---
title: "No bending moment under UDL in parabolic arch"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-no-bm-udl"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-no-bm-udl
site: BCT Engineering
---

# No bending moment under UDL in parabolic arch

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2078 Chaitra, 4 marks; 2072 Asoj, 4 marks (asked 2 times)
- Video: [Lect-100 Show that B. M at any section in Three hinged parabolic arch is zero when subjected to UDL](https://www.youtube.com/watch?v=L7BIs2D2oP4) — Dr. Ashish Agrawal
- 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-no-bm-udl
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

Show that there is no bending moment at any section of a parabolic (three hinged) arch subjected to load uniformly distributed over the horizontal span.

## Short answer

Under a full UDL, the horizontal thrust is wL^2/(8f). The bending moment M_x = w x (L - x)/2 - H y = 0 because the parabolic shape matches the funicular curve, so the arch carries only compression.

## Answer

**Statement:** a three-hinged parabolic arch carrying a UDL $w$ per unit horizontal length over the whole span has zero bending moment at every section.

Let the span be $L$ and the rise $f$, supports at the same level, origin at the left support.

#### Arch equation
$$
y = \frac{4f\,x\,(L-x)}{L^2}
$$

#### Reactions and thrust
By symmetry $R_A = R_B = \dfrac{wL}{2}$. The bending moment at the crown hinge is zero, so, taking moments of the left half about $C$:

$$
H f = \frac{wL}{2}\times\frac{L}{2} - w\,\frac{L}{2}\times\frac{L}{4} = \frac{wL^2}{8} \Rightarrow H = \frac{wL^2}{8f}
$$

#### Bending moment at a section at distance x
$$
\begin{aligned} M_x &= R_A\,x - \frac{w\,x^2}{2} - H\,y \\ &= \frac{wLx}{2} - \frac{wx^2}{2} - \frac{wL^2}{8f}\cdot\frac{4f\,x\,(L-x)}{L^2} \\ &= \frac{w\,x\,(L-x)}{2} - \frac{w\,x\,(L-x)}{2} = 0 \end{aligned}
$$

So $M_x = 0$ for every $x$. The beam bending moment $\dfrac{wx(L-x)}{2}$ is exactly cancelled by $H\,y$, because both are parabolic with the same shape.

**Conclusion:** the parabola is the funicular (equilibrium) shape for a UDL on the horizontal span. The arch is then in pure compression (thrust) with no bending and no radial shear.

## Check yourself

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

1. What is the horizontal thrust of a three-hinged parabolic arch under a UDL w over the full span?
   A. wL^2/(4f)
   B. wL^2/(8f)
   C. wL/(2f)
   D. wL^2/(16f)
   Answer: B. Zero moment at the crown: Hf = (wL/2)(L/2) - w(L/2)(L/4) = wL^2/8, so H = wL^2/(8f). (Key point: Thrust)

2. M_x = R_A x - wx^2/2 - H y. Why does this vanish?
   A. Because R_A = 0
   B. The beam moment w x (L - x)/2 equals H y, both being parabolic with the same shape
   C. Because H is zero
   D. Because the arch is symmetric only
   Answer: B. H y = (wL^2/8f)(4f x (L - x)/L^2) = w x (L - x)/2, equal to the beam moment. (Key point: Moment at the section)

3. Why is the parabola the ideal shape here?
   A. It is the funicular shape for a UDL on the horizontal span, so the arch has pure compression
   B. It has the longest length
   C. It reduces the span
   D. It makes reactions horizontal
   Answer: A. The thrust line coincides with the arch axis, so there is no bending and no radial shear. (Key point: Meaning of result)

## Related in Theory of Structures I

- [Parabolic arch 20 m: two loads and UDL, 5 m section](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-20m-two-loads-udl-right-half) — Theory of Structures I, chapter 6
- [Parabolic arch 100 m: UDL left half and 80 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-80kn-udl-d-20m) — Theory of Structures I, chapter 6, asked 4×
- [BM, shear, thrust at any section, UDL on arch](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-general-section-udl-formula) — Theory of Structures I, chapter 6
- [Arch 160 m: rolling UDL, section 50 m from right](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-160m-udl-30knm-60m-50m) — Theory of Structures I, chapter 6
- [Moment area: mid-span deflection under UDL](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-ss-udl-midspan) — Theory of Structures I, chapter 4
- [ILD for bending moment at section D of arch](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-ild-bm-d-given-arch) — Theory of Structures I, chapter 6

## Related in other subjects

- [Two-hinged arch: horizontal reaction, UDL on half](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-12knm-horizontal-reaction-50m-8m) — Theory of Structures II, chapter 2
- [Two-hinged parabolic arch: horizontal reaction, symmetric loads](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-secant-symmetric-loads-80m) — Theory of Structures II, chapter 2
- [Two-hinged arch: horizontal reaction, BM at C](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-horizontal-reaction-bm-c-50knm) — 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