--- title: "BM, shear, thrust at any section, UDL on arch" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-general-section-udl-formula" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-general-section-udl-formula site: BCT Engineering --- # BM, shear, thrust at any section, UDL on arch Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2076 Baisakh, 6 marks - Video: [Two Hinged Arch with UDL - Horizontal Thrust Derivation and Prove BM at any section is 0](https://www.youtube.com/watch?v=CHvrnnwOCF4) — Stan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-general-section-udl-formula - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question Find the bending moment, radial shear and normal thrust at any section of a three hinged symmetrical parabolic arch loaded with uniformly distributed load w kN/m² over its entire span. Neglect the self-weight of the arch. ## Short answer For a three-hinged symmetric parabolic arch under full span UDL, bending moment and radial shear are zero at all sections. Normal thrust N = H/cosθ, where H = wL²/(8f). N is minimum at the crown (equal to H) and maximum at the springings, H√(1+16f²/L²). ## Answer **Data:** three-hinged symmetrical parabolic arch, span $L$, rise $f$, supports at the same level, origin at the left support. UDL $w$ per unit horizontal length over the whole span (if the load is given per m², multiply by the width of the arch to get $w$ per unit length). Self-weight neglected. #### Arch equation and slope $$ y = \frac{4f\,x\,(L-x)}{L^2}, \qquad \tan\theta = \frac{dy}{dx} = \frac{4f\,(L-2x)}{L^2} $$ #### Reactions and thrust $$ R_A = R_B = \frac{wL}{2}, \qquad H = \frac{M_C}{f} = \frac{1}{f}\left[\frac{wL}{2}\cdot\frac{L}{2} - \frac{wL}{2}\cdot\frac{L}{4}\right] = \frac{wL^2}{8f} $$ #### Bending moment $$ M = \frac{wLx}{2} - \frac{wx^2}{2} - H\,y = \frac{w\,x\,(L-x)}{2} - \frac{wL^2}{8f}\cdot\frac{4f\,x\,(L-x)}{L^2} = 0 $$ #### Radial shear The simple-beam shear at $x$ is $V_b = w\left(\dfrac{L}{2} - x\right)$. $$ V = V_b\cos\theta - H\sin\theta = \cos\theta\left[w\left(\frac{L}{2}-x\right) - \frac{wL^2}{8f}\cdot\frac{4f\,(L-2x)}{L^2}\right] = \cos\theta\left[\frac{w\,(L-2x)}{2} - \frac{w\,(L-2x)}{2}\right] = 0 $$ #### Normal thrust $$ N = V_b\sin\theta + H\cos\theta = H\tan\theta\sin\theta + H\cos\theta = \frac{H}{\cos\theta} $$ $$ N = \frac{wL^2}{8f}\sqrt{1 + \frac{16f^2\,(L-2x)^2}{L^4}} $$ **Results:** $M = 0$, $V = 0$ and $N = \dfrac{H}{\cos\theta}$ at every section. $N$ is smallest at the crown ($N = H = \dfrac{wL^2}{8f}$) and largest at the springings, where $N = H\sqrt{1+\left(\dfrac{4f}{L}\right)^2}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is H for the three-hinged parabolic arch under a full-span UDL w? A. wL^2/(8f) B. wL/(2f) C. wL^2/(4f) D. wL^2/(16f) Answer: A. H = M_C/f = (wL^2/8)/f. (Key point: Thrust) 2. What are the bending moment and radial shear at any section? A. M = wL^2/8, V = wL/2 B. Both zero C. M = 0 and V = H D. M maximum at crown Answer: B. The beam moment is cancelled by H y and the beam shear component is cancelled by H sin(theta), so both vanish. (Key point: Results for M and V) 3. Which expression gives N? A. N = H cos(theta) B. N = H/cos(theta) C. N = H sin(theta) D. N = wL/2 Answer: B. N = V_b sin(theta) + H cos(theta) = H tan(theta) sin(theta) + H cos(theta) = H/cos(theta). (Key point: Normal thrust) 4. Where is the normal thrust smallest and largest? A. Smallest at springings, largest at crown B. Smallest at crown (N = H), largest at springings C. Constant everywhere D. Largest at quarter span Answer: B. At the crown theta = 0 so N = H; at the springings theta is largest, so 1/cos(theta) is largest. 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