---
title: "Circular arch 40 m: sketch, equation, forces at 10 m"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-circular-40m-60kn-4knm"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-circular-40m-60kn-4knm
site: BCT Engineering
---

# Circular arch 40 m: sketch, equation, forces at 10 m

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2072 Magh, 12 marks
- Video: [Three Hinged circular Arch - Formula Derivation For Radius R, Height y and Slope θ](https://www.youtube.com/watch?v=thyM0zJiiHI) — 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-circular-40m-60kn-4knm
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

A three hinged circular arch has span 40 m and rise 5 m. Make a sketch of the arch and give the equation to it. It carries a concentrated load 60 kN at 8 m from the right support and uniformly distributed load 4 kN/m over left half portion. Determine bending moment, radial shear force and normal thrust at a section 10 m from the left support.

## Short answer

Using equilibrium equations for a three-hinged circular arch with span 40 m and rise 5 m under given loads, at 10 m from left support: bending moment = 32.73 kN m, radial shear = 0.98 kN, normal thrust = 131.94 kN.

## Answer

**Data:** three-hinged circular arch, span $L = 40$ m, rise $f = 5$ m. Radius $R = \dfrac{L^2/4+f^2}{2f} = \dfrac{400+25}{10} = 42.5$ m; the centre lies $R - f = 37.5$ m below the springing line, at mid-span.

#### Sketch and equation of the arch
```text
              C (crown hinge)
            .-'''-.            rise f = 5 m
        .-'         '-.
     .-'                 '-.
   A o                       o B
   |<---------- 40 m ------>|
   UDL 4 kN/m on A-C      60 kN at 8 m from B
```

With the origin at $A$ and $x$ measured along the springing line, the circle with centre $(20, -37.5)$ and radius $42.5$ gives

$$
y = \sqrt{42.5^2 - (x-20)^2} - 37.5
$$

Check: $y(0) = \sqrt{1806.25-400} - 37.5 = 0$ and $y(20) = 42.5 - 37.5 = 5$ m.

Loads: UDL 4 kN/m on the left half ($x = 0$ to 20 m); 60 kN at 8 m from $B$ ($x = 32$ m). Section at $x = 10$ m.

#### Step 1: Slope at the section
$$
\tan\theta = \frac{dy}{dx} = \frac{20-x}{\sqrt{R^2-(x-20)^2}}
$$

At the section $x = 10$ m: $y = 3.807$ m, $\tan\theta = 0.2421$, so $\theta = 13.61^\circ$, $\sin\theta = 0.2353$, $\cos\theta = 0.9719$.

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

$$
R_B \times 40 = (4\times 20)\times 10 + 60\times 32 = 2720 \Rightarrow R_B = 68\ \text{kN}
$$

$$
R_A = 140 - 68 = 72\ \text{kN}
$$

#### 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 $f = 5$ m):

$$
H \times 5 = 72\times 20 - (4\times 20)\times 10 = 640 \Rightarrow H = 128\ \text{kN}
$$

#### Step 4: Forces at the section (x = 10 m)
Simple-beam values at the section: $M_b = 520$ kN m, $V_b = 32$ kN.

$$
\begin{aligned} M &= M_b - H\,y = 520 - 128\times 3.807 = 32.73\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = 32\times 0.9719 - 128\times 0.2353 = 0.98\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 32\times 0.2353 + 128\times 0.9719 = 131.94\ \text{kN} \end{aligned}
$$

**Answer: at 10 m from the left support: bending moment $= 32.73$ kN m, radial shear $= 0.98$ kN, normal thrust $= 131.94$ kN.**

## Check yourself

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

1. For a circular arch of span 40 m and rise 5 m, what are the radius and the centre position?
   A. R = 40 m, centre at the crown
   B. R = 42.5 m, centre 37.5 m below the springing line at mid-span
   C. R = 42.5 m, centre at the springing line
   D. R = 25 m, centre 20 m below the springing line
   Answer: B. R = (20^2 + 5^2)/(2x5) = 42.5 m and the centre lies R - f = 37.5 m below the springing line. (Key point: Radius and centre)

2. How is H found?
   A. H = R_A = 72 kN
   B. H x 5 = 72 x 10, so H = 144 kN
   C. H x 5 = R_A x 20 - (4x20)x10 = 640, so H = 128 kN
   D. H = 60 kN, the point load
   Answer: C. Moment at crown hinge C for the left part: 72x20 - 80x10 = 640. Dividing by the rise 5 m gives 128 kN. (Key point: Thrust)

3. How is tan(theta) obtained for the circular arch?
   A. Differentiate: dy/dx = (20 - x)/sqrt(R^2 - (x-20)^2) = 0.2421 at x = 10
   B. Use 4f(L-2x)/L^2 as for a parabola
   C. tan(theta) = y/x
   D. tan(theta) = f/L
   Answer: A. The circular equation is differentiated, giving 10/sqrt(1806.25-100) = 0.2421. (Key point: Slope at the section)

4. How is R_B found?
   A. R_B = 140/2 = 70 kN
   B. R_B x 40 = 60 x 8, so 12 kN
   C. Moments about A with the UDL at 20 m: 55 kN
   D. Moments about A: R_B x 40 = 80x10 + 60x32, giving 68 kN
   Answer: D. The left-half UDL (80 kN) acts at 10 m and the 60 kN load is 32 m from A. 2720/40 = 68 kN. (Key point: Reaction R_B)

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