---
title: "Parabolic arch 90 m: BM, shear, thrust at D"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-90m-180kn-6knm-d"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-90m-180kn-6knm-d
site: BCT Engineering
---

# Parabolic arch 90 m: BM, shear, thrust at D

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2073 Magh, 10 marks
- Video: [SA54: Analysis of Three-Hinged Arches](https://www.youtube.com/watch?v=pJKfOvN36J0) — Dr. Structure
- 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-90m-180kn-6knm-d
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

Determine bending moment, radial shear force and normal thrust at point D of the three hinged parabolic arch shown in figure below. [Figure: span 90 m, rise 6 m; 180 kN vertical load 25 m from the left support; section D 30 m from the left support A; 6 kN/m UDL over the right half.]

## Short answer

At section D (30 m) of a 90 m span, 6 m rise three-hinged parabolic arch with 180 kN at 25 m left and UDL 6 kN/m right half: reactions RA=197.5 kN, H=881.25 kN. With y=5.333 m, θ=5.08°, Mb=5025 kN m, Vb=17.5 kN. Bending moment = 325 kN m, radial shear = −60.59 kN, normal thrust = 879.34 kN.

## Answer

**Data:** parabolic three-hinged arch, $L = 90$ m, $f = 6$ m. 180 kN vertical load at 25 m from $A$; UDL 6 kN/m on the right half ($x = 45$ to 90 m). Section $D$ at $x = 30$ m from $A$.

#### Step 1: Equation of the arch
$$
y = \frac{4f\,x\,(L-x)}{L^2} = \frac{4\times 6\,x\,(90-x)}{90^2}
$$

$$
\tan\theta = \frac{dy}{dx} = \frac{4f\,(L-2x)}{L^2}
$$

At the section $x = 30$ m: $y = 5.333$ m, $\tan\theta = 0.0889$, so $\theta = 5.08^\circ$, $\sin\theta = 0.0885$, $\cos\theta = 0.9961$.

#### Step 2: Vertical reactions
Taking moments about $A$ for the whole arch:

$$
R_B \times 90 = 180\times 25 + (6\times 45)\times 67.5 = 22725 \Rightarrow R_B = 252.5\ \text{kN}
$$

$$
R_A = 450 - 252.5 = 197.5\ \text{kN}
$$

#### Step 3: Horizontal thrust
The bending moment at the crown hinge $C$ is zero. Taking moments of the forces on the left of $C$ about $C$ (rise $f = 6$ m):

$$
H \times 6 = 197.5\times 45 - 180\times 20 = 5287.5 \Rightarrow H = 881.25\ \text{kN}
$$

#### Step 4: Forces at the section (x = 30 m)
Simple-beam values at the section: $M_b = 5025$ kN m, $V_b = 17.5$ kN.

$$
\begin{aligned} M &= M_b - H\,y = 5025 - 881.25\times 5.333 = 325\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 17.5\times 0.9961 - 881.25\times 0.0885 = -60.59\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 17.5\times 0.0885 + 881.25\times 0.9961 = 879.34\ \text{kN} \end{aligned}
$$

**Answer: at $D$: bending moment $= 325$ kN m, radial shear $= -60.59$ kN, normal thrust $= 879.34$ kN.**

## Check yourself

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

1. How is R_B found?
   A. Moments about A: R_B x 90 = 180x25 + (6x45)x67.5, giving 252.5 kN
   B. R_B = 450/2 = 225 kN
   C. Moments about A with the UDL at 45 m: 218 kN
   D. Moments about C for the right half: 270 kN
   Answer: A. The right-half UDL (270 kN) acts at 67.5 m from A. Total is 22725/90 = 252.5 kN. (Key point: R_B)

2. Why does the UDL on the right half not appear in the equation for H taken on the left part?
   A. The UDL has no effect on the thrust
   B. The left part is considered, and the UDL lies entirely on the right part
   C. The UDL acts through the crown hinge
   D. The hinge transfers the UDL to the support
   Answer: B. Taking moments about C of forces on the left only: H x 6 = 197.5x45 - 180x20, so H = 881.25 kN. The right UDL influences H only through R_A. (Key point: Crown condition)

3. What is M at D?
   A. M = R_A x 30 = 5925 kN m
   B. M = H y = 4700 kN m
   C. M = Mb - H y = 5025 - 881.25 x 5.333 = 325 kN m
   D. M = Mb + H y = 9725 kN m
   Answer: C. The simple-beam moment 5025 kN m minus the thrust moment 4700 kN m gives 325 kN m. (Key point: Bending moment at D)

4. tan(theta) at x = 30 m for L = 90 and f = 6 is:
   A. 0 because it is near the crown
   B. 0.5
   C. 4f/L = 0.267
   D. 4f(L-2x)/L^2 = 0.0889
   Answer: D. tan(theta) = 24x30/8100 = 0.0889, so theta = 5.08 degrees. (Key point: Section slope)

## Close variants set in other exams

- [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×

## Related in Theory of Structures I

- [Arch: reactions, BM, thrust and shear at D by ILD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-50m-udl-50kn-ild-d-15m) — 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×
- [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
- [Arch 20 m: reactions, BM, thrust, shear, BMD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-20m-udl-2knm-left-8m) — Theory of Structures I, chapter 6
- [Parabolic arch 80 m: inclined load, forces at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-80m-250kn-inclined-100kn) — Theory of Structures I, chapter 6, asked 2×
- [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

## 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 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
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- [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