--- title: "Arch 100 m rise 12 m: rolling UDL 60 m, section 10 m" subject: "Theory of Structures I" chapter: 6 key: "semester_4/theory_of_structures_i:arch-100m-rise12-udl-60m-10m" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-100m-rise12-udl-60m-10m site: BCT Engineering --- # Arch 100 m rise 12 m: rolling UDL 60 m, section 10 m Theory of Structures I · Chapter 6: Statically Determinate Arches - Asked in: 2076 Baisakh, 10 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-rise12-udl-60m-10m - 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 12 m carries a UDL of intensity 20 kN/m of length 60 m moving from left to right. Calculate horizontal thrust and maximum bending moments at a distance 10 m from the left end; using ILD. ## Short answer Maximum horizontal thrust is 1750 kN. At 10 m from left end, maximum positive moment is +2571 kN m with thrust 531.5 kN; maximum negative moment is -2560 kN m with thrust 1518.5 kN. ## Answer **Data:** parabolic three-hinged arch, $L = 100$ m, $f = 12$ m, $y = 0.0048\,x(100-x)$. A UDL of 20 kN/m, 60 m long, rolls from left to right. Section at $x = 10$ m: $y = 4.32$ m, $\tan\theta = 0.0048\,(100-20) = 0.384$. #### Horizontal thrust $H$ is maximum when the load is symmetrical about the crown (20 m to 80 m): the ILD ordinates are $\dfrac{x'}{2f} = 0.833$ at 20 m and $\dfrac{L}{4f} = 2.083$ at the crown: $$ H_{max} = 20\times\left[2\times\frac{(0.833+2.083)}{2}\times 30\right] = 1750\ \text{kN} $$ #### ILDs for the section at x = 10 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 $\dfrac{L}{4f} = 2.0833$ 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 | | 10 | section D | 0.417 | 7.2 | | 35.71 | neutral point | 1.488 | 0 | | 50 | crown C | 2.083 | -4 | | 100 | B | 0 | 0 | The ILD of $M$ crosses zero at the **neutral point**, 35.71 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 UDL covers the span from $x = 0$ to 35.71 m (the rest of the load is still off the span), which is exactly the region where the ILD is positive. $$ \begin{aligned} M &= w\times(\text{area of ILD}) = 20\times 128.571 = 2571.4\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 20\times 26.573 = 531.46\ \text{kN} \end{aligned} $$ At the section: $y = 4.32$ m, $\theta = 21.01^\circ$, simple-beam shear $V_b = 386.73$ kN. $$ \begin{aligned} V &= V_b\cos\theta - H\sin\theta = 386.73\times0.9335 - 531.46\times0.3585 = 170.51\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 386.73\times0.3585 + 531.46\times0.9335 = 634.78\ \text{kN} \end{aligned} $$ #### Maximum negative bending moment The negative region (35.71 m to 100 m) is longer than the load, so the 60 m load is placed with equal ILD ordinates at its two ends: it covers $x = 36.67$ m to 96.67 m (the whole load is on the span). $$ \begin{aligned} M &= w\times(\text{area of ILD}) = 20\times(-128) = -2560\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 20\times 75.926 = 1518.52\ \text{kN} \end{aligned} $$ At the section: $y = 4.32$ m, $\theta = 21.01^\circ$, simple-beam shear $V_b = 400.0$ kN. $$ \begin{aligned} V &= V_b\cos\theta - H\sin\theta = 400.0\times0.9335 - 1518.52\times0.3585 = -170.94\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 400.0\times0.3585 + 1518.52\times0.9335 = 1560.99\ \text{kN} \end{aligned} $$ **Answer: maximum thrust $= 1750$ kN; maximum bending moment at 10 m from the left end $= +2571$ kN m (load from the support up to the neutral point), with thrust $H = 531.5$ kN, and $-2560$ kN m (negative), with $H = 1518.5$ kN.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 60 m rolling UDL, when is H maximum? A. When the load covers 0 to 60 m B. When the load is symmetrical about the crown (20 m to 80 m) C. When the load ends at the section D. When the load covers the neutral point Answer: B. The ILD of H is a triangle with peak at the crown, so a centred load covers the largest area. H max = 1750 kN. (Key point: Maximum thrust position) 2. What is the peak ordinate of the ILD of H for L = 100 m and f = 12 m? A. L/8f = 1.04 B. L/2f = 4.17 C. L/4f = 2.083 at the crown D. f/L = 0.12 Answer: C. The unit load at the crown gives simple-beam crown moment L/4, divided by f. So 100/48 = 2.083. (Key point: Peak ordinate of H-ILD) 3. For maximum positive moment at the section 10 m from A, how is the UDL placed? A. From A up to the neutral point at 35.71 m B. From the neutral point to B C. Centred at the crown D. From the section to the crown only Answer: A. The M-ILD is positive from A to the neutral point (35.71 m), so loading exactly that part gives the maximum positive moment, 2571 kN m. (Key point: Position for max positive moment at 10 m) 4. Why is the thrust different for the maximum positive and maximum negative cases (531 kN and 1518 kN)? A. H is fixed by the span and rise alone B. H depends on the area of the H-ILD under the load, which is different for the two load positions C. H is zero when the moment is positive D. H is the same but the vertical reaction changes Answer: B. H = w x area of the H-ILD under the loaded length. The negative-moment position is near the crown, where H ordinates are high. 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