--- title: "Beam with two symmetrical loads by moment-area" subject: "Strength of Materials" chapter: 8 key: "semester_4/strength_of_materials_mechanical:symmetrical-beam-two-loads-moment-area" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/symmetrical-beam-two-loads-moment-area site: BCT Engineering --- # Beam with two symmetrical loads by moment-area Strength of Materials · Chapter 8: Deflection of Beams by Moment- area Method - Asked in: year not given, 6 marks - Video: [Moment-Area Method Made Simple | Complete Derivation, Beam Slope & Deflection](https://www.youtube.com/watch?v=kBKLLwhtlfM) — Cipher Spectrum - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/symmetrical-beam-two-loads-moment-area - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-8 ## Question A simply supported beam of span L carries two equal point loads P, each at a distance a from the nearer support. Using the moment-area method, derive expressions for the slope at the supports and the deflection at mid-span. Evaluate them for L = 6 m, a = 2 m, P = 30 kN and EI = 25 000 kN m^2. ## Short answer Slope at supports = Pa(L–a)/(2EI); mid-span deflection = Pa(3L²–4a²)/(24EI). For L=6 m, a=2 m, P=30 kN, EI=25000 kN·m²: slopes = 4.8×10⁻³ rad, maximum deflection = 9.2 mm downward. ## Answer #### Set-up ```text P P | | v____________v A a | L-2a a B R_A = P R_B = P ``` By symmetry $R_A = R_B = P$. Bending moment: rises linearly from 0 at A to $Pa$ at the load, then constant $Pa$ between the loads. The $M/EI$ diagram is a trapezoid, symmetrical about mid-span C. #### Slope at support A By symmetry the slope at mid-span C is zero (the tangent at C is horizontal). By Theorem I, the slope at A equals the area of the $M/EI$ diagram between A and C: $$ \theta_A = \frac1{EI}\left[\frac12\,a\,(Pa) + \left(\frac L2-a\right)(Pa)\right] = \frac{Pa(L-a)}{2EI} $$ #### Deflection at mid-span The deflection at C equals the deviation of A from the tangent at C, i.e. the moment of the area between A and C about A (Theorem II): $$ y_C = \frac{1}{EI}\left[\frac{Pa^2}{2}\cdot\frac{2a}{3} + Pa\left(\frac L2-a\right)\left(\frac{L}{4}+\frac a2\right)\right] $$ The first term is the triangle (centroid $2a/3$ from A); the second is the rectangle (centroid at $a + \frac12(\frac L2 - a) = \frac L4+\frac a2$). $$ y_C = \frac{1}{EI}\left[\frac{Pa^3}{3} + \frac{PaL^2}{8} - \frac{Pa^3}{2}\right] = \boxed{\frac{Pa\,(3L^2-4a^2)}{24EI}} $$ #### Numerical values $$ \theta_A = \frac{30\times2\times(6-2)}{2\times25000} = 4.8\times10^{-3}\ \text{rad} $$ $$ y_C = \frac{30\times2\times(3\times36-4\times4)}{24\times25000} = \frac{60\times92}{600000} = 9.2\times10^{-3}\ \text{m} $$ **Answer: $\theta_A = \theta_B = 4.8\times10^{-3}$ rad $(0.275^\circ)$; maximum deflection at mid-span $= 9.2$ mm downward.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the tangent at mid-span C horizontal? A. The bending moment is zero at C B. The shear force is maximum at C C. The loading and beam are symmetric, so the slope at C is zero D. The deflection is zero at C Answer: C. Symmetry makes the slope at C vanish, so C gives a reference tangent. (Key point: Zero slope by symmetry) 2. How is θA found? A. The moment of the area about A B. The area of the triangle only, Pa²/2EI C. PaL/EI times the span D. The area of the M/EI diagram between A and C, giving Pa(L - a)/2EI Answer: D. Triangle a·Pa/2 plus rectangle (L/2 - a)Pa gives Pa(L - a)/2. (Key point: Slope at support) 3. How is the deflection at C found? A. As the deviation of A from the tangent at C: moment of M/EI area about A B. As the area between A and C C. As the moment of the area about C D. As the slope times L Answer: A. Because the tangent at C is horizontal, yC = t(A/C) = moment of area about A. (Key point: Mid-span deflection method) 4. For L = 6, a = 2, P = 30, EI = 25000, what are θA and yC? A. 9.6×10⁻³ rad and 9.2 mm B. 4.8×10⁻³ rad and 9.2 mm C. 4.8×10⁻³ rad and 18.4 mm D. 2.4×10⁻³ rad and 4.6 mm Answer: B. θA = 60×4/50000 = 4.8×10⁻³ and yC = 60×92/600000 = 9.2×10⁻³ m. 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