Arches — Three-Hinged & Two-Hinged — Structural Analysis Notes

Last reviewed 16 Sept 2026 · 6 min read

How an arch works

An arch is a curved member supported at its ends so that the supports can provide horizontal reactions (thrust) as well as vertical ones. The horizontal thrust $H$ produces a moment that opposes the beam moment caused by the loads:

M_{x} = μ_{x} - H y

where $μ_{x}$ is the bending moment in a simply supported beam of the same span under the same loads, and $y$ is the height of the arch axis at the section. Because $H y$ cancels much of $μ_{x}$ , an arch carries load mainly by compression, with small bending. That is why masonry — strong in compression, weak in tension — could span large openings as arches.

Arch	Hinges	Static indeterminacy
Three-hinged	Two supports + crown	0 (determinate)
Two-hinged	Two supports	1
Fixed (hingeless)	None	3
Tied arch	Tie rod carries the thrust	Depends on hinges

Linear arch and Eddy's theorem

For a given loading, the linear arch (line of thrust, funicular shape) is the arch shape in which the bending moment is zero everywhere — the shape of a hanging cable under the same loads, turned upside down.

For a UDL on the horizontal span, the linear arch is a parabola.
For concentrated loads, it is a polygon.

FormulaEddy's theorem

The bending moment at any section of an arch equals $H$ times the vertical intercept between the actual arch axis and the linear arch (line of thrust):

M_{x} = H \times (intercept)

If the line of thrust lies within the middle third of a masonry arch section, no tension develops.

Three-hinged arches

Unknowns: $V_{A}$ , $H_{A}$ , $V_{B}$ , $H_{B}$ (4). Equations: 3 of equilibrium + $M = 0$ at the crown hinge = 4. So the arch is determinate.

Parabolic arch geometry

With origin at the left support, span $L$ and rise $h$ :

y = \frac{4 h x ( L - x )}{L ^{2}}, tan θ = \frac{d y}{d x} = \frac{4 h ( L - 2 x )}{L ^{2}}

Circular arch geometry

Radius $R$ from span and rise: $R = \frac{L ^{2}}{8 h} + \frac{h}{2}$ . At a horizontal distance $x^{'}$ from the crown, $y = R^{2} - x^{'2} - (R - h)$ .

UDL over the whole span (parabolic three-hinged arch)

$V_{A} = V_{B} = w L /2$ . Moment at crown $= 0$ :

\frac{w L}{2} \cdot \frac{L}{2} - \frac{w L}{2} \cdot \frac{L}{4} - H h = 0 \Rightarrow H = \frac{w L ^{2}}{8 h}

At any section, $μ_{x} = \frac{w x ( L - x )}{2}$ and $H y = \frac{w L ^{2}}{8 h} \cdot \frac{4 h x ( L - x )}{L ^{2}} = \frac{w x ( L - x )}{2}$ .

RememberKey result

A parabolic three-hinged (or two-hinged) arch under a full-span UDL has zero bending moment at every section — it is in pure compression. $H = w L^{2} /8 h$ , i.e. the maximum beam moment divided by the rise.

Normal thrust and radial shear

At a section where the arch axis makes angle $θ$ with the horizontal, resolve $H$ and the vertical shear $V$ (beam shear at that section):

FormulaInternal forces in an arch

N = H cos θ + V sin θ (normal thrust)

Q = V cos θ - H sin θ (radial shear)

M = μ_{x} - H y

Downloaded from civilgyantests.com · Not for resale. Verify every code clause and date against the official source.

Arches — Three-Hinged & Two-Hinged