--- title: "Arch 32 m: two rolling loads, max +/- moment at 8 m" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-32m-two-rolling-loads-8m" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-32m-two-rolling-loads-8m site: BCT Engineering --- # Arch 32 m: two rolling loads, max +/- moment at 8 m Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2075 Bhadra, 10 marks - Video: [CE 202 module 6 part 8 Problems on moving loads and ILD on Arches](https://www.youtube.com/watch?v=efXF5YBKUgI) — BEENA MOL - 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-32m-two-rolling-loads-8m - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question A three hinged symmetric parabolic arch has a span of 32 m and a rise of 8 m. It is subjected to two rolling loads of magnitudes 80 kN and 40 kN separately by a distance of 3 m. The load moves from right to left with 80 kN load leading. Determine the maximum positive moment and negative moment at a section 8 m from the left support. ## Short answer 32 m span, 8 m rise three-hinged parabolic arch with rolling loads 80 kN (leading) and 40 kN, 3 m apart moving right to left. At 8 m: max +BM = 285 kN m (V=46.96 kN, N=98.95 kN) when loads at 8 m and 11 m; max −BM = –225 kN m (V=0, N=125.78 kN) when loads at crown and 19 m. ## Answer **Data:** symmetrical parabolic three-hinged arch, $L = 32$ m, $f = 8$ m, $y = \dfrac{4\times 8\,x(32-x)}{32^2} = \dfrac{x(32-x)}{32}$. Two loads, 80 kN (leading) and 40 kN, 3 m apart, move from right to left (so the 40 kN load is 3 m to the right of the 80 kN load). Section at $x = 8$ m from the left support: $y = 6$ m, $\tan\theta = \dfrac{32-16}{32} = 0.5$, $\theta = 26.57^\circ$. #### ILDs for the section at x = 8 m Thrust ILD: $H = \dfrac{x'}{2f}$ (load left of the crown) or $\dfrac{L-x'}{2f}$ (right of the crown), peak $\dfrac{L}{4f} = 1$ at the crown. Moment ILD: $M = M_b - H\,y$. | x (m) | Point | ILD of H | ILD of M at section | |---|---|---|---| | 0 | A | 0 | 0 | | 8 | section D | 0.5 | 3 | | 12.8 | neutral point | 0.8 | 0 | | 16 | crown C | 1 | -2 | | 32 | B | 0 | 0 | The ILD of $M$ is zero at the neutral point, 12.8 m from $A$; it is positive on one side and negative on the other. #### Maximum positive bending moment The critical position is: 80 kN at 8 m, 40 kN at 11 m (measured from $A$). | Load (kN) | Position (m) | Ordinate of M-ILD (m) | P x y (kN m) | Ordinate of H-ILD | |---|---|---|---|---| | 80 | 8 | 3 | 240 | 0.5 | | 40 | 11 | 1.125 | 45 | 0.688 | $$ M = \sum P\,y = 285\ \text{kN m}, \qquad H = \sum P\,y_H = 67.5\ \text{kN} $$ With $y = 6$ m, $\theta = 26.57^\circ$ and the simple-beam shear $V_b = 86.25$ kN at the section: $$ \begin{aligned} V &= V_b\cos\theta - H\sin\theta = 46.96\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 98.95\ \text{kN} \end{aligned} $$ #### Maximum negative bending moment The critical position is: 80 kN at 16 m, 40 kN at 19 m (measured from $A$). | Load (kN) | Position (m) | Ordinate of M-ILD (m) | P x y (kN m) | Ordinate of H-ILD | |---|---|---|---|---| | 80 | 16 | -2 | -160 | 1 | | 40 | 19 | -1.625 | -65 | 0.812 | $$ M = \sum P\,y = -225\ \text{kN m}, \qquad H = \sum P\,y_H = 112.5\ \text{kN} $$ With $y = 6$ m, $\theta = 26.57^\circ$ and the simple-beam shear $V_b = 56.25$ kN at the section: $$ \begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 125.78\ \text{kN} \end{aligned} $$ **Answer: maximum positive moment $= +285$ kN m (80 kN at the section, 40 kN 3 m to its right); maximum negative moment $= -225$ kN m (80 kN at 16 m, i.e. at the crown, and 40 kN at 19 m).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The loads move from right to left with the 80 kN load leading. Where is the 40 kN load relative to the 80 kN? A. 3 m to the right of the 80 kN load B. 3 m to the left of the 80 kN load C. At the same point D. 3 m to the right, only when it is on the crown Answer: A. Moving left, the leading load is the leftmost, so the 40 kN trails 3 m behind on the right. (Key point: Load order) 2. Where is the heavier load placed for maximum positive moment at the section 8 m from A? A. At the crown (16 m) B. At the section (peak ordinate of the ILD, 3 m), with 40 kN at 11 m C. At the neutral point (12.8 m) D. At the right support Answer: B. The positive ILD peaks at the section. Putting the 80 kN there and the 40 kN at 11 m gives 240 + 45 = 285 kN m. (Key point: Critical position for max positive moment) 3. Where is the 80 kN load placed for maximum negative moment? A. At the section (8 m) B. At the neutral point (12.8 m) C. At the crown (16 m), where the ILD ordinate is -2 m D. At B (32 m) Answer: C. The negative ILD is greatest at the crown (-2 m). With 80 kN at the crown and 40 kN at 19 m, M = -160 - 65 = -225 kN m. (Key point: Critical position for max negative moment) 4. At the neutral point (12.8 m from A) what is the ILD ordinate of the bending moment at the section? A. Zero B. Maximum positive (3 m) C. Maximum negative (-2 m) D. Equal to the H ordinate (0.8) Answer: A. The M-ILD changes sign at the neutral point, so a load there causes no moment at the section. 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