--- title: "Arch 100 m rise 15 m: rolling UDL 10 m, section 60 m" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-100m-rise15-udl-10m-60m-section" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-100m-rise15-udl-10m-60m-section site: BCT Engineering --- # Arch 100 m rise 15 m: rolling UDL 10 m, section 60 m Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2076 Bhadra, 14 marks - Video: [Maximum Bending Moment |Horizantal Thrust |Three Hinged Arch at same level|Structural Analysis](https://www.youtube.com/watch?v=YDYyHvLwoDw) — KK Sir ki classes - 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-100m-rise15-udl-10m-60m-section - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6 ## Question A three hinged parabolic arch of span 100 m and rise 15 m is to be designed to carry a rolling load of 50 kN/m of span 10 m. Determine the maximum bending moment, radial shear, and normal thrust at 60 m from the left support with the help of influence line diagrams. ## Short answer For a three-hinged parabolic arch (span 100 m, rise 15 m) with a 10 m UDL of 50 kN/m, influence lines give at 60 m: max positive BM = +2136 kN m (shear 192 kN, thrust 584.6 kN) with load at 58.8–68.8 m; max negative BM = −1817 kN m (shear −136.6 kN, thrust 784.6 kN) with load at 40.83–50.83 m. ## Answer **Data:** parabolic three-hinged arch, $L = 100$ m, $f = 15$ m, $y = 0.006\,x(100-x)$. Rolling UDL 50 kN/m of length 10 m. Section $D$ at $x = 60$ m (in the right half): $y = 14.4$ m, $\tan\theta = 0.006\,(100-120) = -0.12$, so the tangent falls to the right ($\theta = -6.84^\circ$). For a section in the right half, the neutral point lies between the crown and the section. #### ILDs for the section at x = 60 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 $\dfrac{L}{4f} = 1.6667$ 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 | | 50 | crown C | 1.667 | -4 | | 54.55 | neutral point | 1.515 | 0 | | 60 | section D | 1.333 | 4.8 | | 100 | B | 0 | 0 | The ILD of $M$ crosses zero at the **neutral point**, 54.55 m from $A$ (the vertical through the intersection of line $A$–$C$ extended and line $B$–$D$ 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 = 58.8 m to 68.8 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. $$ \begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(42.72) = 2136\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(12.0667) = 603.33\ \text{kN} \end{aligned} $$ At the section: $y = 14.4$ m, $\theta = -6.84^\circ$, simple-beam shear $V_b = 121$ kN. $$ \begin{aligned} V &= V_b\cos\theta - H\sin\theta = 121\times0.9929 - 603.33\times(-0.1191) = 192.02\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 121\times(-0.1191) + 603.33\times0.9929 = 584.62\ \text{kN} \end{aligned} $$ #### Maximum negative bending moment The UDL covers the region from x = 40.83 m to 50.83 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. $$ \begin{aligned} M &= w\times(\text{area of ILD}) = 50\times(-36.333) = -1816.7\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 50\times(15.2537) = 762.69\ \text{kN} \end{aligned} $$ At the section: $y = 14.4$ m, $\theta = -6.84^\circ$, simple-beam shear $V_b = -229.15$ kN. $$ \begin{aligned} V &= V_b\cos\theta - H\sin\theta = (-229.15)\times0.9929 - 762.69\times(-0.1191) = -136.65\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = (-229.15)\times(-0.1191) + 762.69\times0.9929 = 784.55\ \text{kN} \end{aligned} $$ **Answer: at 60 m from the left support the maximum bending moments are $+2136$ kN m (radial shear $192$ kN, normal thrust $584.6$ kN) and $-1817$ kN m (radial shear $-136.6$ kN, normal thrust $784.6$ kN), with the 10 m UDL in the positions stated above.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a section at 60 m in the right half, where does the neutral point lie? A. Between A and the crown B. At the crown C. Between the crown and the section (54.55 m from A) D. Beyond the section, near B Answer: C. For a section right of the crown, the ILD of M is negative at the crown (-4) and positive at the section (+4.8), so it crosses zero in between. (Key point: Neutral point for a section in right half) 2. Why is tan(theta) negative (-0.12) at x = 60 m? A. The section is past the crown, where the arch tangent falls to the right B. The arch is circular there C. The load rolls from right to left D. The thrust acts to the left Answer: A. dy/dx = 0.006(100-2x) changes sign at the crown, x = 50 m. At x = 60 it is -0.12. (Key point: Slope sign) 3. Where is the 10 m rolling load placed for the maximum positive moment at 60 m? A. Centred on the crown (45 m to 55 m) B. On the positive part of the ILD around the section (58.8 m to 68.8 m) C. Entirely left of the neutral point D. From 90 m to 100 m at B Answer: B. Positive ordinates exist between the neutral point (54.55 m) and B. The 10 m load is placed over the highest part, near the section, giving +2136 kN m. (Key point: Position for maximum positive moment) 4. Where is the load placed for the maximum negative moment? A. Right under the section B. At the right support B C. Over the full span D. On the negative part of the ILD near the crown (40.83 m to 50.83 m) Answer: D. Negative ordinates occur from A to the neutral point and peak at the crown (-4). The 10 m load goes where the area is largest, giving -1816.7 kN m. 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