---
title: "Neutral point in ILD of arch"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-neutral-point"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-neutral-point
site: BCT Engineering
---

# Neutral point in ILD of arch

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2077 Chaitra, 4 marks; 2076 Bhadra, 6 marks; 2075 Bhadra, 4 marks; 2072 Magh, 6 marks; 2071 Magh (old course), 6 marks (asked 5 times)
- 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-neutral-point
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

What is the neutral point in an influence line diagram of an arch? Explain how neutral points are determined in ILD of bending moment (and shear force) of a three hinged arch, and derive the expression for the location of the neutral point.

## Short answer

The neutral point is the load position for zero bending moment at a section. Graphically, intersect line AD (support to section) with line BC (other support to crown) and drop a vertical to the span. For a parabolic arch, s0 = L^2/(3L - 2a).

## Answer

#### Neutral point
For a section $D$ of a three-hinged arch, the **neutral point** is the position of a unit load on the span for which the bending moment at $D$ is zero. In the ILD of $M_D$ it is the point where the influence line crosses the base line (ordinate zero). A load on one side of it produces a positive moment and on the other side a negative moment at $D$; this is why a rolling load must be put on different sides of the neutral point for the maximum positive and negative moments.

#### Graphical determination (section D in the left half)
When a load $P$ acts on the left half of the arch (on the segment $DC$), the right portion $CB$ carries no load, so it is a two-force member: the reaction at $B$ acts along the line $BC$. For $M_D = 0$ the reaction at $A$ must act along the line $AD$ (its moment about $D$ is then zero). The load $P$, $R_A$ and $R_B$ are in equilibrium, so the three lines must be concurrent.

1. Join $B$ to the crown hinge $C$ and extend it.
2. Join $A$ to the section $D$ and extend it.
3. Let the two lines meet at $E$.
4. The vertical through $E$ meets the span at the **neutral point** $N$.

```text
            C
         .-' '-.        E  (intersection of AD and BC)
      D-'       '-.   .'
     /             '-B.
    A    . . . N . . . . 
         neutral point (vertical below E)
```

#### Derivation of the position
Let the span be $L$, the rise $f$, and let the section be at $x_D = a$ with ordinate $y_D$. Take a unit load at a distance $s$ from $A$ with $a < s < L/2$.

$$
R_A = \frac{L-s}{L}, \qquad H = \frac{M_C}{f} = \frac{R_B\,(L/2)}{f} = \frac{s}{2f}
$$

$$
M_D = R_A\,a - H\,y_D = \frac{(L-s)\,a}{L} - \frac{s\,y_D}{2f}
$$

Putting $M_D = 0$:

$$
a - s\left(\frac{a}{L} + \frac{y_D}{2f}\right) = 0 \Rightarrow s_0 = \frac{a}{\dfrac{a}{L}+\dfrac{y_D}{2f}} = \frac{2f}{\dfrac{2f}{L}+\dfrac{y_D}{a}}
$$

This is the same as the intersection of the lines $y = \dfrac{y_D}{a}x$ (line $AD$) and $y = \dfrac{2f}{L}(L-x)$ (line $BC$).

For a **parabolic** arch, $y_D = \dfrac{4fa(L-a)}{L^2}$, so

$$
s_0 = \frac{L^2}{3L-2a}
$$

**Example:** $L = 100$ m, $f = 10$ m, $a = 30$ m: $s_0 = \dfrac{100^2}{300-60} = 41.67$ m from $A$. A load to the left of $s_0$ gives positive $M_D$, a load to the right gives negative $M_D$ (this is confirmed by the ILD ordinates $+8.4$ m at $D$ and $-6$ m at the crown).

For the **radial shear** at $D$ the neutral point is found in the same way: it is the load position for which $V_D = R_A\cos\theta - H\sin\theta = 0$, i.e. where the reaction at $A$ is parallel to the tangent at $D$ (line through $A$ parallel to the tangent at $D$ meets line $BC$ at the point vertically above the neutral point).

## Check yourself

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

1. What is the neutral point in the ILD of an arch for a section D?
   A. Position of the unit load for which the moment at D is zero
   B. The crown of the arch
   C. The point of maximum moment
   D. The foot of the section D
   Answer: A. It is where the influence line crosses the base line, so loads on either side give moments of opposite sign. (Key point: Definition)

2. For a load on the left half, why does R_B act along BC?
   A. Because B is a roller
   B. Because the unloaded right half with hinge C and B is a two-force member
   C. Because the thrust is vertical
   D. Because C is fixed
   Answer: B. Hinge C has zero moment and CB has no load, so only two forces act on it and they must be collinear along BC. (Key point: Two-force member)

3. For M_D = 0, what is the direction of R_A, and how is the neutral point found?
   A. R_A along AC; vertical through A
   B. R_A along AD; the lines AD and BC meet at E, and the vertical through E gives N
   C. R_A horizontal; point below C
   D. R_A along BD; mid-span
   Answer: B. R_A must pass through D for zero moment there; the load, R_A and R_B are concurrent at E = AD intersect BC. (Key point: Graphical construction)

4. For a parabolic arch, the neutral point is s0 = L^2/(3L - 2a). For L = 100 m, a = 30 m, s0 = ?
   A. 33.3 m
   B. 50 m
   C. 41.67 m
   D. 30 m
   Answer: C. s0 = 10000/(300 - 60) = 41.67 m from A. (Key point: Parabolic arch formula)

## Related in Theory of Structures I

- [Arch: reactions, BM, thrust and shear at D by ILD](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-50m-udl-50kn-ild-d-15m) — Theory of Structures I, chapter 6
- [ILD for bending moment at section D of arch](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-ild-bm-d-given-arch) — Theory of Structures I, chapter 6
- [Circular arch: BM, thrust, shear at D, BM diagram](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-d-20knm-10kn-8m) — Theory of Structures I, chapter 6
- [Define influence line diagram and its uses](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Adefine-ild-uses) — Theory of Structures I, chapter 5, asked 5×
- [No bending moment under UDL in parabolic arch](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-no-bm-udl) — Theory of Structures I, chapter 6, asked 2×
- [Overhang beam ILD for shear force at C](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aoverhang-ild-sf-c-2071o) — Theory of Structures I, chapter 5

## Related in other subjects

- [Define shear force and bending moment](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Adefine-shear-force-bending-moment) — Strength of Materials, chapter 1, asked 2×
- [Two-hinged arch: BM, thrust, radial shear, support yield](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-thrust-shear-yield-support) — Theory of Structures II, chapter 2
- [ILD: bending moment at D, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-d-ab-bc-2) — Theory of Structures II, chapter 4
- [ILD: bending moment at B, two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aild-bm-b-two-span-8) — Theory of Structures II, chapter 4