---
title: "Circular arch: BM, thrust, shear at D, BM diagram"
subject: "Theory of Structures I"
chapter: 6
key: "semester_4/theory_of_structures_i:arch-circular-d-20knm-10kn-8m"
source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/arch-circular-d-20knm-10kn-8m
site: BCT Engineering
---

# Circular arch: BM, thrust, shear at D, BM diagram

Theory of Structures I · Chapter 6: Statically Determinate Arches

- Asked in: 2070 Magh, 12 marks
- Video: [Three Hinged Circular Arch - Problem No 1](https://www.youtube.com/watch?v=57BRqIgDUHg) — 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-circular-d-20knm-10kn-8m
- Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-6

## Question

Determine B.M, normal thrust, radial shear at point D of circular arch as shown in figure below. Also draw bending moment diagram. [Figure: three hinged circular arch, A and B hinge supports; crown hinge C with a rise of 5 m; AD = 4 m, D to C horizontally 4 m, C to B 8 m as dimensioned; 20 kN/m UDL on the left part; 10 kN horizontal load at 3 m from the crown. Details partly unclear.]

## Short answer

Bending moment at D is 67.71 kN m, normal thrust is 69.89 kN, and radial shear is 6.48 kN.

## Answer

**Assumptions (details partly unclear in the figure):** three-hinged circular arch, $A$ and $B$ hinged, crown hinge $C$ at $x = 8$ m with rise $f = 5$ m; $AD = 4$ m, $DC = 4$ m, $CB = 8$ m, so the span is $L = 16$ m and $D$ is at $x = 4$ m. UDL 20 kN/m on the left part $AC$ ($x = 0$ to 8 m). The 10 kN horizontal load acts to the right at the arch point 3 m (horizontally) to the right of the crown ($x = 11$ m).

Radius: $R = \dfrac{L^2/4+f^2}{2f} = \dfrac{64+25}{10} = 8.9$ m. Equation: $y = \sqrt{8.9^2-(x-8)^2} - 3.9$. At $D$: $y_D = 4.05$ m, $\tan\theta = \dfrac{8-4}{\sqrt{8.9^2-16}} = 0.5031$, $\theta = 26.71^\circ$. At the 10 kN load, $y(11) = 4.479$ m.

#### Reactions
Let $H_A$ act to the right at $A$ and $H_B$ to the left at $B$.

**Horizontal equilibrium:** $H_A - H_B + 10 = 0 \Rightarrow H_B = H_A + 10$.

**Moments about $A$:**
$$
V_B\times 16 = (20\times 8)\times 4 + 10\times 4.479 \Rightarrow V_B = \frac{684.79}{16} = 42.8\ \text{kN}, \qquad V_A = 160 - 42.8 = 117.2\ \text{kN}
$$

**Hinge condition** (left part, moments about $C$; the 10 kN load is on the right part):
$$
V_A\times 8 - H_A\times 5 - 160\times 4 = 0 \Rightarrow H_A = \frac{117.2\times 8 - 640}{5} = 59.52\ \text{kN}, \qquad H_B = 69.52\ \text{kN}
$$

#### Forces at D
$$
\begin{aligned} M &= V_A\times 4 - 20\times 4\times 2 - H_A\,y_D = 468.8 - 160 - 59.52\times 4.05 = 67.71\ \text{kN m}\\ V_b &= V_A - 20\times 4 = 37.2\ \text{kN}\\ V &= V_b\cos\theta - H_A\sin\theta = 6.48\ \text{kN}\\ N &= V_b\sin\theta + H_A\cos\theta = 69.89\ \text{kN} \end{aligned}
$$

#### Bending moment diagram
$M = V_A\,x - H_A\,y - \text{(moments of the loads)}$ at various sections:

| x (m) | y (m) | M (kN m) |
|---|---|---|
| 0 | 0 | 0 |
| 2 | 2.673 | 35.28 |
| 4 | 4.05 | 67.71 |
| 6 | 4.772 | 59.15 |
| 8 | 5 | 0 |
| 10 | 4.772 | -72.05 |
| 11 | 4.479 | -97.4 |
| 12 | 4.05 | -110.39 |
| 14 | 2.673 | -100.26 |
| 16 | 0 | 0 |

The bending moment is continuous at the 10 kN load (the load has no moment arm about its own point); only the slope of the curve changes there. $M = 0$ at $A$, $C$ and $B$ (hinges). The largest sagging moment is about $+70.1$ kN m at $x \approx 4.6$ m (under the UDL), and the largest hogging moment is about $-113.1$ kN m at $x \approx 12.7$ m (in the right half).

```text
 BMD (positive = sagging, scale approx.)
 x (m)   0    4     8   11    12.7    16
 M       0   67.7  0  -97.4  -113.1   0
```

**Answer: at $D$: bending moment $= 67.71$ kN m, normal thrust $= 69.89$ kN, radial shear $= 6.48$ kN.**

## Check yourself

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

1. For span 16 m and rise 5 m, what is the radius of the circular arch?
   A. R = (L^2/4 + f^2)/(2f) = 89/10 = 8.9 m
   B. R = L^2/(8f) = 6.4 m
   C. R = L/2 + f = 13 m
   D. R = sqrt(8^2 + 5^2) = 9.43 m
   Answer: A. From the chord geometry, R = (half-span^2 + rise^2)/(2 x rise) = (64 + 25)/10 = 8.9 m. (Key point: Radius of the circular arch)

2. The 10 kN horizontal load acts to the right. With H_A acting right at A and H_B left at B, what is the relation?
   A. H_A = H_B
   B. H_B = H_A - 10
   C. H_A + H_B = 10
   D. H_B = H_A + 10
   Answer: D. Sum of horizontal forces: H_A - H_B + 10 = 0, so the two thrusts are no longer equal (H_B = H_A + 10). (Key point: Horizontal equilibrium with a horizontal load)

3. Why is only the left part used (moments about C) to get H_A = 59.52 kN?
   A. Because the left support has no vertical reaction
   B. Because the 10 kN load lies on the right part, so the left part contains only V_A, H_A and the UDL
   C. Because the moment at the crown is maximum
   D. Because circular arches need no hinge condition
   Answer: B. M_C = 0 for the left free body: V_A x 8 - H_A x 5 - 160 x 4 = 0, which involves no unknown from the right side. (Key point: Hinge condition for H_A)

4. How is the bending moment at D (x = 4 m, y = 4.05 m) found, and where is it zero in the BMD?
   A. M = H_A y only; zero at D
   B. M = V_A x - UDL moment - H_A y = 67.71 kN m; zero at the hinges A, C and B
   C. M = V_A x + H_A y; zero at the crown only
   D. M = V_B x 12 - H_B y; zero at the load point
   Answer: B. M = 468.8 - 160 - 59.52 x 4.05 = 67.71 kN m, and the moment is zero at the three hinges. (Key point: Bending moment at D and in the BMD)

## Related in Theory of Structures I

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