---
title: "Arch 20 m: ILDs and check by equilibrium"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-20m-rise4-ild-check-4kn"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-20m-rise4-ild-check-4kn
site: BCT Engineering
---

# Arch 20 m: ILDs and check by equilibrium

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2071 Bhadra, 13 marks
- 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-20m-rise4-ild-check-4kn
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

A three hinged symmetrical parabolic arch of span 20 m and rise 4 m is with a point load of magnitude 4 kN at 4 m distance from the left hinge. First, draw influence line diagram (ILD) for bending moment (BM), radial shear (RS), and normal thrust (NT) for the section where the point load is and then calculate the values of BM, RS and NT at the section using the ILDs. Also check these values of internal forces at the section by first principle using equilibrium equations.

## Short answer

Using influence line ordinates for the section at the load (4 m from left support), with a 4 kN point load: M = 7.68 kN m, radial shear = 2.019 kN, normal thrust = 3.188 kN. Equilibrium check confirms these values.

## Answer

**Data:** symmetrical parabolic three-hinged arch, $L = 20$ m, $f = 4$ m: $y = \dfrac{4\times 4\,x(20-x)}{400} = 0.04\,x\,(20-x)$. A 4 kN point load acts at $x = 4$ m from the left hinge (support) $A$, and the internal forces are required at the section where the load acts. The section is taken just to the left of the load, so the load itself is not part of the free body. At $x = 4$: $y = 2.56$ m, $\tan\theta = 0.04\,(20-8) = 0.48$, $\theta = 25.64^\circ$, $\sin\theta = 0.4327$, $\cos\theta = 0.9015$.

#### Influence lines (unit load at x')
- $R_A = \dfrac{20-x'}{20}$, $H = \dfrac{x'}{2f} = \dfrac{x'}{8}$ for $x'\le 10$ and $\dfrac{20-x'}{8}$ for $x'\ge 10$ (peak $\dfrac{L}{4f} = 1.25$ at the crown).
- $M = M_b - H\,y$, with $M_b = \dfrac{x'(20-4)}{20} = 0.8x'$ ($x'\le 4$) or $\dfrac{4(20-x')}{20}$ ($x'\ge 4$).
- $V = V_b\cos\theta - H\sin\theta$ and $N = V_b\sin\theta + H\cos\theta$, with $V_b = -\dfrac{x'}{20}$ (load left of the section) or $\dfrac{20-x'}{20}$ (load right of the section).

| Unit load at x' (m) | H | M (m) | V (radial) | N |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 4 | 0.5 | 1.92 | 0.5049 | 0.7969 |
| 10 | 1.25 | -1.2 | -0.0902 | 1.3433 |
| 15 | 0.625 | -0.6 | -0.0451 | 0.6716 |
| 20 | 0 | 0 | 0 | 0 |

(The row at 4 m is for the load just to the right of the section.)

#### Values from the ILDs for P = 4 kN at the section
$$
\begin{aligned} M &= 4\times 1.92 = 7.68\ \text{kN m}\\ V &= 4\times 0.5049 = 2.019\ \text{kN}\\ N &= 4\times 0.7969 = 3.188\ \text{kN} \end{aligned}
$$

#### Check by first principles (equilibrium)
$$
R_B = \frac{4\times 4}{20} = 0.8\ \text{kN}, \qquad R_A = 3.2\ \text{kN}
$$

Moment at the crown hinge (right part): $H f = R_B\times 10 = 8 \Rightarrow H = \dfrac{8}{4} = 2\ \text{kN}$.

$$
\begin{aligned} M &= R_A\times 4 - H\times y = 12.8 - 2\times 2.56 = 7.68\ \text{kN m}\\ V &= R_A\cos\theta - H\sin\theta = 3.2(0.9015) - 2(0.4327) = 2.019\ \text{kN}\\ N &= R_A\sin\theta + H\cos\theta = 3.2(0.4327) + 2(0.9015) = 3.188\ \text{kN} \end{aligned}
$$

The ILD values and the equilibrium values agree.

**Answer: $M = 7.68$ kN m, radial shear $V = 2.019$ kN, normal thrust $N = 3.188$ kN at the section.**

## Check yourself

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

1. Why is the section taken just on the left of the 4 kN load when finding the internal forces there?
   A. So that the load is not part of the free body and the values are unambiguous
   B. Because the load always causes zero bending moment at its own point
   C. Because the left support is a hinge and the right is not
   D. Because the thrust is zero to the left of the load
   Answer: A. At a point load the shear and normal force jump; taking the section just left keeps the load out of the free body so a single value results. (Key point: Section just left of the load)

2. For the 4 kN load at 4 m, R_B = 0.8 kN. How is the thrust H found from the crown hinge?
   A. H x 4 = 0.8 x 10 giving H = 2 kN (moment about crown, right part)
   B. H = R_B x 4 / 10 = 0.32 kN
   C. H x 4 = 3.2 x 10 giving H = 8 kN
   D. H = 4 x 4 / 20 = 0.8 kN
   Answer: A. Taking moments about the crown hinge for the right part gives H f = R_B x 10, so H = 8/4 = 2 kN. (Key point: Horizontal thrust by hinge condition)

3. Which expression gives the bending moment at x = 4 m (y = 2.56 m) from first principles?
   A. M = R_B x 4 - H x y
   B. M = R_A x 4 + H x y
   C. M = R_A x 4 - H x y = 12.8 - 5.12 = 7.68 kN m
   D. M = R_A x 4 = 12.8 kN m
   Answer: C. The arch moment is the simple-beam moment minus the thrust moment H y, giving 12.8 - 2 x 2.56 = 7.68 kN m. (Key point: Bending moment at the section)

4. At the section with theta = 25.64 deg, how are radial shear V and normal thrust N obtained from R_A = 3.2 kN and H = 2 kN?
   A. V = R_A sin(theta) + H cos(theta); N = R_A cos(theta) - H sin(theta)
   B. V = R_A cos(theta) - H sin(theta); N = R_A sin(theta) + H cos(theta)
   C. V = R_A - H; N = R_A + H
   D. V = H cos(theta); N = R_A cos(theta)
   Answer: B. Resolve the forces normal and parallel to the arch tangent: V = R_A cos(theta) - H sin(theta) = 2.019 kN and N = R_A sin(theta) + H cos(theta) = 3.188 kN, matching the ILD values. (Key point: Radial shear and normal thrust formulas)

## Related in Theory of Structures I

- [Parabolic arch 100 m: UDL left half and 80 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-100m-80kn-udl-d-20m) — Theory of Structures I, chapter 6, asked 4×
- [Circular arch ILDs at 3 m, then internal forces](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-12m-ild-3m-section) — Theory of Structures I, chapter 6, asked 2×
- [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
- [Circular arch: two rolling loads 20 kN and 30 kN](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-circular-100m-20kn-30kn-25m) — Theory of Structures I, chapter 6, asked 2×
- [Parabolic arch 20 m: two loads and UDL, 5 m section](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Aarch-20m-two-loads-udl-right-half) — Theory of Structures I, chapter 6
- [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×

## Related in other subjects

- [Two-hinged parabolic arch: BM at D](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-point-d-30-50) — Theory of Structures II, chapter 2
- [Two-hinged parabolic arch: horizontal reaction, symmetric loads](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-secant-symmetric-loads-80m) — 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
- [Two-hinged arch: BM at 10 m intervals](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aarch-bm-80kn-d-i-ic) — Theory of Structures II, chapter 2