---
title: "Circular arch 120 m: two rolling loads, 30 m section"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-circular-120m-8kn-12kn"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-circular-120m-8kn-12kn
site: BCT Engineering
---

# Circular arch 120 m: two rolling loads, 30 m section

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2068 Bhadra, 16 marks
- Video: [SA55: Analysis of a Three-hinged Arch Bridge](https://www.youtube.com/watch?v=WNAeZAfiseQ) — Dr. Structure
- 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-120m-8kn-12kn
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

A three hinged circular arch has a span of 120 m and a rise of 15 m. Two point loads of 8 kN and 12 kN, spaced 10 m apart, roll over the arch from left to right with 8 kN load leading. Using the influence line diagram, find the maximum bending moments at a section 30 m from the left support. Also find normal thrust and radial shear at the same section corresponding to the maximum bending moment.

## Short answer

Maximum positive bending moment is +177.27 kN m with normal thrust 16.49 kN and radial shear 0.12 kN; maximum negative moment is -146.36 kN m with normal thrust 38.48 kN and radial shear 0.29 kN.

## Answer

**Data:** three-hinged circular arch, $L = 120$ m, $f = 15$ m. Radius $R = \dfrac{60^2+15^2}{2\times 15} = 127.5$ m. Two loads, 8 kN (leading) and 12 kN, 10 m apart, roll from left to right (the 12 kN load is 10 m behind the 8 kN load). Section at $x = 30$ m: $y = \sqrt{127.5^2-30^2} - 112.5 = 11.42$ m.

#### ILDs for the section at x = 30 m
Thrust ILD: $H = \dfrac{x'}{2f}$ (load left of the crown) or $\dfrac{L-x'}{2f}$ (right of the crown), peak $\dfrac{L}{4f} = 2$ 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 |
| 30 | section D | 1 | 11.08 |
| 47.57 | neutral point | 1.586 | 0 |
| 60 | crown C | 2 | -7.841 |
| 120 | B | 0 | 0 |

The ILD of $M$ is zero at the neutral point, 47.57 m from $A$; it is positive on one side and negative on the other.

#### Maximum positive bending moment
The critical position is: 8 kN at 30 m, 12 kN at 20 m (measured from $A$).

| Load (kN) | Position (m) | Ordinate of M-ILD (m) | P x y (kN m) | Ordinate of H-ILD |
|---|---|---|---|---|
| 8 | 30 | 11.08 | 88.64 | 1 |
| 12 | 20 | 7.386 | 88.64 | 0.667 |

$$
M = \sum P\,y = 177.27\ \text{kN m}, \qquad H = \sum P\,y_H = 16\ \text{kN}
$$

With $y = 11.42$ m, $\theta = 13.61^\circ$ and the simple-beam shear $V_b = 4$ kN at the section:

$$
\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0.12\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 16.49\ \text{kN} \end{aligned}
$$

#### Maximum negative bending moment
The critical position is: 8 kN at 70 m, 12 kN at 60 m (measured from $A$).

| Load (kN) | Position (m) | Ordinate of M-ILD (m) | P x y (kN m) | Ordinate of H-ILD |
|---|---|---|---|---|
| 8 | 70 | -6.534 | -52.27 | 1.667 |
| 12 | 60 | -7.841 | -94.09 | 2 |

$$
M = \sum P\,y = -146.36\ \text{kN m}, \qquad H = \sum P\,y_H = 37.33\ \text{kN}
$$

With $y = 11.42$ m, $\theta = 13.61^\circ$ and the simple-beam shear $V_b = 9.33$ kN at the section:

$$
\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0.29\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 38.48\ \text{kN} \end{aligned}
$$

**Answer: at 30 m from the left support: maximum positive bending moment $=+177.27$ kN m with normal thrust $16.49$ kN and radial shear $0.12$ kN; maximum negative bending moment $=-146.36$ kN m with normal thrust $38.48$ kN and radial shear $0.29$ kN.**

## Check yourself

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

1. What is the radius for a circular arch of span 120 m and rise 15 m?
   A. R = 60^2/(8 x 15) = 30 m
   B. R = (60^2 + 15^2)/(2 x 15) = 127.5 m
   C. R = 120 x 15 / 2 = 900 m
   D. R = 60 + 15 = 75 m
   Answer: B. R = (half-span^2 + rise^2)/(2 x rise) = 3825/30 = 127.5 m. (Key point: Radius of the circular arch)

2. What does the neutral point (47.57 m from A) on the M ILD indicate?
   A. The ILD of H is maximum
   B. The influence ordinate of M is zero, with positive ordinates on one side and negative on the other
   C. The section where the radial shear is zero
   D. The location of the crown hinge
   Answer: B. Unit load at the neutral point produces no bending moment at the section; the sign of the ordinate changes there. (Key point: Meaning of the neutral point)

3. For maximum positive moment, where are the loads placed?
   A. 12 kN at the crown and 8 kN at 70 m
   B. Both loads at mid-span
   C. 8 kN at the section (30 m) with 12 kN 10 m behind it at 20 m
   D. 8 kN at 10 m and 12 kN at 30 m
   Answer: C. The largest positive ordinate is under the section; the heavier 12 kN load trails the 8 kN load, so the 8 kN leads at 30 m and 12 kN is at 20 m, giving M = 177.27 kN m. (Key point: Critical position for maximum positive moment)

4. For maximum negative moment, the loads are at 70 m (8 kN) and 60 m (12 kN). Why?
   A. Because the heavier load sits at the crown where the negative ordinate of the M ILD is greatest
   B. Because H is zero there
   C. Because the loads must be next to the right support
   D. Because the shear is maximum there
   Answer: A. The negative ordinate is largest near the crown (-7.841), so the heavier 12 kN load goes there with 8 kN following, giving M = -146.36 kN m. (Key point: Maximum negative moment)

## Close variants set in other exams

- [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×

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