Three-Hinged Arches

Three-Hinged Arches

$Learning Objectives$

Explain why the crown hinge makes a planar three-hinged arch statically determinate.
Calculate vertical support reactions and horizontal thrust.
Determine local normal force, shear, and bending moment at an arch section.
Treat a section coincident with a point load using an explicit left-face or right-face convention.
Compare a funicular parabolic profile with a valid minor circular-arc profile.
Explain why ideal hinge-compatible support movement does not create unloaded secondary restraint forces.

Three-hinged arch reactions and crown condition

The two supports and crown hinge define a statically determinate arch. The zero crown moment supplies the equation needed to determine horizontal thrust.

$Model assumptions and sign convention$

The simulations use a planar arch with hinges at both supports and at the crown. Supports are at equal elevation except in the imposed-movement scenario. External loads are vertical. Horizontal thrust is reported as a compressive action directed inward at both supports. Section normal force is reported positive in compression; local shear follows the displayed tangent-normal axes.

Section forces are reported from the isolated left segment. The local tangent $+t$ follows increasing $x$ along the arch and the displayed counter-clockwise normal is $+n$ . Normal force $N$ is positive in compression. Shear $V$ is positive along $+n$ on the cut face of the isolated left segment.

At a section exactly coincident with a point load, the force resultants are discontinuous. The simulation therefore requires an explicit $a^-$ limit that excludes the point load or $a^+$ limit that includes it. The bending moment remains continuous because the point load has zero lever arm at the cut.

$Three-Hinged Arch$

A curved rigid-body system with two support hinges and one internal hinge. The internal hinge transmits force but no bending moment, providing the additional equilibrium condition needed to solve the four planar support-reaction components.

Parabolic arch profile

Symmetric parabola with span L and crown rise h.

y(x)=\frac{4h}{L^2}x(L-x)

Arch bending moment

Beam-equivalent moment reduced by the horizontal-thrust contribution.

M_{\mathrm{arch}}(x)=M_{\mathrm{beam}}(x)-Hy(x)

Variables

Symbol	Description	Unit
$H$	Horizontal thrust	kN
$y(x)$	Arch ordinate above the support chord	m
$M_{\mathrm{beam}}$	Moment in the equivalent simply supported beam	kN·m

Moving point-load reactions

Solve the equivalent simply supported beam reactions first. The zero moment at the crown then determines the horizontal thrust from either isolated half of the arch. The implementation verifies both vertical-force equilibrium and the crown moment residual.

Simulation

Controls

20.0 m

5.0 m

100 kN

5.00 m

Results

75.000 kN

25.000 kN

Horizontal thrust

50.000 kN

Crown moment residual

0.000000000 kN·m

support and crown equilibrium verified

A_y=P(L-a)/L,\quad B_y=Pa/L,\quad H=M_{beam}(L/2)/h

$Interpretation question$

Why does horizontal thrust approach zero when the point load is placed directly at a support?

Crown-hinge horizontal thrust

For a symmetric parabolic arch carrying a full-span load uniform per horizontal metre, the selected arch profile is funicular. The horizontal thrust satisfies the zero crown-moment condition, and substituting the parabolic ordinate into $M_{\mathrm{beam}}-Hy$ produces a zero bending-moment residual throughout the span.

Horizontal thrust under full-span horizontal UDL

Symmetric level-support parabolic arch.

H=\frac{wL^2}{8h}

Simulation

Controls

20.0 m

5.0 m

12.0 kN/m

Results

Support vertical

120.000 kN

Horizontal thrust

120.000 kN

Parabolic moment residual

0.000000000 kN·m

funicular UDL equilibrium verified

H=\frac{wL^2}{8h}

$Interpretation question$

Why does increasing rise reduce the horizontal thrust for fixed span and loading?

Section normal force, shear, and bending moment

At a selected cut, calculate the global horizontal and vertical force components and then rotate them into axes tangent and normal to the arch. The tangent angle follows from the derivative of the profile. The simulation reconstructs the original global force components from $N$ and $V$ and reports the transformation residual as an independent check.

When the cut coordinate equals the point-load coordinate, select the $a^-$ or $a^+$ limit deliberately. The two limiting sections have the same bending moment but different force resultants because the concentrated load is excluded from the left limit and included in the right limit.

Tangent slope and local resultants

Resolve the section force into tangent and normal directions.

\tan\theta=\frac{dy}{dx},\qquad N=H\cos\theta+V_g\sin\theta,\qquad V=H\sin\theta-V_g\cos\theta

Simulation

Controls

20.0 m

5.0 m

100 kN

5.00 m

6.00 m

Results

Normal compression N

37.139 kN

Left-face shear V

41.781 kN

Bending moment M

140.000 kN·m

Force transform residual

0.000000000 kN

left-segment section equilibrium verified

N=H\cos\theta+V_g\sin\theta,\quad V=H\sin\theta-V_g\cos\theta

$Interpretation question$

Why can an arch carry substantial compression even where its bending moment is small?

Parabolic and circular shape comparison

A parabolic profile is funicular for a full-span load uniform over the horizontal projection, so the ideal bending moment is zero throughout. A circular profile generally has a different ordinate and therefore develops bending under the same loading and horizontal thrust.

The comparison uses a single-valued minor circular arc passing through the two supports and crown. This representation requires

0<h\le\frac{L}{2}.

A rise greater than $L/2$ would require a major arc and is intentionally rejected rather than drawn with the wrong branch of the circle.

Simulation

Controls

20.0 m

5.0 m

12.0 kN/m

6.00 m

Results

Parabolic moment

0.000000000 kN·m

Circular ordinate

4.343 m

Circular moment

-17.126 kN·m

Circle radius

12.500 m

shared span/rise geometry verified

M(x)=M_{beam}(x)-Hy_{shape}(x)

$Interpretation question$

Why does matching the crown rise alone not make a circular arch funicular for the selected loading?

Temperature and support movement

An unloaded ideal three-hinged arch can change configuration through its hinges when temperature changes or a support settles, so these imposed movements do not create redundant secondary restraint forces. This statement does not mean that loaded reactions remain unchanged; equilibrium must be recalculated for the altered geometry.

The simulation uses an explicit kinematic compatibility model. Support A receives the prescribed downward settlement, support B remains fixed, and both arch halves receive the same free thermal scale $1+\alpha\Delta T$ . The changed crown $C'$ is the upper intersection that satisfies the thermally scaled hinge-to-crown half-chord length from both supports. Each original half-profile is then mapped to its moved support and the compatible crown by a same-shape similarity transformation.

The green changed arch is drawn at true physical scale. Detached dashed visibility arrows may be magnified when millimetre-scale movements would otherwise be unreadable; the exact magnification factor is printed beside each arrow. The familiar quantity $\alpha\Delta T L$ is shown only as a free chord-length reference and is not interpreted as a support displacement or restraint force.

Imposed-movement compatibility

Uniform free thermal strain changes the compatible half-arch geometry without creating a redundant unloaded restraint force.

$\varepsilon_T=\alpha\Delta T,\qquad \ell_h'=(1+\varepsilon_T)\ell_h,\qquad R_{\text{secondary}}=0$

Simulation

Controls

20.0 m

5.0 m

12.0 µ/°C

35 °C

0.030 m

Results

Thermal strain

420.000 µ

Free chord reference

8.400 mm

Moved crown

(9.992, 4.995) m

Secondary restraint

0 kN in unloaded ideal model

hinge-compatible geometry verified

Green geometry is true-scale compatibility geometry. Red/blue dashed visibility arrows are detached annotations; when magnified, the exact multiplier is printed beside the arrow.

\varepsilon_T=\alpha\Delta T,\quad \ell_h'=(1+\varepsilon_T)\ell_h,\quad R_{secondary}=0

$Interpretation question$

Why can a three-hinged arch accommodate imposed movement without the secondary forces that arise in a two-hinged arch?

$Three-hinged arch analysis procedure$

STEP-BY-STEP

Draw the entire-arch free-body diagram and solve the vertical reactions.
Cut the arch at the crown hinge.
Apply zero moment about the crown to determine horizontal thrust.
For a section cut, isolate the left segment, state the $+t/+n$ basis, and choose the $a^-$ or $a^+$ limit when the cut coincides with a point load.
Calculate $M_{\mathrm{beam}}-Hy$ .
Differentiate the profile to obtain the tangent angle.
Resolve global force components into local normal and shear components.
Reconstruct the global force components from $N$ and $V$ .
Verify vertical equilibrium, crown moment, transformation residual, and the stated sign convention.

$Limits of the model$

The simulations are rigid-body statics models. They do not calculate elastic deflection, buckling, material stress, second-order effects, foundation capacity, load combinations, or design-code compliance. The zero secondary-force statement applies only to the unloaded ideal hinge-compatible model. The circular comparison is geometric and does not imply equal stiffness or equal material response.

Three-Hinged Arch Analysis Workflow

Solve a planar three-hinged arch from whole-arch equilibrium and the internal-hinge zero-moment condition, then evaluate section forces and funicular behavior without assuming level supports unless stated.

Confirm arch geometry, hinge locations, support elevations, and loading → Draw the whole-arch free-body diagram; Draw the whole-arch free-body diagram → Apply whole-arch force and moment equilibrium; Apply whole-arch force and moment equilibrium → Isolate one side at the internal hinge and impose Mhinge = 0; Isolate one side at the internal hinge and impose Mhinge = 0 → Solve the remaining support reactions, including horizontal thrust; Solve the remaining support reactions, including horizontal thrust → Whole-arch and hinge equilibrium residuals acceptable?; Whole-arch and hinge equilibrium residuals acceptable? — No → Correct support model, load resultants, geometry, or signs; Whole-arch and hinge equilibrium residuals acceptable? — Yes → Section forces required?; Correct support model, load resultants, geometry, or signs → Draw the whole-arch free-body diagram; Section forces required? — Yes → Choose the cut face and explicitly include or exclude coincident point loads; Section forces required? — No → Profile is funicular for the actual loading and support geometry?; Choose the cut face and explicitly include or exclude coincident point loads → Compute M directly; for level supports and vertical loading, M = Mbeam − Hy; Compute M directly; for level supports and vertical loading, M = Mbeam − Hy → Resolve section forces into local tangent-normal N and V; Resolve section forces into local tangent-normal N and V → Profile is funicular for the actual loading and support geometry?; Profile is funicular for the actual loading and support geometry? — Yes → Verify near-zero bending moment along the funicular profile; Profile is funicular for the actual loading and support geometry? — No → Nonzero bending is generally expected away from hinges; Verify near-zero bending moment along the funicular profile → Verify hinge moments, section equilibrium, and coordinate transformation; Nonzero bending is generally expected away from hinges → Verify hinge moments, section equilibrium, and coordinate transformation; Verify hinge moments, section equilibrium, and coordinate transformation → Arch equilibrium verified

Confirm arch geometry, hinge locations, support elevations, and loading: terminator
Draw the whole-arch free-body diagram: process
Apply whole-arch force and moment equilibrium: process
Isolate one side at the internal hinge and impose Mhinge = 0: process
Solve the remaining support reactions, including horizontal thrust: process
Whole-arch and hinge equilibrium residuals acceptable?: decision
Correct support model, load resultants, geometry, or signs: process
Section forces required?: decision
Choose the cut face and explicitly include or exclude coincident point loads: process
Compute M directly; for level supports and vertical loading, M = Mbeam − Hy: process
Resolve section forces into local tangent-normal N and V: process
Profile is funicular for the actual loading and support geometry?: decision
Verify near-zero bending moment along the funicular profile: process
Nonzero bending is generally expected away from hinges: process
Verify hinge moments, section equilibrium, and coordinate transformation: process
Arch equilibrium verified: terminator

Key Takeaways

The crown hinge supplies a zero-moment condition that makes the ideal planar arch determinate.
Vertical reactions can be obtained from the equivalent simply supported beam.
Horizontal thrust follows from crown equilibrium.
Arch moment equals beam-equivalent moment minus $Hy$ .
Section force resultants use the isolated-left-segment convention, with $N$ positive in compression, $V$ positive along the displayed $+n$ , and explicit $a^-/a^+$ limits at concentrated-load discontinuities.
A funicular profile minimizes bending only for its matching load distribution.
A minor circular-arc comparison requires $h\le L/2$ .
Ideal three-hinged arches can accommodate compatible support movement and uniform thermal strain without redundant unloaded restraint force; the changed geometry must still satisfy hinge compatibility.

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