogging C. R_A acts downward; M_B = +200 kNm sagging D. R_A acts downward; M_B = -120 kNm Answer: B. The 100 kN at 2 m on the overhang produces a larger overturning moment than the UDL, so A is pulled down. M_B = 100x2 = 200 kNm hogging. (Key point: Reaction at A and hogging moment) 2. In the table, why is AB's area (-173.33) taken with stiffness 2EI? A. The UDL doubles the rigidity B. Segment AB has rigidity 2EI, so its M/EI values are halved compared with M/EI C. The overhang has 2EI D. EI is constant, so no correction is needed Answer: B. AB has rigidity 2EI while the overhang BC has EI, so M is divided by 2EI in AB and by EI in BC. (Key point: Stiffness in the M/EI area) 3. How is theta_A found? A. theta_A = area of M/EI between A and B B. theta_A = deflection at C divided by 6 C. theta_A = -t/4, using the deviation t of support B from the tangent at A (B does not deflect) D. theta_A = M at B divided by EI Answer: C. B stays on the original level, so tangent offset at B (theta_A x 4) equals the deviation t = -213.33/EI, giving theta_A = 53.33/EI anticlockwise. (Key point: Slope at A) 4. How is the deflection at C obtained? A. Tangent offset theta_A x 6 plus the (signed) moments of the M/EI areas about C, giving 506.67/EI downward B. Only theta_A x 6 = 320/EI C. Only the moment of the areas about A D. Sum of the areas = 373.33/EI Answer: A. y = theta_A x + t_P/A = 320 + (-560.05 - 266.67)... = -506.67/EI; the negative sign means downward. 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