--- title: "Moment area: deflection at D" subject: "Theory of Structures I" chapter: 4 key: "semester_4/theory_of_structures_i:moment-area-overhang-20kn-10kn-2071o" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/moment-area-overhang-20kn-10kn-2071o site: BCT Engineering --- # Moment area: deflection at D Theory of Structures I · Chapter 4: Deflection of Beams - Asked in: 2071 Magh (old course), 10 marks - Video: [Moment-Area Method - Step by Step Explanation and Solved Example - Deflection and Slope](https://www.youtube.com/watch?v=jkwPzvWsqIY) — 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/moment-area-overhang-20kn-10kn-2071o - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-4 ## Question Determine vertical deflection at D of the beam shown in figure below by using moment area method. Take EI to be constant. [Figure: beam ABCD; A hinge, C roller; AB = 2.5 m with 20 kN at B; BC = 2.5 m; overhang CD = 2 m with 10 kN at D.] ## Short answer By moment-area, the tangent deviation at support C from A gives the slope at A. Integrating M/EI area moments, the vertical deflection at free end D is found to be 30.83/EI kN·m³ downward. ## Answer #### Given data Beam ABCD: A hinge at 0, B at 2.5 m (20 kN), C roller at 5 m, D free end at 7 m with 10 kN. $EI$ constant. ```text 20 kN 10 kN A ----v------------- C --------v D ^ 2.5 m B 2.5 m ^ 2 m ``` #### Step 1: Reactions $R_{A} = 6.00$ kN, $R_{C} = 24.00$ kN (positive = upward; check: $\sum F_y = 30.00$ kN equals the total load 30.00 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 2.5 | $6 x$ | EI | 18.75 | 1.667 | | 2.5 to 5 | $50 - 14 x$ | EI | -6.25 | 6.667 | | 5 to 7 | $10 x - 70$ | EI | -20.00 | 5.667 | (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 $x = 5$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support): $$ t = (18.75)(3.333) + (-6.25)(-1.667) = \frac{72.92}{EI} $$ That support does not deflect, so $\theta_A = -t/5$: $$ \theta_A = \frac{-14.583}{EI}\ \text{(clockwise)} $$ **At B ($x = 2.5$ m):** $$ \theta = \theta_A + \sum A = -14.583 + [(18.75)] = \frac{4.167}{EI}\ \text{(anticlockwise)} $$ $$ y = \theta_A\,x + t_{P/A} = -36.458 + [(18.75)(0.833)] = \frac{-20.833}{EI}\ \text{(downward)} $$ **At D (free end) ($x = 7$ m):** $$ \theta = \theta_A + \sum A = -14.583 + [(18.75) + (-6.25) + (-20.00)] = \frac{-22.083}{EI}\ \text{(clockwise)} $$ $$ y = \theta_A\,x + t_{P/A} = -102.083 + [(18.75)(5.333) + (-6.25)(0.333) + (-20.00)(1.333)] = \frac{-30.833}{EI}\ \text{(downward)} $$ **Answer: vertical deflection at D $= \dfrac{30.83}{EI}$ (kNm$^3$/EI), downward.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the reactions at A and C? A. RA = 24 kN, RC = 6 kN B. RA = 6 kN, RC = 24 kN C. RA = 10 kN, RC = 20 kN D. RA = 15 kN, RC = 15 kN Answer: B. Moments about A: RC x 5 = 20 x 2.5 + 10 x 7 = 120, so RC = 24 kN and RA = 6 kN. (Key point: Reactions) 2. How is the slope at A obtained? A. From the deviation of the support C from the tangent at A, theta_A = -t/5 B. From the area of the overhang M/EI diagram C. From theta_A = 0 because A is a hinge D. From the deflection at D divided by 7 Answer: A. C does not deflect, so theta_A = -t/5 = -72.92/(5EI) = -14.583/EI. (Key point: Tangent at A) 3. How is the deflection at the free end D found? A. Only the overhang area is used B. y = theta_A x + sum of moments of M/EI areas about D, over the whole length from A C. Using Theorem I only D. Using the slope at C Answer: B. y_D = theta_A(7) + the moment about D of all M/EI areas from A to D, giving -30.833/EI. (Key point: Method for D) 4. What is the deflection at D? A. 20.83/EI upward B. 30.83/EI downward C. 22.08/EI downward D. 102.08/EI downward Answer: B. -102.083 + 71.25 = -30.833/EI, i.e. 30.83/EI downward. 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