--- title: "Pure bending; deflection of SS beam under UDL" subject: "Strength of Materials" chapter: 7 key: "semester_3/strength_of_materials:pure-bending-udl-deflection" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/pure-bending-udl-deflection site: BCT Engineering --- # Pure bending; deflection of SS beam under UDL Strength of Materials · Chapter 7: Theory of Flexure - Asked in: 2079 Bhadra, 8 marks - Video: [Mechanics of Materials: Lesson 62 - Slope and Deflection Beam Bending Introduction](https://www.youtube.com/watch?v=vd5tpO3eRdg) — Jeff Hanson - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/pure-bending-udl-deflection - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7 ## Question What is pure bending? Explain with suitable example. Determine the maximum deflection in a simply supported beam AB of length L, carrying uniformly distributed load of intensity w kN/m over whole span. ## Short answer Pure bending occurs when a beam segment is under constant bending moment with zero shear, e.g., the middle portion of a simply supported beam with two symmetric point loads. Maximum deflection for a simply supported beam under full-span UDL w is 5wL⁴/(384EI) downward at mid-span. ## Answer #### Pure bending **Pure bending** is the bending of a beam under a constant bending moment with zero shear force over the length considered. Only normal (bending) stresses act on the cross-section; there is no shear stress. Example: a simply supported beam carrying two equal loads $P$ placed symmetrically at distance $a$ from each support. Between the loads the shear force is zero and $M = Pa$ is constant, so that middle portion is in pure bending. ```text P P |<--a-->| | v v A ^----------------------^ B SFD: +P | 0 | -P BMD: rises to Pa, flat, falls ``` #### Maximum deflection of SS beam under UDL Take A as origin, span $L$, load $w$ per unit length. Reactions $R_A = R_B = wL/2$. Bending moment at distance $x$: $$ EI\frac{d^2y}{dx^2} = M = \frac{wL}{2}x - \frac{w x^2}{2} $$ Integrate once (slope): $$ EI\frac{dy}{dx} = \frac{wL x^2}{4} - \frac{w x^3}{6} + C_1 $$ Integrate again (deflection): $$ EI\,y = \frac{wL x^3}{12} - \frac{w x^4}{24} + C_1 x + C_2 $$ Boundary conditions: at $x=0$, $y=0$ gives $C_2 = 0$. At $x=L$, $y=0$: $$ 0 = \frac{wL^4}{12} - \frac{wL^4}{24} + C_1 L \;\Rightarrow\; C_1 = -\frac{wL^3}{24} $$ So $$ y = \frac{w}{24EI}\left(2Lx^3 - x^4 - L^3 x\right) $$ By symmetry the slope is zero and deflection is maximum at mid-span $x = L/2$: $$ y_{max} = \frac{w}{24EI}\left(\frac{2L^4}{8} - \frac{L^4}{16} - \frac{L^4}{2}\right) = -\frac{5wL^4}{384EI} $$ **Answer:** $y_{max} = \dfrac{5wL^4}{384EI}$ downward, at mid-span. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What defines pure bending? A. Constant bending moment with zero shear force B. Constant shear force with zero moment C. Zero moment and shear D. Maximum shear Answer: A. Pure bending has constant M and V = 0, so only normal stresses act. (Key point: Pure bending) 2. Which portion of a simply supported beam with two symmetric equal loads is in pure bending? A. Between the supports and loads B. At the supports C. Between the two loads D. No portion Answer: C. Between the loads the shear is zero and M = Pa is constant. (Key point: Example) 3. How are C1 and C2 found for the simply supported beam under UDL? A. y max at ends B. slope = 0 at x = 0 and y = 0 at L C. slope = 0 at both ends D. y = 0 at x = 0 and x = L Answer: D. C2 = 0 from y(0) = 0, and y(L) = 0 gives C1 = -wL^3/24. (Key point: Boundary conditions) 4. Where and how much is the maximum deflection? A. 5wL^4/(384EI) at mid-span B. wL^4/(48EI) at mid-span C. 5wL^4/(384EI) at quarter span D. wL^4/(8EI) at the support Answer: A. By symmetry slope is zero at mid-span, giving 5wL^4/384EI downward. 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