---
title: "Arch 160 m: rolling UDL, section 50 m from right"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-160m-udl-30knm-60m-50m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-160m-udl-30knm-60m-50m
site: BCT Engineering
---

# Arch 160 m: rolling UDL, section 50 m from right

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2069 Bhadra, 16 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-160m-udl-30knm-60m-50m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

A three hinged parabolic arch has a span of 160 m and a rise of 25 m. A uniformly distributed load of intensity 30 kN/m of length 60 m rolls over the arch from left to right. Using the influence line diagram, find the maximum bending moment at a section 50 m from the right support. Also find normal thrust and radial shear at the section corresponding to the maximum bending moment.

## Short answer

Maximum positive bending moment is +12870 kN m with normal thrust 1258.1 kN and radial shear 228 kN; maximum negative moment is -11410 kN m with normal thrust 2149.5 kN and radial shear -168.5 kN.

## Answer

**Data:** three-hinged symmetrical parabolic arch, $L = 160$ m, $f = 25$ m, $y = \dfrac{4\times 25\,x(160-x)}{160^2} = 0.00390625\,x\,(160-x)$. UDL 30 kN/m, length 60 m, rolls from left to right. The section is 50 m from the right support, i.e. at $x = 110$ m: $y = 21.484$ m, $\tan\theta = 0.00390625\,(160-220) = -0.2344$, so the tangent slopes downward to the right ($\theta = -13.19^\circ$).

For a section in the right half, the neutral point lies between the crown and the section; a UDL on the part of the span to its left gives a hogging (negative) moment, and one on the part to its right gives a sagging (positive) moment, as the ILD below shows.

#### ILDs for the section at x = 110 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 $\dfrac{L}{4f} = 1.6$ 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 |
| 80 | crown C | 1.6 | -9.375 |
| 92.63 | neutral point | 1.347 | 0 |
| 110 | section D | 1 | 12.891 |
| 160 | B | 0 | 0 |

The ILD of $M$ crosses zero at the **neutral point**, 92.63 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 = 94.53 m to 154.53 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.

$$
\begin{aligned} M &= w\times(\text{area of ILD}) = 30\times(429.016) = 12870.5\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 30\times(42.564) = 1276.92\ \text{kN} \end{aligned}
$$

At the section: $y = 21.484$ m, $\theta = -13.19^\circ$, simple-beam shear $V_b = -65.06$ kN.

$$
\begin{aligned} V &= V_b\cos\theta - H\sin\theta = (-65.06)\times0.9736 - 1276.92\times(-0.2282) = 228.04\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = (-65.06)\times(-0.2282) + 1276.92\times0.9736 = 1258.08\ \text{kN} \end{aligned}
$$

#### Maximum negative bending moment
The UDL covers the region from x = 28.18 m to 88.18 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.

$$
\begin{aligned} M &= w\times(\text{area of ILD}) = 30\times(-380.327) = -11409.8\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 30\times(68.4778) = 2054.33\ \text{kN} \end{aligned}
$$

At the section: $y = 21.484$ m, $\theta = -13.19^\circ$, simple-beam shear $V_b = -654.52$ kN.

$$
\begin{aligned} V &= V_b\cos\theta - H\sin\theta = (-654.52)\times0.9736 - 2054.33\times(-0.2282) = -168.48\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = (-654.52)\times(-0.2282) + 2054.33\times0.9736 = 2149.49\ \text{kN} \end{aligned}
$$

**Answer: at 50 m from the right support, the maximum bending moment is $+12870$ kN m (UDL covering about 94.5 m to 154.5 m), with normal thrust $1258.1$ kN and radial shear $228$ kN; the maximum negative moment is $-11410$ kN m (UDL covering about 28.2 m to 88.2 m), with normal thrust $2149.5$ kN and radial shear $-168.5$ kN.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For a three-hinged arch of span 160 m and rise 25 m, what is the peak ordinate of the ILD of H and where is it?
   A. 6.4 at the supports
   B. 1.6 at the crown (L/4f)
   C. 3.2 at the section
   D. 0.8 at mid-span
   Answer: B. H = x'/(2f) for a unit load left of the crown, giving L/(4f) = 160/100 = 1.6 at the crown. (Key point: Peak of the H ILD)

2. How should the 60 m UDL be placed for the maximum positive bending moment?
   A. Over the whole span
   B. Centred on the crown
   C. Over the region of the ILD where ordinates are positive (about 94.5 m to 154.5 m)
   D. Next to the left support
   Answer: C. The moment equals w times the area of the ILD under the load, so it is maximised by covering the positive-ordinate part; the load here lies right of the neutral point. (Key point: Loading for maximum positive moment)

3. After finding H from the ILD area, how are V and N at the section found?
   A. V = V_b cos(theta) - H sin(theta); N = V_b sin(theta) + H cos(theta), using theta at the section and the simple-beam shear V_b
   B. V = H and N = V_b
   C. V = M / y and N = H only
   D. V = V_b only; N = H / cos(theta) only
   Answer: A. Resolve V_b and H along the arch tangent at the section (here theta = -13.19 deg), giving V = 228 kN and N = 1258 kN for the maximum positive case. (Key point: N and V at the section)

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