---
title: "Two-hinged arch: horizontal reaction, UDL on half"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:arch-12knm-horizontal-reaction-50m-8m"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/arch-12knm-horizontal-reaction-50m-8m
site: BCT Engineering
---

# Two-hinged arch: horizontal reaction, UDL on half

Theory of Structures II · Chapter 2: Force method

- Asked in: 2069 Asar, 5 marks
- Video: [Two Hinged Parabolic Arch - Problem 1 - Structural Analysis 2](https://www.youtube.com/watch?v=7pq6DGfRrBQ) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/arch-12knm-horizontal-reaction-50m-8m
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

A two hinged symmetrical parabolic arch of secant variation cross section having span 50 m and rise 8 m is loaded with a uniformly distributed load of 12 kN/m extending from the left hand support to the centre of the arch as shown in figure below. Determine the horizontal reaction at the support.

## Short answer

For secant variation I = Ic secθ, horizontal thrust H = ∫M0 y dx / ∫y² dx. With half-span UDL 12 kN/m, integrating M0 expressions yields H = 400000 / 1706.67 = 234.38 kN inward at each support.

## Answer

For a two-hinged arch the thrust $H$ makes the horizontal displacement of the support zero. With $I=I_c\sec\theta$ (so $ds/I=dx/I_c$):
$$
H=\frac{\int_0^L M_0\,y\,dx}{\int_0^L y^2\,dx}
$$

#### Data
$L=50$ m, $h=8$ m, $w=12$ kN/m on the left half (0 to 25 m). Parabola: $y=\dfrac{4hx(L-x)}{L^2}=\dfrac{32x(50-x)}{2500}$.

#### Reactions and $M_0$
Total load $=12\times25=300$ kN acting at 12.5 m from A.
$$
V_A=\frac{300\times(50-12.5)}{50}=225\ \text{kN},\qquad V_B=75\ \text{kN}
$$
- For $0\le x\le25$: $M_0=225x-6x^2$
- For $25\le x\le50$: $M_0=75(50-x)$

#### Integrals
$$
\int_0^{50}M_0\,y\,dx=400000\ \text{kN m}^3,\qquad \int_0^{50}y^2dx=\frac{8h^2L}{15}=\frac{8\times64\times50}{15}=1706.67\ \text{m}^3
$$
$$
H=\frac{400000}{1706.67}=234.375\ \text{kN}
$$

Check: by symmetry a UDL on the half span gives half of the full-span thrust, $\tfrac12\cdot\dfrac{wL^2}{8h}=\tfrac12\times\dfrac{12\times2500}{64}=234.375$ kN. ✓

**Answer: horizontal reaction $H=234.38$ kN at each support (inward).** Vertical reactions: $V_A=225$ kN and $V_B=75$ kN.

## Check yourself

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

1. In a two-hinged arch, what condition determines the horizontal thrust H?
   A. The vertical deflection at the crown is zero
   B. The horizontal displacement of one support is zero
   C. The bending moment at the crown is zero
   D. The sum of vertical reactions equals the load
   Answer: B. The hinge supports cannot move apart, so the horizontal displacement of the support due to loads and H together is zero. (Key point: Condition for the thrust H)

2. With I = Ic sec(theta), which expression gives H?
   A. H = integral(M0 y dx) / integral(y^2 dx)
   B. H = integral(M0 dx) / integral(y dx)
   C. H = integral(y dx) / integral(M0 y dx)
   D. H = integral(M0^2 dx) / integral(y^2 dx)
   Answer: A. Since ds/I = dx/Ic, minimising strain energy gives H = integral(M0 y dx) / integral(y^2 dx). (Key point: Formula for H)

3. What is H for the UDL on the left half, and how can it be checked?
   A. 468.75 kN, twice the full-span thrust
   B. 234.375 kN, half of the full-span thrust wL^2/8h
   C. 117.19 kN, a quarter of wL^2/8h
   D. 300 kN, equal to the total load
   Answer: B. By superposition the half-span UDL gives half the full-span thrust: 0.5 x 12 x 2500/64 = 234.375 kN. (Key point: Result and check)

## Related in Theory of Structures II

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