---
title: "Arch: reactions, BM, thrust and shear at D by ILD"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-50m-udl-50kn-ild-d-15m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-50m-udl-50kn-ild-d-15m
site: BCT Engineering
---

# Arch: reactions, BM, thrust and shear at D by ILD

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2071 Magh, 15 marks
- Video: [ILD for Arches | Structural Analysis | Lecture 25 | English](https://www.youtube.com/watch?v=y1rTKKHFWQU) — Letz Prepare
- 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-50m-udl-50kn-ild-d-15m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

In the three hinged parabolic arch loaded as shown below, determine reactions at supports and also find bending moment, normal thrust and radial shear force at section D 15 m far from A. Draw influence line diagram for bending moment and normal thrust at that point and again determine the bending moment and normal thrust at D by using the ILD. [Figure: parabolic three hinged arch, span 50 m, rise 10 m at the crown C; C is 30 m from A (and 20 m from B as dimensioned); D at 15 m from A; 20 kN/m UDL over the left portion A to C; 50 kN vertical load 20 m from B.]

## Short answer

Reactions: V_A = 440 kN, V_B = 210 kN, H = 437.5 kN. At section D (15 m from A) with UDL and point load: M = 675 kN m, N = 459.35 kN, V = 0 kN. ILD method yields same results.

## Answer

**Data and assumptions:** parabolic three-hinged arch, span $L = 50$ m, rise $f = 10$ m (the parabola is symmetrical about mid-span, $y = \dfrac{4\times 10\,x(50-x)}{50^2} = 0.016\,x(50-x)$). As dimensioned, the crown hinge $C$ is 30 m from $A$ and 20 m from $B$, where $y_C = 9.6$ m. UDL 20 kN/m on $AC$ ($x = 0$ to 30 m); 50 kN vertical at 20 m from $B$, i.e. at $C$ ($x = 30$ m). Section $D$ at $x = 15$ m: $y_D = 8.4$ m, $\tan\theta = 0.016\,(50-30) = 0.32$, $\theta = 17.74^\circ$.

#### Reactions
**Vertical reactions** (moments about $A$): the loads are 600 kN (UDL, at 15 m) and 50 kN (at 30 m).

$$
V_B\times 50 = 600\times 15 + 50\times 30 = 10500 \Rightarrow V_B = 210\ \text{kN}, \qquad V_A = 650 - 210 = 440\ \text{kN}
$$

**Horizontal thrust** (hinge at $C$, left part, load at $C$ itself has no moment about $C$):

$$
H\times 9.6 = V_A\times 30 - 600\times 15 = 4200 \Rightarrow H = 437.5\ \text{kN}
$$

(Check from the right part: $H\times 9.6 = V_B\times 20 = 4200$.)

#### Forces at D
$$
\begin{aligned} M &= V_A\times 15 - 20\times 15\times 7.5 - H\,y_D = 6600 - 2250 - 437.5\times 8.4 = 675\ \text{kN m}\\ V_b &= V_A - 20\times 15 = 140\ \text{kN}\\ V &= V_b\cos\theta - H\sin\theta = 140(0.9523) - 437.5(0.3047) = 0\ \text{kN}\\ N &= V_b\sin\theta + H\cos\theta = 140(0.3047) + 437.5(0.9523) = 459.35\ \text{kN} \end{aligned}
$$

#### Influence lines for M and N at D (unit load at x')
- Thrust (hinge at $x_C = 30$, $y_C = 9.6$): for $x'\le 30$ the right part $CB$ is a two-force member, so $H = \dfrac{R_B(L-x_C)}{y_C} = \dfrac{x'}{50}\cdot\dfrac{20}{9.6} = \dfrac{x'}{24}$; for $x'\ge 30$, $H = \dfrac{R_A\,x_C}{y_C} = \dfrac{50-x'}{16}$ (peak $1.25$ at $C$).
- $M = M_b - H\,y_D$ and $N = V_b\sin\theta + H\cos\theta$, where $V_b$ is the simple-beam shear.

| Unit load at x' (m) | H | M at D (m) | N at D |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 15 | 0.625 | 5.25 | 0.5038 |
| 15 | 0.625 | 5.25 | 0.8086 |
| 30 | 1.25 | -4.5 | 1.3124 |
| 40 | 0.625 | -2.25 | 0.6562 |
| 50 | 0 | 0 | 0 |

The N-ILD is discontinuous at $D$ by $\sin\theta$ (the two rows at 15 m are the load just left and just right of $D$).

#### M and N at D from the ILDs
UDL on $AC$ (area of ILD from 0 to 30 m) plus the 50 kN load at $C$ (ordinate at 30 m):

$$
\begin{aligned} M &= 20\times(45) + 50\times(-4.5) = 675\ \text{kN m}\\ N &= 20\times(19.687) + 50\times(1.3124) = 459.35\ \text{kN} \end{aligned}
$$

These agree with the values from direct equilibrium.

**Answer: reactions $V_A = 440$ kN, $V_B = 210$ kN, $H = 437.5$ kN (at both supports). At $D$: $M = 675$ kN m, $N = 459.35$ kN, $V = 0$ kN.**

## Check yourself

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

1. What gives H = 437.5 kN?
   A. Moment about A of all loads divided by the rise 10 m
   B. Total load divided by the span
   C. Sum of moments of the left part about C is zero: H x 9.6 = V_A x 30 - 600 x 15 = 4200
   D. Moment about D equals zero
   Answer: C. At the hinge C the bending moment is zero; taking the left part (the 50 kN load at C has no moment about C) gives H = 4200/9.6 = 437.5 kN. (Key point: Thrust from the crown hinge)

2. At D, V_b = 140 kN, theta = 17.74 deg, H = 437.5 kN. What is the radial shear V?
   A. Zero, because V_b cos(theta) equals H sin(theta)
   B. 140 kN, equal to V_b
   C. 437.5 kN, equal to H
   D. 459.35 kN
   Answer: A. V = V_b cos(theta) - H sin(theta) = 140(0.9523) - 437.5(0.3047) which is about 0, while N = 459.35 kN carries the force. (Key point: Radial shear at D)

3. Why does the influence line for normal thrust at D jump at D?
   A. Because the thrust H is discontinuous at D
   B. Because the arch has a hinge at D
   C. Because the bending moment changes sign at D
   D. Because the simple-beam shear V_b jumps by 1 at D, so N = V_b sin(theta) + H cos(theta) jumps by sin(theta)
   Answer: D. H is continuous but V_b jumps by unity when the unit load crosses the section, so N jumps by sin(theta). (Key point: N-ILD discontinuity)

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