---
title: "Arch 100 m: rolling UDL 60 m, section at 30 m"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-100m-rolling-udl-60m-25"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-100m-rolling-udl-60m-25
site: BCT Engineering
---

# Arch 100 m: rolling UDL 60 m, section at 30 m

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2080 Chaitra, 16 marks
- Video: [Influence Line Diagrams for Simply Supported Beams - Problem No 2 (UDL length less than beam span)](https://www.youtube.com/watch?v=vIXTV9x7lsg) — 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-rolling-udl-60m-25
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

In a three hinged symmetrical parabolic arch of span 100 m and central rise 10 m, a UDL load 25 kN/m and length 60 m moves from left to right. Determine maximum moment and normal thrusts at a section 30 m from left support.

## Short answer

Using ILDs, maximum positive moment is +4375 kN m with normal thrust 1083.2 kN; maximum negative moment is -4375 kN m with normal thrust 2081.5 kN at the 30 m section.

## Answer

**Data:** three-hinged symmetrical parabolic arch, $L = 100$ m, $f = 10$ m, $y = \dfrac{4f\,x(L-x)}{L^2} = 0.004\,x\,(100-x)$. A UDL of 25 kN/m and 60 m length moves from left to right. Section $D$ at $x = 30$ m: $y_D = 8.4$ m, $\tan\theta = 0.004\,(100-60) = 0.16$, $\theta = 9.09^\circ$. (When the load covers only the first or the last part of the span, part of the 60 m load is off the span.)

#### ILDs for the section at x = 30 m
The ILD of the horizontal thrust is $H = \dfrac{x'}{2f}$ for a unit load at $x'$ left of the crown and $\dfrac{L-x'}{2f}$ right of it (peak $\dfrac{L}{4f} = 2.5$ at the crown). The ILD of the bending moment at the section is $M = M_b - H\,y$.

| x (m) | Point | ILD of H | ILD of M at section |
|---|---|---|---|
| 0 | A | 0 | 0 |
| 30 | section D | 1.5 | 8.4 |
| 41.67 | neutral point | 2.083 | 0 |
| 50 | crown C | 2.5 | -6 |
| 100 | B | 0 | 0 |

The ILD of $M$ crosses zero at the **neutral point**, 41.67 m from $A$ (vertically below the intersection of line $AD$ produced and line $BC$). The ordinates are positive on one side of it and negative on the other.

#### Maximum positive bending moment
The UDL covers the span from $x = 0$ to 41.67 m (the rest of the load is still off the span), which is exactly the region where the ILD is positive.

$$
\begin{aligned} M &= w\times(\text{area of ILD}) = 25\times 175 = 4375\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 25\times 43.403 = 1085.07\ \text{kN} \end{aligned}
$$

At the section: $y = 8.4$ m, $\theta = 9.09^\circ$, simple-beam shear $V_b = 74.65$ kN.

$$
\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 74.65\times0.9874 - 1085.07\times0.158 = -97.72\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 74.65\times0.158 + 1085.07\times0.9874 = 1083.24\ \text{kN} \end{aligned}
$$

#### Maximum negative bending moment
The UDL covers the span from $x = 41.67$ m to 100 m (the rest of the load is off the span beyond $B$), which is exactly the region where the ILD is negative.

$$
\begin{aligned} M &= w\times(\text{area of ILD}) = 25\times(-175) = -4375\ \text{kN m}\\ H &= w\times(\text{area of ILD of }H) = 25\times 81.597 = 2039.93\ \text{kN} \end{aligned}
$$

At the section: $y = 8.4$ m, $\theta = 9.09^\circ$, simple-beam shear $V_b = 425.35$ kN.

$$
\begin{aligned} V &= V_b\cos\theta - H\sin\theta = 425.35\times0.9874 - 2039.93\times0.158 = 97.72\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 425.35\times0.158 + 2039.93\times0.9874 = 2081.51\ \text{kN} \end{aligned}
$$

#### Maximum horizontal thrust (for information)
The ILD of $H$ is positive everywhere; the thrust is greatest when the 60 m load is symmetrical about the crown (20 m to 80 m): $H_{max} = 25\times 105 = 2625$ kN.

**Answer: maximum bending moment at 30 m from the left support is $+4375$ kN m (UDL from the left support up to the neutral point at 41.67 m), with normal thrust $1083.2$ kN, and $-4375$ kN m (UDL from the neutral point to the right support), with normal thrust $2081.5$ kN.**

## Check yourself

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

1. What is the neutral point of the ILD of bending moment at a section?
   A. The point of maximum ordinate (the section itself)
   B. The crown, where H is greatest
   C. The point where the ILD ordinate is zero and the sign changes (41.67 m from A here)
   D. The point where the ILD of H is zero
   Answer: C. At the neutral point a load causes no moment at the section. Loads on one side give positive moment and loads on the other side give negative moment. (Key point: Neutral point)

2. For a section at 30 m, where must the 60 m UDL be placed for the maximum positive moment?
   A. Covering the span from A up to the neutral point (0 to 41.67 m)
   B. Covering the whole span from 0 to 100 m
   C. Covering the span from the neutral point to B
   D. Covering 20 m to 80 m, symmetrical about the crown
   Answer: A. Maximum positive moment needs the load over the entire positive part of the ILD, which stops at the neutral point. Loading beyond it would reduce the moment. (Key point: Position for maximum positive moment)

3. What is the peak ordinate of the ILD of H, and where is it?
   A. L/8f = 1.25 at the section
   B. L/4f = 2.5 at the crown
   C. f/L = 0.1 at the crown
   D. L/2f = 5 at the supports
   Answer: B. H = M_C/f, and a unit load at the crown gives simple-beam crown moment L/4. So peak H = L/4f = 100/40 = 2.5. (Key point: ILD of horizontal thrust)

4. How do the maximum positive and negative moments at 30 m compare, and how are they calculated?
   A. Positive is 4375 kN m but negative is zero because the load is off the span
   B. Both equal H x y, computed from the maximum thrust only
   C. Positive is 4375 kN m and negative is the same sign with the load at the crown
   D. Both are 4375 kN m in magnitude (+ and -); each is w times the area of the ILD over the loaded length
   Answer: D. M = w x (area of the M-ILD under the load). Here the positive and negative areas happen to be 175 and -175, giving +4375 and -4375 kN m. (Key point: Maximum moments)

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