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.

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.

At a section exactly coincident with a point load, the internal shear and normal-force resultants are discontinuous. The section-force simulation therefore requires an explicit left face that excludes the point load or right face 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)=4hL2x(L−x)y(x)=\frac{4h}{L^2}x(L-x)

Arch bending moment

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

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

Variables

SymbolDescriptionUnit
HHHorizontal thrustkN
y(x)y(x)Arch ordinate above the support chordm
MbeamM_{\mathrm{beam}}Moment in the equivalent simply supported beamkN¡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.

Advanced engineering statics simulation

Three-Hinged Arch Engineering Suite

Reactions, crown thrust, section resultants, shape effects, and imposed-movement compatibility.

Solve vertical support reactions and crown-compatible horizontal thrust for a moving point load.

Horizontal span
20.0 m
m
10.0–50.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Arch rise
5.0 m
m
1.5–12.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Point load
100 kN
kN
1–250

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Load position
5.0 m
m
0.0–20.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Three-hinged arch equilibriumPin supports A and B plus the moment-releasing crown hinge C form a determinate system.ACBP 100.0 kNAy 75.0By 25.0HHH = 50.000 kN from M_C = 0
Ay
75.000 kN
By
25.000 kN
Horizontal thrust
50.000 kN
support and crown equilibrium verified

Concept question: Predict H from the equivalent-beam crown moment divided by the rise.

Model scope and verification

Scope: Planar, equal-elevation, pin-connected three-hinged arch unless the imposed-movement scenario is selected. Normal force is positive in compression. A section coincident with the point load uses the explicitly selected left or right face. Circular comparison is restricted to a single-valued minor arc with h ≤ L/2. The UDL acts vertically and is uniform per horizontal metre.

Acceptance check: Solve the equivalent simply supported beam reactions, enforce zero crown moment for H, use M_arch = M_beam − Hy, draw the cut normal to the arch tangent, and reconstruct global section-force components from N and V.

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 Mbeam−HyM_{\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=wL28hH=\frac{wL^2}{8h}

Advanced engineering statics simulation

Three-Hinged Arch Engineering Suite

Reactions, crown thrust, section resultants, shape effects, and imposed-movement compatibility.

Enforce zero crown moment for a symmetric parabolic arch under a vertical UDL uniform over the horizontal projection.

Horizontal span
20.0 m
m
10.0–50.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Arch rise
5.0 m
m
1.5–12.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Vertical UDL per horizontal metre
12.0 kN/m
kN/m
0.5–40.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Three-hinged arch equilibriumPin supports A and B plus the moment-releasing crown hinge C form a determinate system.ACBdownward w = 12 kN/m of horizontal projectionHHH = wL²/(8h) = 120.000 kN
Support vertical
120.000 kN
Horizontal thrust
120.000 kN
Parabolic moment residual
0.000000000 kN¡m
funicular UDL residual verified

Concept question: Predict the horizontal thrust under the full-span vertical UDL.

Model scope and verification

Scope: Planar, equal-elevation, pin-connected three-hinged arch unless the imposed-movement scenario is selected. Normal force is positive in compression. A section coincident with the point load uses the explicitly selected left or right face. Circular comparison is restricted to a single-valued minor arc with h ≤ L/2. The UDL acts vertically and is uniform per horizontal metre.

Acceptance check: Solve the equivalent simply supported beam reactions, enforce zero crown moment for H, use M_arch = M_beam − Hy, draw the cut normal to the arch tangent, and reconstruct global section-force components from N and V.

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 NN and VV and reports the transformation residual as an independent check.

When the cut coordinate equals the point-load coordinate, select the left or right face deliberately. The two faces have the same bending moment but different force resultants because the concentrated load lies between them.

Tangent slope and local resultants

Resolve the section force into tangent and normal directions.

tan⁡θ=dydx,N=Hcos⁡θ+Vgsin⁡θ,V=Vgcos⁡θ−Hsin⁡θ\tan\theta=\frac{dy}{dx},\qquad N=H\cos\theta+V_g\sin\theta,\qquad V=V_g\cos\theta-H\sin\theta

Advanced engineering statics simulation

Three-Hinged Arch Engineering Suite

Reactions, crown thrust, section resultants, shape effects, and imposed-movement compatibility.

Cut the arch normal to its tangent and resolve internal resultants with an explicit discontinuity convention.

Horizontal span
20.0 m
m
10.0–50.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Arch rise
5.0 m
m
1.5–12.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Point load
100 kN
kN
1–250

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Load position
5.0 m
m
0.0–20.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Section coordinate
6.0 m
m
0.0–20.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Three-hinged arch equilibriumPin supports A and B plus the moment-releasing crown hinge C form a determinate system.ACBP 100.0 kNtnx = 6 mLOCAL SECTION RESULTANTSN = 37.14 kN compressionV = -41.78 kNM = 140 kN¡mCoincident load uses the right face.
Normal compression
37.139 kN
θ = 21.8°
Local shear
-41.781 kN
Bending moment
140.000 kN¡m
section equilibrium and transform verified

Concept question: Predict M = Mbeam − Hy at the selected section and selected cut face.

Model scope and verification

Scope: Planar, equal-elevation, pin-connected three-hinged arch unless the imposed-movement scenario is selected. Normal force is positive in compression. A section coincident with the point load uses the explicitly selected left or right face. Circular comparison is restricted to a single-valued minor arc with h ≤ L/2. The UDL acts vertically and is uniform per horizontal metre.

Acceptance check: Solve the equivalent simply supported beam reactions, enforce zero crown moment for H, use M_arch = M_beam − Hy, draw the cut normal to the arch tangent, and reconstruct global section-force components from N and V.

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≤L2.0<h\le\frac{L}{2}.

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

Advanced engineering statics simulation

Three-Hinged Arch Engineering Suite

Reactions, crown thrust, section resultants, shape effects, and imposed-movement compatibility.

Compare the funicular parabolic profile with a valid minor circular arc under the same vertical UDL.

Horizontal span
20.0 m
m
10.0–50.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Arch rise
5.0 m
m
1.5–10.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Vertical UDL per horizontal metre
12.0 kN/m
kN/m
0.5–40.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Section coordinate
6.0 m
m
0.0–20.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Three-hinged arch equilibriumPin supports A and B plus the moment-releasing crown hinge C form a determinate system.ACBdownward w = 12 kN/m of horizontal projectionPARABOLA: FUNICULAR FOR MATCHING UDLCIRCLE: ORDINATE MISMATCH CREATES BENDINGM circular = -17.126 kN·mMinor circular arc requires h ≤ L/2.
Parabolic moment
0.000000000 kN¡m
Funicular for the selected full-span vertical UDL.
Circular ordinate
4.343 m
Circular moment
-17.126 kN¡m
profile comparison geometry verified

Concept question: Predict the circular-profile bending moment at the selected section.

Model scope and verification

Scope: Planar, equal-elevation, pin-connected three-hinged arch unless the imposed-movement scenario is selected. Normal force is positive in compression. A section coincident with the point load uses the explicitly selected left or right face. Circular comparison is restricted to a single-valued minor arc with h ≤ L/2. The UDL acts vertically and is uniform per horizontal metre.

Acceptance check: Solve the equivalent simply supported beam reactions, enforce zero crown moment for H, use M_arch = M_beam − Hy, draw the cut normal to the arch tangent, and reconstruct global section-force components from N and V.

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.

Positive settlement is defined and drawn downward. Thermal movement uses the free span change αΔTL\alpha\Delta T L and may be positive for heating or negative for cooling.

Free thermal span change

Unrestrained linear thermal change used by the compatibility model.

ΔL=αΔTL\Delta L=\alpha\Delta T L

Advanced engineering statics simulation

Three-Hinged Arch Engineering Suite

Reactions, crown thrust, section resultants, shape effects, and imposed-movement compatibility.

Visualize downward settlement and free thermal change in an unloaded ideal three-hinged arch.

Horizontal span
20.0 m
m
10.0–50.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Arch rise
5.0 m
m
1.5–12.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Thermal coefficient
12.0 ¾/°C
¾/°C
5.0–25.0

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Temperature change
35 °C
°C
-50–80

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Downward support settlement
0.030 m
m
0.000–0.100

Drag for exploration or enter an exact value. Press Enter to apply and Escape to restore.

Three-hinged arch equilibriumPin supports A and B plus the moment-releasing crown hinge C form a determinate system.ACBUNLOADED IDEAL HINGE-COMPATIBLE MODELDownward settlement = 30 mmFree thermal change = 8.400 mmSecondary restraint force = 0Loaded reactions require reanalysis in the changed geometry.
Free thermal change
8.400 mm
Downward settlement
30 mm
no unloaded secondary force
hinge-compatible movement model verified

Concept question: Predict αΔTL for the unloaded hinge-compatible model.

Model scope and verification

Scope: Planar, equal-elevation, pin-connected three-hinged arch unless the imposed-movement scenario is selected. Normal force is positive in compression. A section coincident with the point load uses the explicitly selected left or right face. Circular comparison is restricted to a single-valued minor arc with h ≤ L/2. The UDL acts vertically and is uniform per horizontal metre.

Acceptance check: Solve the equivalent simply supported beam reactions, enforce zero crown moment for H, use M_arch = M_beam − Hy, draw the cut normal to the arch tangent, and reconstruct global section-force components from N and V.

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

  1. Draw the entire-arch free-body diagram and solve the vertical reactions.
  2. Cut the arch at the crown hinge.
  3. Apply zero moment about the crown to determine horizontal thrust.
  4. For a section cut, state whether a coincident point load is excluded or included.
  5. Calculate Mbeam−HyM_{\mathrm{beam}}-Hy.
  6. Differentiate the profile to obtain the tangent angle.
  7. Resolve global force components into local normal and shear components.
  8. Reconstruct the global force components from NN and VV.
  9. 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.

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 HyHy.
  • Section force resultants require an explicit cut-face convention at concentrated-load discontinuities.
  • A funicular profile minimizes bending only for its matching load distribution.
  • A minor circular-arc comparison requires h≤L/2h\le L/2.
  • Ideal three-hinged arches can accommodate imposed movement without redundant unloaded restraint force.
PrevNext
Quiz Me