--- title: "SS beam 6 m: maximum and point deflection" subject: "Strength of Materials" chapter: 7 key: "semester_3/strength_of_materials:ss-beam-6m-max-deflection-20kn" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/ss-beam-6m-max-deflection-20kn site: BCT Engineering --- # SS beam 6 m: maximum and point deflection Strength of Materials · Chapter 7: Theory of Flexure - Asked in: 2068 Baisakh (old course), 8 marks - Video: [Deflection of Beams Problem | Macaulay's Method | simply supported beam | GATE](https://www.youtube.com/watch?v=Qt-lls7jV5I) — Michael Thomas Rex F - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/ss-beam-6m-max-deflection-20kn - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7 ## Question A simply supported beam of 6m span is subjected to a concentrated load of 20kN at a distance of 4m from the left support. Calculate (i) The position and the value of maximum deflection (ii) Deflection under the point load. [Figure: simply supported beam with two 20 kN loads drawn at 2 m and 4 m from the left support (2 m spacing); figure shows 20 kN at 2 m, 2 m and 2 m.] ## Short answer Maximum deflection is 77.42/EI at 3.266 m from the left support; deflection under the 20 kN load is 71.11/EI (EI in kNm², result in m). ## Answer **Reading of the question:** the text states one 20 kN load at 4 m from the left support, so a single load is used. $EI$ is not given, so the answer is in terms of $EI$ (kNm²). $L = 6$ m, $a = 4$ m, $b = 2$ m. ```text 20 kN v A ^---------------^---- B |<---4 m--->|<2m>| ``` #### Reactions $R_A = 20\times 2/6 = 6.67$ kN, $R_B = 13.33$ kN. #### Position of maximum deflection The maximum deflection lies in the longer segment (A to load), where slope is zero. Measured from A: $$ x = \sqrt{\frac{L^2 - b^2}{3}} = \sqrt{\frac{36-4}{3}} = 3.266\ \text{m} $$ This is less than 4 m, so it lies in the longer segment. #### Maximum deflection $$ y_{max} = \frac{P\,b\,x\,(L^2 - b^2 - x^2)}{6EIL} = \frac{20\times2\times3.266\,(36-4-10.667)}{6\times6\,EI} = \frac{77.42}{EI}\ \text{(kNm}^3\text{)} $$ #### Deflection under the load $$ y_C = \frac{P\,a^2 b^2}{3EIL} = \frac{20\times16\times4}{3\times6\,EI} = \frac{71.11}{EI}\ \text{(kNm}^3\text{)} $$ **Answer: (i) maximum deflection $77.42/EI$ at 3.266 m from the left support; (ii) deflection under the load $71.11/EI$** (with $EI$ in kNm², result in m). *If the figure is read as two 20 kN loads at 2 m and 4 m, the beam is symmetrical: maximum deflection at mid-span $= Pa(3L^2-4a^2)/24EI = 153.33/EI$, and under each load $= 133.33/EI$.* ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What load case does the answer use? A. Two 20 kN loads at 2 m and 4 m B. A 20 kN load at mid span C. A single 20 kN load at 4 m from the left support, with EI left symbolic D. EI given as 200 GPa x section Answer: C. The question text states one load, so the answer uses a = 4 m, b = 2 m and gives results in terms of EI. (Key point: Reading of the question) 2. Where does the maximum deflection occur? A. Under the load at 4 m B. At mid span, 3 m from A C. At x = sqrt((L^2 - a^2)/3) from A D. In the longer segment at x = sqrt((L^2 - b^2)/3) = 3.266 m from A, where slope is zero Answer: D. Maximum deflection is where slope is zero, which lies in the longer segment at 3.266 m from A. (Key point: Position of maximum deflection) 3. What is the deflection under the point load? A. P a^2 b^2/(3EIL) = 71.11/EI B. 77.42/EI C. 153.33/EI D. P a b/(4EI) = 40/EI Answer: A. Using y_C = P a^2 b^2/(3EIL) = 20x16x4/(3x6 EI) = 71.11/EI. (Key point: Deflection under load) 4. How do the maximum deflection and deflection under the load compare? A. They are equal B. Maximum (77.42/EI) is slightly larger than under the load (71.11/EI) C. Deflection under the load is larger D. Maximum deflection is zero Answer: B. Maximum deflection is at slope zero, not at the load, and is slightly larger. 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