--- title: "Circular arch 100 m: rolling UDL 25 m, 15 m section" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-circular-100m-rolling-50knm-25m" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-circular-100m-rolling-50knm-25m site: BCT Engineering --- # Circular arch 100 m: rolling UDL 25 m, 15 m section Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2077 Chaitra, 12 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-100m-rolling-50knm-25m - 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 100 m and a rise of 10 m. It is subjected to a rolling load of 50 kN/m of span 25 m moving from left to right. Determine maximum horizontal thrust and maximum bending moment at 15 m from left support with the help of influence line diagram. Also determine absolute maximum bending moment. ## Short answer Maximum horizontal thrust is 2734.4 kN. At 15 m from left support, maximum bending moment is +7283 kN m and minimum is -5519 kN m; absolute maximum bending moment is about 8061 kN m. ## Answer **Data:** three-hinged symmetrical circular arch, $L = 100$ m, $f = 10$ m. Radius $R = \dfrac{50^2+10^2}{2\times 10} = 130$ m. Rolling UDL 50 kN/m, length 25 m, moving left to right. Section at $x = 15$ m: $y_D = \sqrt{130^2-35^2} - 120 = 5.2$ m. #### Maximum horizontal thrust The ILD of $H$ is a triangle with peak $\dfrac{L}{4f} = 2.5$ at the crown. $H$ is maximum when the load is placed symmetrically about the crown (from 37.5 m to 62.5 m), where the ordinates are $1.875$ at the ends and $2.5$ at the middle: $$ \text{Area} = 2\times\frac{(1.875+2.5)}{2}\times 12.5 = 54.6875\ \text{m}, \qquad H_{max} = 50\times 54.6875 = 2734.4\ \text{kN} $$ #### 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 $\dfrac{L}{4f} = 2.5$ 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 | 8.85 | | 36.59 | neutral point | 1.829 | 0 | | 50 | crown C | 2.5 | -5.5 | | 100 | B | 0 | 0 | The ILD of $M$ crosses zero at the **neutral point**, 36.59 m from $A$ (vertically below the intersection of line $AD$ produced and line $BC$). The ordinates are positive on one side of it and negative on the other. #### Maximum positive bending moment The 25 m load is shorter than the positive region, so it is placed with equal ILD ordinates at its two ends: it covers $x = 4.75$ m to 29.75 m (the whole load is on the span). $$ \begin{aligned} M &= w\times(\text{area of ILD}) = 50\times 145.66 = 7283\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times 21.5625 = 1078.13\ \text{kN} \end{aligned} $$ At the section: $y = 5.2$ m, $\theta = 15.62^\circ$, simple-beam shear $V_b = 521.88$ kN. $$ \begin{aligned} V &= V_b\cos\theta - H\sin\theta = 521.88\times0.9631 - 1078.13\times0.2692 = 212.34\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 521.88\times0.2692 + 1078.13\times0.9631 = 1178.82\ \text{kN} \end{aligned} $$ #### Maximum negative bending moment In the same way, the load is placed with equal end ordinates in the negative region: it covers $x = 44.71$ m to 69.71 m (the whole load is on the span). $$ \begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(-110.388) = -5519.4\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times 52.088 = 2604.38\ \text{kN} \end{aligned} $$ At the section: $y = 5.2$ m, $\theta = 15.62^\circ$, simple-beam shear $V_b = 534.86$ kN. $$ \begin{aligned} V &= V_b\cos\theta - H\sin\theta = 534.86\times0.9631 - 2604.38\times0.2692 = -186.07\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 534.86\times0.2692 + 2604.38\times0.9631 = 2652.22\ \text{kN} \end{aligned} $$ #### Absolute maximum bending moment The absolute maximum is found by repeating the above for every section and every position of the UDL (the 25 m load fully on the span). The largest value occurs at a section about 22 m from the support, with the load covering about 7.9 m to 32.9 m of the span: $$ M_{abs} \approx 8061\ \text{kN m (sagging, at } x \approx 22.2\ \text{m}) $$ (The largest hogging moment anywhere is about 6382 kN m, near $x \approx 24$ m, which is smaller.) **Answer: $H_{max} = 2734.4$ kN; $M_{max}$ at 15 m $=+7283$ kN m (and $-5519$ kN m); absolute maximum bending moment $\approx 8061$ kN m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the radius of the circular arch found for span 100 m and rise 10 m? A. R = 100^2/(8 x 10) = 125 m B. R = 50 + 10 = 60 m C. R = (50^2 + 10^2)/(2 x 10) = 130 m D. R = (100^2 + 10^2)/(2 x 10) = 505 m Answer: C. From the geometry of a circular segment, R = (half-span^2 + rise^2)/(2 x rise) = 130 m. (Key point: Radius of circular arch) 2. Where should the 25 m rolling UDL be placed for the maximum horizontal thrust? A. Symmetrically about the crown (37.5 m to 62.5 m) B. From the left support (0 to 25 m) C. Covering the neutral point of the section D. Ending exactly at the crown (25 m to 50 m) Answer: A. The ILD of H is a triangle peaking at the crown, so the largest area under a 25 m length is when it is centred on the crown. H max = 50 x 54.6875 = 2734.4 kN. (Key point: Position for maximum thrust) 3. Where is the UDL placed for the maximum positive moment at the 15 m section, and what is it? A. Centred on the crown, M = +7283 kN m B. On the positive part of the ILD, near the section (4.75 to 29.75 m), M = +7283 kN m C. From the neutral point to B, M = +7283 kN m D. On the span from 0 to 25 m, M = -5519 kN m Answer: B. The positive ILD ordinates lie between A and the neutral point (36.59 m). The load is placed over the highest part of that region and M = w x area = 50 x 145.66 = 7283 kN m. (Key point: Max positive moment at 15 m) 4. How is the absolute maximum bending moment found? A. It is simply the maximum moment at the 15 m section B. It equals H max x rise C. It occurs at the crown where the thrust is greatest D. By repeating the ILD calculation for every section and every load position, then taking the largest (about 8061 kN m at x about 22.2 m) Answer: D. The maximum at one section is not the absolute maximum. Checking all sections shows the largest sagging moment is about 8061 kN m near x = 22 m. (Key point: Absolute maximum moment) ## Close variants set in other exams - [Circular arch 50 m: rolling UDL, 15 m section](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-50m-rolling-50knm-10m) — 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 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 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: 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 - [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× - [Circular arch: two rolling loads 20 kN and 30 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-100m-20kn-30kn-25m) — Theory of Structures I, chapter 6, asked 2× ## 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