---
title: "Parabolic arch 100 m: UDL left half and 80 kN"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-100m-80kn-udl-d-20m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-100m-80kn-udl-d-20m
site: BCT Engineering
---

# Parabolic arch 100 m: UDL left half and 80 kN

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2075 Baisakh, 12 marks
- Video: [Three Hinged Parabolic Arch - Problem No 2](https://www.youtube.com/watch?v=qu1ZVSr75Wg) — Stan Academy
- 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-100m-80kn-udl-d-20m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

In the three hinged parabolic arch shown in figure below determine bending moment, normal thrust and radial shear force at section D. [Figure: parabolic three hinged arch, span 100 m, rise 10 m at the crown C; section D at 20 m horizontally from the left support A; 50 kN/m UDL over the left half; 80 kN vertical load 25 m from the right support B.]

## Short answer

For a three-hinged parabolic arch (span 100 m, rise 10 m) with 50 kN/m UDL left half and 80 kN at 25 m from right, at section D (20 m from left): y=6.4 m, θ=13.5°. Reactions: RA=1895 kN, H=3225 kN. Bending moment = 7260 kN m, shear = 117.66 kN, thrust = 3344.82 kN.

## Answer

**Data:** parabolic three-hinged arch, $L = 100$ m, rise $f = 10$ m at the crown $C$. UDL 50 kN/m on the left half ($x = 0$ to 50 m); 80 kN vertical load 25 m from $B$ ($x = 75$ m). Section $D$ at $x = 20$ m from $A$.

#### Step 1: Equation of the arch
$$
y = \frac{4f\,x\,(L-x)}{L^2} = \frac{4\times 10\,x\,(100-x)}{100^2}
$$

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

At the section $x = 20$ m: $y = 6.4$ m, $\tan\theta = 0.24$, so $\theta = 13.5^\circ$, $\sin\theta = 0.2334$, $\cos\theta = 0.9724$.

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

$$
R_B \times 100 = (50\times 50)\times 25 + 80\times 75 = 68500 \Rightarrow R_B = 685\ \text{kN}
$$

$$
R_A = 2580 - 685 = 1895\ \text{kN}
$$

#### 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 $f = 10$ m):

$$
H \times 10 = 1895\times 50 - (50\times 50)\times 25 = 32250 \Rightarrow H = 3225\ \text{kN}
$$

#### Step 4: Forces at the section (x = 20 m)
Simple-beam values at the section: $M_b = 27900$ kN m, $V_b = 895$ kN.

$$
\begin{aligned} M &= M_b - H\,y = 27900 - 3225\times 6.4 = 7260\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 895\times 0.9724 - 3225\times 0.2334 = 117.66\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 895\times 0.2334 + 3225\times 0.9724 = 3344.82\ \text{kN} \end{aligned}
$$

**Answer: at $D$: bending moment $= 7260$ kN m, normal thrust $= 3344.82$ kN, radial shear $= 117.66$ kN.**

## Check yourself

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

1. How is H calculated?
   A. H = R_B = 685 kN
   B. H x 10 = R_A x 50 - (50x50)x25, so H = 3225 kN
   C. H x 10 = R_B x 50, so H = 3425 kN
   D. H x 6.4 = R_A x 20, so H = 5921 kN
   Answer: B. Moment at C of left-side forces: 1895x50 - 2500x25 = 32250. Dividing by f = 10 gives 3225 kN. Do not use the 80 kN load, it is on the right. (Key point: Thrust from crown condition)

2. How are y and tan(theta) at x = 20 m obtained?
   A. y = f = 10 m and tan(theta) = 0 as at the crown
   B. y = 0.24 m and tan(theta) = 6.4
   C. From y = 4f x(L-x)/L^2 = 6.4 m and its derivative tan(theta) = 4f(L-2x)/L^2 = 0.24
   D. y = x/L x f = 2 m and tan(theta) = 0.1
   Answer: C. The parabola gives y = 40x20x80/10000 = 6.4 m. Differentiating gives the slope 0.24, so theta = 13.5 degrees. (Key point: Ordinate and slope at D)

3. What is the bending moment at D?
   A. M = H y = 20640 kN m
   B. M = Mb + H y = 48540 kN m
   C. M = R_A x 20 = 37900 kN m
   D. M = Mb - H y = 27900 - 3225 x 6.4 = 7260 kN m
   Answer: D. Simple-beam moment minus the thrust moment: 27900 - 20640 = 7260 kN m. (Key point: Moment at D)

## Close variants set in other exams

- [Parabolic arch 90 m: BM, shear, thrust at D](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-90m-180kn-6knm-d) — Theory of Structures I, chapter 6, asked 2×
- [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×
- [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×

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## Related in other subjects

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