---
title: "Two-hinged parabolic arch: horizontal reaction, symmetric loads"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:arch-secant-symmetric-loads-80m"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/arch-secant-symmetric-loads-80m
site: BCT Engineering
---

# Two-hinged parabolic arch: horizontal reaction, symmetric loads

Theory of Structures II · Chapter 2: Force method

- Asked in: 2065 Shrawan, 8 marks
- Video: [Two Hinged Parabolic Arch (Problem No 6) With Two Point Loads](https://www.youtube.com/watch?v=zyloYX9LE-M) — Stan Academy
- 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-secant-symmetric-loads-80m
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

For the parabolic two hinged arch loaded symmetrically with concentrated loads as shown in the figure, determine the horizontal reaction. Use secant variation of moment of inertia. [Figure: two-hinged parabolic arch AB, span 8 x 10 m = 80 m, rise 5 m; vertical loads 4 kN, 10 kN, 8 kN, 10 kN, 8 kN, 10 kN, 4 kN symmetric about the crown.]

## Short answer

For a two-hinged parabolic arch with secant I variation, H = ∫ M₀ y dx / ∫ y² dx = 131.78 kN (inward) at each support. Symmetric vertical reactions V_A = V_B = 27 kN.

## Answer

**Method:** a two-hinged arch is indeterminate to the first degree; the horizontal thrust $H$ is the redundant. Using the condition that the horizontal displacement of the support is zero (strain energy, bending only):
$$
H=\frac{\int M_0\,y\,\dfrac{ds}{EI}}{\int y^2\,\dfrac{ds}{EI}}
$$
With the secant variation $I=I_c\sec\theta$, $ds/I=dx/I_c$, so
$$
H=\frac{\int_0^{L}M_0\,y\,dx}{\int_0^{L}y^2\,dx}
$$

#### Data
$L=80$ m, rise $h=5$ m, parabolic axis $y=\dfrac{4hx(L-x)}{L^2}=\dfrac{20\,x(80-x)}{6400}$. Loads (kN) at 10 m spacing: 4, 10, 8, 10, 8, 10, 4 at $x=10,20,\ldots,70$ m (total 54 kN).

#### Reactions
By symmetry $V_A=V_B=27$ kN.

#### Simple-beam moment $M_0$ and ordinate $y$
| x (m) | y (m) | $M_0$ (kNm) | $M=M_0-Hy$ (kNm) |
|---|---|---|---|
| 0 | 0.000 | 0.0 | 0.00 |
| 10 | 2.188 | 270.0 | -18.26 |
| 20 | 3.750 | 500.0 | 5.83 |
| 30 | 4.688 | 630.0 | 12.29 |
| 40 | 5.000 | 680.0 | 21.11 |
| 50 | 4.688 | 630.0 | 12.29 |
| 60 | 3.750 | 500.0 | 5.83 |
| 70 | 2.188 | 270.0 | -18.26 |
| 80 | 0.000 | 0.0 | 0.00 |

$M_0$ is linear between the loads, for example $M_0(10)=27\times10=270$, $M_0(20)=27\times20-4\times10=500$, $M_0(30)=27\times30-4\times20-10\times10=630$, $M_0(40)=680$.

#### Integrals
$$
\int_0^{80}M_0\,y\,dx=140562.5\ \text{kN m}^3,\qquad \int_0^{80}y^2dx=\frac{8h^2L}{15}=\frac{8\times25\times80}{15}=1066.67\ \text{m}^3
$$
$$
H=\frac{140562.5}{1066.67}=131.78\ \text{kN}
$$

The bending moment at the crown is $M_C=680-131.78\times5=21.11$ kNm (small compared with $M_0$, which shows how effectively the thrust reduces the moments).

**Answer: horizontal reaction $H=131.78$ kN at each support (inward); $V_A=V_B=27$ kN.**

## Check yourself

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

1. For the two-hinged arch, what is the redundant and the condition used to find it?
   A. Vertical reaction V_A; zero rotation at A
   B. Bending moment at the crown; zero deflection at the crown
   C. Horizontal thrust H; zero horizontal displacement of the support
   D. Axial force in the arch; zero vertical displacement
   Answer: C. A two-hinged arch is indeterminate to the first degree. H is the redundant, found from zero relative horizontal movement of the hinges. (Key point: Redundant and condition)

2. With I = Ic sec(theta), what simplification occurs in the expression for H?
   A. ds/I becomes dx/Ic, so H = Integral(M0 y dx)/Integral(y^2 dx)
   B. H becomes independent of the loads
   C. The integrals vanish and H = 0
   D. ds/I becomes dy/Ic so H uses the rise only
   Answer: A. Since ds = dx sec(theta), ds/I = dx/Ic. The factors of EI and Ic cancel, giving a simple ratio of integrals over x. (Key point: Effect of secant variation of I)

3. For a parabolic arch of span L and rise h, what is the integral of y^2 dx over the span (L = 80 m, h = 5 m)?
   A. 8h^2 L/15 = 1066.67 m^3
   B. h^2 L/3 = 666.67 m^3
   C. 2hL/3 = 266.67 m^3
   D. h^2 L = 2000 m^3
   Answer: A. For a parabola, the integral of y^2 dx is 8 h^2 L/15 = 8 x 25 x 80/15 = 1066.67 m^3. (Key point: Integral of y^2)

4. What is the bending moment at the crown, where M0 = 680 kNm, y = 5 m and H = 131.78 kN?
   A. 680 kNm
   B. 21.11 kNm
   C. -21.11 kNm
   D. 658.9 kNm
   Answer: B. M = M0 - H y = 680 - 131.78 x 5 = 21.11 kNm. The thrust greatly reduces the moment from the simple-beam value. (Key point: Moment in the arch)

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