---
title: "Arch equation given: BM, thrust, shear at 5 m"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-y-equation-30knm-right-half"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-y-equation-30knm-right-half
site: BCT Engineering
---

# Arch equation given: BM, thrust, shear at 5 m

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2081 Chaitra, 8 marks
- Video: [Unsymmetrical Arches| Normal Thrust n Radial Shear And Max BM | Structural Analysis - I | Hindi |](https://www.youtube.com/watch?v=oSo6xE7KVpA) — TheCivilGuy
- 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-y-equation-30knm-right-half
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

The equation of a three-hinged arch, with origin at its left support, is $y = x - \frac{x^2}{40}$. The span of the arch is 40 m. Determine bending moment, normal thrust and radial shear at section 5 m from right support, when the arch is carrying a uniformly distributed load of 30 kN/m for the right half.

## Short answer

Parabolic arch with rise 10 m, section at x = 35 m. Bending moment = 562.5 kN m, normal thrust = 420 kN, radial shear = -60 kN.

## Answer

**Data:** $y = x - \dfrac{x^2}{40}$ with the origin at $A$, span $L = 40$ m. Comparing with $y = \dfrac{4f\,x\,(L-x)}{L^2} = \dfrac{4f}{L}x - \dfrac{4f}{L^2}x^2$: $\dfrac{4f}{L} = 1 \Rightarrow f = 10$ m (the arch is parabolic with a rise of 10 m). UDL 30 kN/m on the right half ($x = 20$ to 40 m); the section is 5 m from $B$, i.e. at $x = 35$ m.

#### Step 1: Equation and slope
$$
y = x - \frac{x^2}{40}
$$

$$
\tan\theta = \frac{dy}{dx} = 1 - \frac{x}{20}
$$

At the section $x = 35$ m: $y = 4.375$ m, $\tan\theta = -0.75$, so $\theta = -36.87^\circ$, $\sin\theta = -0.6$, $\cos\theta = 0.8$.

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

$$
R_B \times 40 = (30\times 20)\times 30 = 18000 \Rightarrow R_B = 450\ \text{kN}
$$

$$
R_A = 600 - 450 = 150\ \text{kN}
$$

#### 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):

$$
H \times 10 = 150\times 20 = 3000 \Rightarrow H = 300\ \text{kN}
$$

#### Step 4: Forces at the section (x = 35 m)
Simple-beam values at the section: $M_b = 1875$ kN m, $V_b = -300$ kN.

$$
\begin{aligned} M &= M_b - H\,y = 1875 - 300\times 4.375 = 562.5\ \text{kN m}\\ V &= V_b\cos\theta - H\sin\theta = (-300)\times 0.8 - 300\times (-0.6) = -60\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = (-300)\times (-0.6) + 300\times 0.8 = 420\ \text{kN} \end{aligned}
$$

**Answer: at 5 m from the right support: $M = 562.5$ kN m, normal thrust $N = 420$ kN, radial shear $V = -60$ kN.**

## Check yourself

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

1. The arch is y = x - x^2/40 with span 40 m. How is the rise found?
   A. Put x = 40 in the equation, so f = 0
   B. Compare with y = 4f x(L-x)/L^2, so 4f/L = 1 and f = 10 m
   C. Differentiate once, so f = 1 m
   D. Set y = x/40, so f = 40 m
   Answer: B. Matching the coefficient of x gives 4f/L = 1, so f = 10 m. This is a parabolic arch with a 10 m rise. (Key point: Rise from the equation)

2. How is H found for the UDL on the right half?
   A. Use H = R_A = 150 kN since the UDL is symmetric
   B. Take moments about A for the whole arch: H = 450 kN
   C. Take moments about the crown hinge of forces on the left part: H x 10 = 150 x 20, so H = 300 kN
   D. Use H = wL/8f for the full UDL
   Answer: C. Bending moment at the crown hinge is zero. With R_A = 150 kN and no load on the left half, H x 10 = 150 x 20, so H = 300 kN. (Key point: Horizontal thrust)

3. Which expressions give the radial shear and normal thrust at the section?
   A. V = Vb cos(theta) - H sin(theta); N = Vb sin(theta) + H cos(theta)
   B. V = Vb sin(theta) + H cos(theta); N = Vb cos(theta) - H sin(theta)
   C. V = Vb - H; N = H
   D. V = Vb cos(theta) + H sin(theta); N = Vb sin(theta) - H cos(theta)
   Answer: A. Resolving the simple-beam shear and the thrust perpendicular and along the tangent gives these forms. Here V = -60 kN and N = 420 kN. (Key point: Section forces)

4. What is the bending moment at 5 m from the right support?
   A. M = Mb + H y = 3187.5 kN m
   B. M = Mb - H y = 1875 - 300 x 4.375 = 562.5 kN m
   C. M = Mb = 1875 kN m
   D. M = H y = 1312.5 kN m
   Answer: B. The arch moment is the simple-beam moment minus the moment of the thrust, H x y. The ordinate at x = 35 m is 4.375 m. (Key point: Bending moment at x = 35 m)

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