--- title: "Cantilever deflection by integration and superposition" subject: "Strength of Materials" chapter: 7 key: "semester_4/strength_of_materials_mechanical:cantilever-udl-point-load-deflection" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/cantilever-udl-point-load-deflection site: BCT Engineering --- # Cantilever deflection by integration and superposition Strength of Materials · Chapter 7: Deflection of Beams by Integration Method - Asked in: year not given, 3+3 marks - Video: [Superposition is WAY Better than Double Integration for Beam Deflection Problems](https://www.youtube.com/watch?v=9K6YXZK5pN4) — Brian Bernard - 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/cantilever-udl-point-load-deflection - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-7 ## Question (a) Using the double integration method, derive the slope and deflection at the free end of a cantilever of length L carrying a uniformly distributed load w over its whole length. (b) A cantilever of length 3 m, E = 200 GPa and I = 8 x 10^7 mm^4, carries a UDL of 10 kN/m over the entire span and a point load of 15 kN at the free end. Using the principle of superposition, find the slope and deflection at the free end. ## Short answer Free-end slope for UDL: wL³/(6EI), deflection wL⁴/(8EI). Superposition: For EI=16000 kN·m², total slope=7.03×10⁻³ rad, total deflection=14.8 mm downward. ## Answer #### (a) Integration Take the origin at the fixed end A and $x$ towards the free end B. The bending moment at a section is hogging, from the load on the part to its right: $$ M = -\frac{w(L-x)^2}{2} $$ $$ EI\frac{d^2y}{dx^2} = -\frac{w(L-x)^2}{2} $$ $$ EI\frac{dy}{dx} = \frac{w(L-x)^3}{6} + C_1, \qquad EI\,y = -\frac{w(L-x)^4}{24} + C_1x + C_2 $$ At the fixed end $x = 0$: slope $= 0$ and $y = 0$. $$ C_1 = -\frac{wL^3}{6}, \qquad C_2 = \frac{wL^4}{24} $$ At the free end $x = L$: $$ EI\,\theta_B = -\frac{wL^3}{6} \;\Rightarrow\; |\theta_B| = \frac{wL^3}{6EI}, \qquad EI\,y_B = -\frac{wL^4}{6}+\frac{wL^4}{24} = -\frac{wL^4}{8} \;\Rightarrow\; |y_B| = \frac{wL^4}{8EI} $$ The negative signs mean the free end slopes and deflects downward. #### (b) Superposition Total deflection is the sum of the deflections due to each load acting alone (valid for linear elastic behaviour and small deflections). $EI = 200\times10^9\times8\times10^{-5} = 16\times10^6\ \text{N m}^2 = 16\,000\ \text{kN m}^2$. | Load | Slope at free end | Deflection at free end | |---|---|---| | UDL $w=10$ kN/m | $\dfrac{wL^3}{6EI} = \dfrac{10\times27}{6\times16000} = 2.81\times10^{-3}$ rad | $\dfrac{wL^4}{8EI} = \dfrac{10\times81}{8\times16000} = 6.33$ mm | | Point load $P=15$ kN | $\dfrac{PL^2}{2EI} = \dfrac{15\times9}{2\times16000} = 4.22\times10^{-3}$ rad | $\dfrac{PL^3}{3EI} = \dfrac{15\times27}{3\times16000} = 8.44$ mm | | **Total** | $7.03\times10^{-3}$ rad | **14.77 mm** | **Answer: slope $= 7.03\times10^{-3}$ rad $(0.40^\circ)$, deflection $= 14.8$ mm downward at the free end.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With the origin at the fixed end, what is the bending moment at section x for a cantilever with UDL w over L? A. M = +w(L - x)²/2, sagging B. M = -wx²/2, hogging C. M = -w(L - x)²/2, hogging D. M = -wL(L - x), hogging Answer: C. The load on the part to the right of the section, w(L - x), acts at its centre and hogs the beam. (Key point: Bending moment) 2. What are the free-end slope and deflection for the UDL cantilever? A. wL³/3EI and wL⁴/8EI B. wL³/6EI and wL⁴/3EI C. wL²/2EI and wL³/3EI D. wL³/6EI and wL⁴/8EI Answer: D. Applying y = y' = 0 at the wall and putting x = L gives |θB| = wL³/6EI and |yB| = wL⁴/8EI. (Key point: Free-end results) 3. For L = 3 m, EI = 16 000 kN m², what are the deflections from the 10 kN/m UDL and the 15 kN point load? A. 6.33 mm and 8.44 mm B. 8.44 mm and 6.33 mm C. 12.7 mm and 4.22 mm D. 6.33 mm and 5.63 mm Answer: A. wL⁴/8EI = 10×81/(8×16000) = 6.33 mm and PL³/3EI = 15×27/(3×16000) = 8.44 mm. (Key point: Superposition numbers) 4. What does superposition require, and what is the total deflection? A. Any material behaviour; total 14.8 mm downward B. Linear elastic behaviour and small deflections; total 14.8 mm downward C. Linear elastic behaviour and small deflections; total 8.44 mm downward D. Linear elastic behaviour; the larger of the two, 8.44 mm Answer: B. Effects of each load add when response is linear: 6.33 + 8.44 = 14.77 mm. 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