---
title: "Parabolic arch 80 m: inclined load, forces at D"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-80m-250kn-inclined-100kn"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-80m-250kn-inclined-100kn
site: BCT Engineering
---

# Parabolic arch 80 m: inclined load, forces at D

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2071 Magh (old course), 10 marks
- Video: [Finding Support Reactions and Drawing Bending Moment Diagram of Three Hinged Parabolic Arch](https://www.youtube.com/watch?v=JMKcU2L89PA) — Master Civil Engineering
- 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-80m-250kn-inclined-100kn
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

A three hinged parabolic arch having span 80 m and central rise 8 m is loaded as shown in figure below. Determine bending moment, normal thrust and radial shear force at section D. [Figure: three hinged parabolic arch ACB; section D at 15 m horizontally from the left support A; 250 kN load inclined at 30° to the horizontal acting 10 m from the crown hinge C on the left half; 100 kN vertical load 15 m from the right support B.]

## Short answer

Using equilibrium and hinge condition for a three-hinged parabolic arch under inclined and vertical loads, at section D (15 m from left support): bending moment = 109.77 kN m, normal thrust = 225.32 kN, radial shear = 22.61 kN.

## Answer

**Data and assumptions:** parabolic three-hinged arch, span $L = 80$ m, rise $f = 8$ m, $y = \dfrac{4\times 8\,x\,(80-x)}{80^2} = 0.005\,x\,(80-x)$. The 250 kN inclined load acts at $x = 30$ m (10 m left of the crown), pointing downward and to the right at 30° to the horizontal (assumed). Its components: horizontal $250\cos30^\circ = 216.51$ kN (to the right), vertical $250\sin30^\circ = 125$ kN (downward). 100 kN vertical load at 15 m from $B$ ($x = 65$ m). Section $D$ at $x = 15$ m.

Heights of the loads: $y(30) = 7.5$ m, $y(65) = 4.875$ m; at $D$: $y = 4.875$ m, $\tan\theta = 0.005\,(80-30) = 0.25$, $\theta = 14.04^\circ$.

Let the thrusts be $H_A$ (to the right at $A$) and $H_B$ (to the left at $B$), and the vertical reactions $V_A$, $V_B$.

#### Reactions
**Horizontal equilibrium:**
$$
H_A - H_B + 216.51 = 0 \Rightarrow H_B = H_A + 216.51
$$

**Moments about $A$:**
$$
V_B\times 80 = 100\times 65 + 125\times 30 + 216.51\times 7.5 \Rightarrow V_B = \frac{11873.8}{80} = 148.42\ \text{kN}
$$

$$
V_A = 225 - 148.42 = 76.58\ \text{kN}
$$

**Hinge condition** (moment about $C$ of the forces on the left part, $C$ at $(40, 8)$):
$$
V_A\times 40 - H_A\times 8 - 125\times 10 - 216.51\times(8-7.5) = 0
$$

$$
H_A = \frac{76.58\times 40 - 1250 - 216.51\times 0.5}{8} = 213.11\ \text{kN}, \qquad H_B = 429.61\ \text{kN}
$$

#### Forces at D (left part, 15 m from A, no load on it)
$$
\begin{aligned} M &= V_A\times 15 - H_A\times y_D = 76.58\times 15 - 213.11\times 4.875 = 109.77\ \text{kN m}\\ V &= V_A\cos\theta - H_A\sin\theta = 76.58(0.9701) - 213.11(0.2425) = 22.61\ \text{kN}\\ N &= V_A\sin\theta + H_A\cos\theta = 76.58(0.2425) + 213.11(0.9701) = 225.32\ \text{kN} \end{aligned}
$$

**Answer: at $D$: bending moment $= 109.77$ kN m (sagging), normal thrust $= 225.32$ kN, radial shear $= 22.61$ kN.**

## Check yourself

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

1. How is the 250 kN load at 30 degrees to the horizontal treated?
   A. Resolved into horizontal 250cos30 = 216.51 kN and vertical 250sin30 = 125 kN
   B. As a vertical 250 kN load
   C. Horizontal 125 kN and vertical 216.51 kN
   D. Ignored in the moment about A
   Answer: A. The horizontal component has a moment about A and about C (arm 8 - 7.5 m) and also affects horizontal equilibrium. (Key point: Components of inclined load)

2. What is the horizontal equilibrium equation here?
   A. H_A = H_B
   B. H_A - H_B + 216.51 = 0, so H_B = H_A + 216.51
   C. H_A + H_B = 216.51
   D. H_B = 0
   Answer: B. Unlike a vertical-load case the thrusts are no longer equal, because the horizontal component of the inclined load must be balanced. (Key point: Horizontal equilibrium)

3. What is included in the moment about C of the left part?
   A. Only V_A and H_A
   B. Only the vertical loads
   C. V_A x 40, H_A x 8, the 125 kN vertical x 10, and the horizontal 216.51 kN x 0.5 m arm
   D. The 100 kN load at 65 m as well
   Answer: C. All forces on the left of C count, including the horizontal component of the inclined load (load height 7.5 m, crown height 8 m). (Key point: Hinge condition)

4. Why are M, V and N at D found using only V_A and H_A?
   A. Because the loads are zero at D
   B. Because the reaction at B equals zero
   C. Because the horizontal load acts at D
   D. D is at 15 m, on the left of all loads, so only the reactions at A act on that part
   Answer: D. The first load is at 30 m, so the left part up to D carries no load: M = 76.58x15 - 213.11x4.875 = 109.77 kN m. (Key point: Section D forces)

## 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×

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