--- title: "Macaulay's method for an off-centre load" subject: "Strength of Materials" chapter: 7 key: "semester_4/strength_of_materials_mechanical:simply-supported-beam-macaulay-point-load" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/simply-supported-beam-macaulay-point-load site: BCT Engineering --- # Macaulay's method for an off-centre load Strength of Materials · Chapter 7: Deflection of Beams by Integration Method - Asked in: year not given, 8 marks - Video: [MACAULAY METHOD FOR SLOPE & DEFLECTION SOLVED EXAMPLE 03](https://www.youtube.com/watch?v=XvMPERAfJ-4) — Structural Funda by:- KheNiSiR - 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/simply-supported-beam-macaulay-point-load - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-7 ## Question A simply supported beam of span 6 m carries a point load of 30 kN at 2 m from the left support A. Take EI = 12 000 kN m^2. Using Macaulay's method, determine (a) the slopes at the supports, (b) the deflection under the load, (c) the position and value of the maximum deflection. ## Short answer Reactions 20 kN, 10 kN. Macaulay integration gives support slopes θA = –5.56×10⁻³ rad, θB = 4.44×10⁻³ rad. Deflection under load: 8.89 mm downward. Maximum deflection 9.68 mm downward at 2.73 m from A. ## Answer #### Reactions $$ R_A = \frac{30\times4}{6} = 20\ \text{kN}, \qquad R_B = 10\ \text{kN} $$ #### Macaulay equations (origin at A, $y$ upward, sagging $M$ positive) $$ EI\frac{d^2y}{dx^2} = 20x - 30\langle x-2\rangle $$ $$ EI\frac{dy}{dx} = 10x^2 - 15\langle x-2\rangle^2 + C_1 $$ $$ EI\,y = \frac{10x^3}{3} - 5\langle x-2\rangle^3 + C_1x + C_2 $$ Boundary conditions: - $x = 0$, $y = 0$ gives $C_2 = 0$. - $x = 6$, $y = 0$: $\dfrac{10\times216}{3} - 5\times4^3 + 6C_1 = 0$, so $720 - 320 + 6C_1 = 0$, so $C_1 = -66.67$. #### (a) Slopes At A ($x=0$): $$ \theta_A = \frac{C_1}{EI} = \frac{-66.67}{12000} = -5.56\times10^{-3}\ \text{rad (clockwise)} $$ At B ($x=6$): $$ EI\theta_B = 10(36) - 15(16) - 66.67 = 53.33 \;\Rightarrow\; \theta_B = 4.44\times10^{-3}\ \text{rad} $$ #### (b) Deflection under the load ($x = 2$) $$ EI\,y_C = \frac{10\times8}{3} - 0 - 66.67\times2 = 26.67 - 133.33 = -106.67 $$ $$ y_C = \frac{-106.67}{12000} = -8.89\times10^{-3}\ \text{m} = 8.89\ \text{mm downward} $$ #### (c) Maximum deflection Maximum deflection occurs where the slope is zero. The longer segment is CB, so test $x>2$: $$ 10x^2 - 15(x-2)^2 - 66.67 = 0 \;\Rightarrow\; -5x^2 + 60x - 126.67 = 0 $$ $$ x^2 - 12x + 25.33 = 0 \;\Rightarrow\; x = 6 - \sqrt{36-25.33} = 6 - 3.266 = 2.734\ \text{m} $$ (this lies beyond 2 m, so it is valid.) $$ EI\,y_{max} = \frac{10(2.734)^3}{3} - 5(0.734)^3 - 66.67(2.734) = 68.12 - 1.98 - 182.27 = -116.1 $$ $$ y_{max} = \frac{-116.1}{12000} = -9.68\times10^{-3}\ \text{m} $$ **Answer: $\theta_A = 5.56\times10^{-3}$ rad, $\theta_B = 4.44\times10^{-3}$ rad; $y_{under\ load} = 8.89$ mm; $y_{max} = 9.68$ mm downward at 2.73 m from A.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the Macaulay moment equation for the 6 m beam with 30 kN at 2 m from A? A. EI y'' = 10x - 30 B. EI y'' = 20x + 30 C. EI y'' = 20x - 30 D. EI y'' = 20x - 30 Answer: D. RA = 30×4/6 = 20 kN; the load term starts at x = 2. (Key point: Macaulay moment equation) 2. How are C1 and C2 found? A. x = 0, y = 0 gives C2 = 0; x = 6, y = 0 gives C1 = -66.67 B. x = 0, slope = 0 gives C1 = 0; x = 6, y = 0 gives C2 C. x = 2, y = 0 and x = 6, y = 0 D. Slope = 0 at both supports Answer: A. Deflection is zero at both supports; the slope is not. (Key point: Finding constants) 3. What is the deflection under the load? A. 9.68 mm downward B. 8.89 mm downward, from EI y = 26.67 - 133.33 at x = 2 C. 5.56 mm downward D. 8.89 mm upward Answer: B. At x = 2 the bracket term is zero: EI y = 10(8)/3 - 66.67(2) = -106.67, giving 8.89 mm down. (Key point: Deflection under load) 4. Where is the maximum deflection found, and where is it? A. At the load point, 2 m from A B. At mid-span, 3 m from A C. Where slope is zero, at 2.73 m from A in segment CB D. Where moment is zero Answer: C. Slope zero at x = 2.734 m gives y = 9.68 mm, larger than under the load. 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