---
title: "Two-hinged arch: horizontal reaction, BM at C"
subject: "Theory of Structures II"
chapter: 2
key: "semester_5/theory_of_structures_ii:arch-horizontal-reaction-bm-c-50knm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/arch-horizontal-reaction-bm-c-50knm
site: BCT Engineering
---

# Two-hinged arch: horizontal reaction, BM at C

Theory of Structures II · Chapter 2: Force method

- Asked in: 2070 Chaitra, 10 marks
- Video: [Two Hinged Parabolic Arch - Problem 2 - Structural Analysis 2](https://www.youtube.com/watch?v=HXJWvLuR7tM) — 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-horizontal-reaction-bm-c-50knm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-2

## Question

Determine the horizontal reaction in the two hinged parabolic arch shown in figure below. Also determine the bending moment at C. ($I = I_c \sec\theta$). [Figure: two-hinged parabolic arch AB, span 60 m (30 m + 30 m), rise 5 m at crown C; UDL 50 kN/m over the right half CB.]

## Short answer

For the two-hinged parabolic arch with UDL on right half, horizontal thrust H = 2250 kN. Bending moment at crown C is zero due to symmetry of loading and parabolic shape.

## Answer

A two-hinged arch has one redundant, the horizontal reaction (thrust) $H$. With $I = I_c\sec\theta$, $ds/I = dx/I_c$, so

$$
H = \frac{\int M_0\,y\,dx}{\int y^2\,dx}
$$

#### Data and profile
Span $L = 60$ m, rise $h = 5$ m, UDL 50 kN/m on the right half CB ($30 \le x \le 60$, with A at the origin).

$$
y = \frac{4h\,x(L-x)}{L^2} = \frac{x(60-x)}{180}
$$

#### Reactions of the simple beam
UDL resultant $= 50\times30 = 1500$ kN at $x = 45$ m.

$$
V_B = \frac{1500(45)}{60} = 1125\ \text{kN},\qquad V_A = 1500 - 1125 = 375\ \text{kN}
$$

#### Free bending moment
- $0 \le x \le 30$: $M_0 = 375\,x$
- $30 \le x \le 60$: $M_0 = 375\,x - 25(x-30)^2$

#### Integrals
$$
\begin{aligned}
\int_0^{60}y^2dx &= \frac{8h^2L}{15} = \frac{8(25)(60)}{15} = 800 \\
\int_0^{30}M_0y\,dx &= 703125, \qquad \int_{30}^{60}M_0y\,dx = 1096875 \\
\int_0^{60}M_0y\,dx &= 1800000
\end{aligned}
$$

#### Horizontal reaction
$$
H = \frac{1800000}{800} = 2250.0\ \text{kN}
$$

#### Bending moment at the crown C ($x = 30$ m)
$$
\begin{aligned}
M_0 &= V_A(30) = 375(30) = 11250\ \text{kNm} \\
y_C &= h = 5\ \text{m} \\
M_C &= M_0 - H\,y_C = 11250 - 2250.0(5) = 0.0\ \text{kNm}
\end{aligned}
$$

The crown moment is zero for this loading, which is a known property of a parabolic two-hinged arch carrying a UDL over half the span.

**Answer: $H = 2250.0$ kN; $M_C = 0.0$ kNm.**

## Check yourself

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

1. For a two-hinged arch with I = Ic sec θ, how is H found?
   A. H = ∫M0 y dx / ∫y^2 dx
   B. H = ∫M0 dx / ∫y dx
   C. H = wL^2/8h always
   D. H = V_A + V_B
   Answer: A. Because ds/I = dx/Ic, the compatibility condition reduces to this ratio. (Key point: Thrust formula)

2. For a parabolic arch, ∫y^2dx over the span equals:
   A. h^2 L/3
   B. 8 h^2 L/15 = 800 here
   C. 4hL/3
   D. h L/2
   Answer: B. 8(25)(60)/15 = 800. (Key point: Integral ∫y^2dx)

3. With ∫M0 y dx = 1,800,000 and ∫y^2dx = 800, H is:
   A. 1125 kN
   B. 375 kN
   C. 2250 kN
   D. 1500 kN
   Answer: C. 1,800,000/800 = 2250 kN. (Key point: Value of H)

## Related in Theory of Structures II

- [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 parabolic arch: BM at D](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-point-d-30-50) — Theory of Structures II, chapter 2
- [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 10 m intervals](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-80kn-d-i-ic) — 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

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- [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×
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