--- title: "Circular arch 50 m: rolling UDL, 15 m section" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-circular-50m-rolling-50knm-10m" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-circular-50m-rolling-50knm-10m site: BCT Engineering --- # Circular arch 50 m: rolling UDL, 15 m section Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2073 Bhadra, 10 marks - 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-circular-50m-rolling-50knm-10m - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question A three hinged symmetrical circular arch has a span 50 m and a rise of 10 m. It is subjected to a rolling load of 50 kN/m span 10 m moving from left to right. Determine maximum bending moment, radial shear and normal thrust at 15 m from left support with the help of influence line diagram. ## Short answer Span 50 m, rise 10 m circular three-hinged arch (R=36.25 m) with rolling UDL 50 kN/m over 10 m, left to right. At 15 m: max positive BM = +1534.3 kN m (shear −79.8 kN, thrust 307.5 kN) with load at 7.7–17.7 m; max negative BM = −1345.6 kN m (shear 60.0 kN, thrust 570.7 kN) with load at 23.49–33.49 m. ## Answer **Data:** symmetrical circular three-hinged arch, $L = 50$ m, $f = 10$ m, $R = \dfrac{25^2+10^2}{2\times 10} = 36.25$ m. Rolling UDL 50 kN/m, 10 m long, from left to right. Section at $x = 15$ m: $y = \sqrt{36.25^2-10^2} - 26.25 = 8.593$ m. #### ILDs for the section at x = 15 m The ILD of the horizontal thrust is $H = \dfrac{x'}{2f}$ for a unit load at $x'$ left of the crown and $\dfrac{L-x'}{2f}$ right of it (peak $\dfrac{L}{4f} = 1.25$ at the crown). The ILD of the bending moment at the section is $M = M_b - H\,y$. | x (m) | Point | ILD of H | ILD of M at section | |---|---|---|---| | 0 | A | 0 | 0 | | 15 | section D | 0.75 | 4.055 | | 20.56 | neutral point | 1.028 | 0 | | 25 | crown C | 1.25 | -3.242 | | 50 | B | 0 | 0 | The ILD of $M$ crosses zero at the **neutral point**, 20.56 m from $A$ (the vertical through the intersection of line $A$–$D$ extended and line $B$–$C$ extended). The ordinates are positive on one side of it and negative on the other. #### Maximum positive bending moment The UDL covers the region from x = 7.7 m to 17.7 m (the whole load is on the span); this lies in the positive region of the ILD, with equal ordinates at the two ends of the load. $$ \begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(30.687) = 1534.3\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(6.352) = 317.58\ \text{kN} \end{aligned} $$ At the section: $y = 8.593$ m, $\theta = 16.01^\circ$, simple-beam shear $V_b = 8.13$ kN. $$ \begin{aligned} V &= V_b\cos\theta - H\sin\theta = 8.13\times0.9612 - 317.58\times0.2759 = -79.79\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 8.13\times0.2759 + 317.58\times0.9612 = 307.50\ \text{kN} \end{aligned} $$ #### Maximum negative bending moment The UDL covers the region from x = 23.49 m to 33.49 m (the whole load is on the span); this lies in the negative region of the ILD, with equal ordinates at the two ends of the load. $$ \begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(-26.912) = -1345.6\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(10.641) = 532.05\ \text{kN} \end{aligned} $$ At the section: $y = 8.593$ m, $\theta = 16.01^\circ$, simple-beam shear $V_b = 215.1$ kN. $$ \begin{aligned} V &= V_b\cos\theta - H\sin\theta = 215.1\times0.9612 - 532.05\times0.2759 = 59.98\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 215.1\times0.2759 + 532.05\times0.9612 = 570.74\ \text{kN} \end{aligned} $$ **Answer: maximum positive bending moment $= +1534.3$ kN m (radial shear $-79.8$ kN, normal thrust $307.5$ kN); maximum negative bending moment $= -1345.6$ kN m (radial shear $59.98$ kN, normal thrust $570.7$ kN).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For span 50 m and rise 10 m, what is the radius? A. 25 m B. 36.25 m C. 62.5 m D. 40 m Answer: B. R = (25^2 + 10^2)/(2 x 10) = 725/20 = 36.25 m. (Key point: Radius) 2. What is the peak of the H-ILD? A. L/4f = 1.25 at the crown B. L/2f = 2.5 C. f/L = 0.2 D. 1.0 at the section Answer: A. For L = 50 and f = 10, the peak is 50/40 = 1.25 at the crown. (Key point: Peak of the thrust ILD) 3. Where is the 10 m UDL placed for maximum positive moment at 15 m? A. Centred on the crown B. From the neutral point to B C. Over the highest part of the positive region near the section (7.7 m to 17.7 m) D. Covering the whole span Answer: C. Positive ordinates lie between A and the neutral point (20.56 m); the load is placed where the area is greatest, giving +1534.3 kN m. (Key point: Position for maximum positive moment) 4. Where is the UDL placed for maximum negative moment? A. Between A and the section B. Right at the section C. Over the neutral point D. On the negative region around and past the crown (23.49 m to 33.49 m) Answer: D. The ILD is negative from the neutral point to B with the peak at the crown, so the load goes there: M = -1345.6 kN m. (Key point: Position for maximum negative moment) ## Close variants set in other exams - [Circular arch 100 m: rolling UDL 25 m, 15 m section](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-100m-rolling-50knm-25m) — Theory of Structures I, chapter 6, asked 2× ## Related in Theory of Structures I - [Arch 100 m rise 15 m: rolling UDL 10 m, section 60 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-rise15-udl-10m-60m-section) — Theory of Structures I, chapter 6 - [Arch 160 m: rolling UDL, section 50 m from right](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-160m-udl-30knm-60m-50m) — Theory of Structures I, chapter 6 - [Arch 100 m rise 12 m: rolling UDL 60 m, section 10 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-rise12-udl-60m-10m) — Theory of Structures I, chapter 6 - [Arch 100 m: rolling UDL 60 m, section at 30 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-rolling-udl-60m-25) — Theory of Structures I, chapter 6 - [Circular arch 120 m: two rolling loads, 30 m section](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-120m-8kn-12kn) — Theory of Structures I, chapter 6, asked 2× - [Parabolic arch 100 m: UDL left half and 80 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-80kn-udl-d-20m) — Theory of Structures I, chapter 6, asked 4× ## Related in other subjects - [Two-hinged arch: horizontal reaction, UDL on half](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-12knm-horizontal-reaction-50m-8m) — Theory of Structures II, chapter 2 - [Two-hinged arch: BM at section x-x](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-section-xx-60-85) — Theory of Structures II, chapter 2 - [Two-hinged arch: BM at 10 m intervals](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-80kn-d-i-ic) — Theory of Structures II, chapter 2 - [Two-hinged arch: BM, thrust, radial shear, support yield](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-thrust-shear-yield-support) — Theory of Structures II, chapter 2