---
title: "Circular arch: two rolling loads 20 kN and 30 kN"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-circular-100m-20kn-30kn-25m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-circular-100m-20kn-30kn-25m
site: BCT Engineering
---

# Circular arch: two rolling loads 20 kN and 30 kN

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2074 Bhadra, 16 marks
- Video: [Three Hinged Arch with BMD and ILD](https://www.youtube.com/watch?v=bI-ts0j_jjo) — Sumit Paudel
- 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-100m-20kn-30kn-25m
- 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 100 m and a rise of 10 m. Two point loads of 20 kN and 30 kN, spaced 5 m apart, roll over the arch from left to right with 20 kN load leading. Using the influence line diagram, find the maximum bending moments at a section 25 m from the left support. Also find normal thrust and radial shear at the same section corresponding to the maximum bending moment.

## Short answer

Span 100 m, rise 10 m circular three-hinged arch (R=130 m) with 20 kN and 30 kN rolling loads (5 m apart, 20 kN leading, left to right): at 25 m, max +BM=408.5 kN m (N=55.7 kN, V=−1.74 kN) loads at 25 m, 20 m; max −BM=−308.8 kN m (N=122.4 kN, V=0.48 kN) loads at 55 m, 50 m.

## Answer

**Data:** symmetrical circular three-hinged arch, $L = 100$ m, $f = 10$ m, $R = 130$ m. Two loads, 20 kN (leading) and 30 kN, 5 m apart, roll from left to right (the 30 kN load is 5 m behind the 20 kN load). Section at $x = 25$ m: $y = \sqrt{130^2-25^2} - 120 = 7.574$ m.

#### ILDs for the section at x = 25 m
Thrust ILD: $H = \dfrac{x'}{2f}$ (load left of the crown) or $\dfrac{L-x'}{2f}$ (right of the crown), peak $\dfrac{L}{4f} = 2.5$ 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 |
| 25 | section D | 1.25 | 9.283 |
| 39.77 | neutral point | 1.988 | 0 |
| 50 | crown C | 2.5 | -6.434 |
| 100 | B | 0 | 0 |

The ILD of $M$ is zero at the neutral point, 39.77 m from $A$; it is positive on one side and negative on the other.

#### Maximum positive bending moment
The critical position is: 20 kN at 25 m, 30 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 |
|---|---|---|---|---|
| 20 | 25 | 9.283 | 185.66 | 1.25 |
| 30 | 20 | 7.426 | 222.79 | 1 |

$$
M = \sum P\,y = 408.46\ \text{kN m}, \qquad H = \sum P\,y_H = 55\ \text{kN}
$$

With $y = 7.574$ m, $\theta = 11.09^\circ$ and the simple-beam shear $V_b = 9$ kN at the section:

$$
\begin{aligned} V &= V_b\cos\theta - H\sin\theta = -1.74\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 55.7\ \text{kN} \end{aligned}
$$

#### Maximum negative bending moment
The critical position is: 20 kN at 55 m, 30 kN at 50 m (measured from $A$).

| Load (kN) | Position (m) | Ordinate of M-ILD (m) | P x y (kN m) | Ordinate of H-ILD |
|---|---|---|---|---|
| 20 | 55 | -5.79 | -115.81 | 2.25 |
| 30 | 50 | -6.434 | -193.01 | 2.5 |

$$
M = \sum P\,y = -308.82\ \text{kN m}, \qquad H = \sum P\,y_H = 120\ \text{kN}
$$

With $y = 7.574$ m, $\theta = 11.09^\circ$ and the simple-beam shear $V_b = 24$ kN at the section:

$$
\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 0.48\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 122.38\ \text{kN} \end{aligned}
$$

**Answer: maximum positive moment at 25 m $= +408.5$ kN m with $N = 55.7$ kN and $V = -1.74$ kN; maximum negative moment $= -308.8$ kN m with $N = 122.4$ kN and $V = 0.48$ kN.**

## Check yourself

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

1. What is the radius of this circular arch (span 100 m, rise 10 m)?
   A. 125 m
   B. 130 m
   C. 100 m
   D. 65 m
   Answer: B. R = (50^2 + 10^2)/(2 x 10) = 130 m. The ordinate at x = 25 m then follows from y = sqrt(R^2 - (x-50)^2) - 120. (Key point: Radius)

2. The 20 kN load leads while the loads roll left to right. Where is the 30 kN load relative to the 20 kN?
   A. 5 m behind (to the left of) the 20 kN load
   B. 5 m ahead (to the right)
   C. At the same point
   D. 10 m behind
   Answer: A. Moving right, the leading load is on the right and the following 30 kN load is 5 m to its left. (Key point: Load order)

3. Which position gives the maximum positive moment at 25 m?
   A. 20 kN at the crown and 30 kN at 45 m
   B. 30 kN at the neutral point (39.77 m)
   C. 20 kN at the section (25 m) and 30 kN at 20 m
   D. Both loads right of the crown
   Answer: C. The positive ILD peaks at the section (9.283). Placing a load there and the other close beside it gives 185.66 + 222.79 = 408.46 kN m. (Key point: Critical position for max positive moment)

4. Where are the loads for maximum negative moment?
   A. 20 kN at the crown and 30 kN at 45 m
   B. 30 kN at the section
   C. Both loads left of the neutral point
   D. 30 kN at the crown (50 m) and 20 kN at 55 m
   Answer: D. The negative ILD peaks at the crown (-6.434). The heavier 30 kN goes at the crown with 20 kN 5 m ahead, giving -308.82 kN m. (Key point: Critical position for max negative moment)

## Close variants set in other exams

- [Circular arch 120 m: two rolling loads, 30 m section](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-120m-8kn-12kn) — Theory of Structures I, chapter 6, asked 2×

## Related in Theory of Structures I

- [Arch 100 m rise 15 m: rolling UDL 10 m, section 60 m](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-rise15-udl-10m-60m-section) — Theory of Structures I, chapter 6
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