---
title: "Circular arch ILDs at 3 m, then internal forces"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-12m-ild-3m-section"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-12m-ild-3m-section
site: BCT Engineering
---

# Circular arch ILDs at 3 m, then internal forces

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2070 Bhadra, 10 marks; 2066 Kartik, 10 marks (asked 2 times)
- 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-12m-ild-3m-section
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

A three-hinged symmetrical circular arch is of 12 m span and 4 m rise. Draw influence line diagram for bending moment, radial shear and normal thrust in the section at distance of 3 m from the left support. Use the diagrams to determine these internal forces in the section when the left half of the span is loaded with a uniformly distributed load of intensity 20 kN/m and a vertical concentrated load of magnitude 40 kN at a distance of 3 m from the right support.

## Short answer

Using influence line diagrams for the section 3 m from the left support, the UDL on the left half and the 40 kN point load at 9 m from left give internal forces: bending moment 14.02 kN m, radial shear 7.79 kN, and normal thrust 71.69 kN. Horizontal thrust is 60 kN.

## Answer

**Data:** three-hinged symmetrical circular arch, $L = 12$ m, $f = 4$ m. Radius $R = \dfrac{L^2/4+f^2}{2f} = \dfrac{36+16}{8} = 6.5$ m (centre 2.5 m below the springing level). Section $D$ at $x = 3$ m from $A$:

$$
y = \sqrt{6.5^2-(3-6)^2} - 2.5 = 3.266\ \text{m}, \qquad \tan\theta = \frac{6-3}{5.766} = 0.5203, \ \theta = 27.49^\circ
$$

#### Influence lines (unit load at $x'$)
- Thrust: $H = \dfrac{x'}{2f} = \dfrac{x'}{8}$ for $x'\le 6$; $\dfrac{12-x'}{8}$ for $x'\ge 6$ (peak $0.75$ at the crown).
- Bending moment: $M = M_b - H\,y$, with $M_b = \dfrac{x'(12-3)}{12}$ for $x'\le 3$ and $\dfrac{3(12-x')}{12}$ for $x'\ge 3$.
- Radial shear: $V = V_b\cos\theta - H\sin\theta$ with $V_b = -\dfrac{x'}{12}$ (load left of $D$) or $\dfrac{12-x'}{12}$ (load right of $D$).
- Normal thrust: $N = V_b\sin\theta + H\cos\theta$.

| Load at x' (m) | H | M (m) | V (radial) | N |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 3- | 0.375 | 1.0251 | -0.3949 | 0.2173 |
| 3+ | 0.375 | 1.0251 | 0.4923 | 0.6788 |
| 6 | 0.75 | -0.9497 | 0.0974 | 0.8961 |
| 9 | 0.375 | -0.4749 | 0.0487 | 0.4481 |
| 12 | 0 | 0 | 0 | 0 |

The M-ILD is positive near $D$ and negative towards the crown; the V-ILD has a jump of $\cos\theta = 0.8871$ at $D$; the N-ILD has a jump of $\sin\theta = 0.4615$ at $D$ (shown by the two rows at 3 m).

#### Loading
UDL 20 kN/m on the left half ($x' = 0$ to 6 m) and a 40 kN point load at 3 m from the right support ($x' = 9$ m).

Each quantity $= 20\times(\text{area of ILD from 0 to 6}) + 40\times(\text{ordinate at 9 m})$. The areas are obtained by the trapezoidal rule on each straight segment (using the values immediately left and right of $D$):

| Quantity | Area (0-6 m) | Ordinate at 9 m | Total |
|---|---|---|---|
| M | 1.6509 | -0.4749 | 14.02 |
| V | 0.2922 | 0.0487 | 7.79 |
| N | 2.6883 | 0.4481 | 71.69 |
| H | 2.25 | 0.375 | 60 |

$$
\begin{aligned} M &= 14.02\ \text{kN m}\\ V &= 7.79\ \text{kN}\\ N &= 71.69\ \text{kN}\\ H &= 60\ \text{kN} \end{aligned}
$$

**Check (direct statics):** $R_A = 100$ kN, $R_B = 60$ kN, $H = 60$ kN; $M = 100\times 3 - 20\times 3\times 1.5 - 60\times 3.266 = 14.02$ kN m.

**Answer: bending moment $14.02$ kN m (sagging), radial shear $7.79$ kN, normal thrust $71.69$ kN at the section 3 m from the left support.**

## Check yourself

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

1. For L = 12 m and f = 4 m, what is the radius?
   A. 6.5 m
   B. 9 m
   C. 5 m
   D. 4.5 m
   Answer: A. R = (L^2/4 + f^2)/(2f) = (36 + 16)/8 = 6.5 m. (Key point: Radius of the circular arch)

2. What is the peak ordinate of the thrust H influence line, and where?
   A. 0.375 at D
   B. 0.75 at the crown
   C. 1.5 at the crown
   D. 0.5 at A
   Answer: B. H = x'/(2f) for a load on the left half, so at the crown x' = 6 gives 6/8 = 0.75. (Key point: Thrust ILD)

3. How is radial shear computed from the ILD?
   A. V = V_b cos(theta) - H sin(theta)
   B. V = V_b sin(theta) + H cos(theta)
   C. V = M_b - H y
   D. V = H tan(theta)
   Answer: A. The beam shear is resolved perpendicular to the tangent, minus the component of H. (Key point: Radial shear expression)

4. With R_A = 100 kN, H = 60 kN, y = 3.266 m, what is M at D?
   A. 14.02 kN m
   B. 300 kN m
   C. 196 kN m
   D. 105 kN m
   Answer: A. M = 100 x 3 - 20 x 3 x 1.5 - 60 x 3.266 = 300 - 90 - 195.96 = 14.02 kN m. (Key point: Load effect and check)

## Related in Theory of Structures I

- [Arch 20 m: ILDs and check by equilibrium](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-20m-rise4-ild-check-4kn) — Theory of Structures I, chapter 6
- [Circular arch: BM, thrust, shear at D, BM diagram](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-d-20knm-10kn-8m) — Theory of Structures I, chapter 6
- [Circular arch: BM, shear, thrust at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-60m-d-500kn) — Theory of Structures I, chapter 6, asked 2×
- [Arch 40 m: ILDs for thrust, BM at 8 m, shear](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-40m-rise10-ild-thrust-bm-shear) — Theory of Structures I, chapter 6
- [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×
- [Circular arch 50 m: rolling UDL, 15 m section](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-50m-rolling-50knm-10m) — Theory of Structures I, chapter 6, asked 2×

## Related in other subjects

- [Two-hinged arch: BM, thrust, radial shear, support yield](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-thrust-shear-yield-support) — Theory of Structures II, chapter 2
- [Two-hinged arch: BM at section x-x](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-section-xx-60-85) — Theory of Structures II, chapter 2
- [ILD: shear at section x-x, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-shear-section-xx-two-span) — Theory of Structures II, chapter 4
- [ILD: bending moment at D, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-d-ab-bc-2) — Theory of Structures II, chapter 4