Three-Hinged Arches - Theory & Concepts

Learning Objectives

  • Define the characteristics and geometric components of a three-hinged arch.
  • Understand why a three-hinged arch is statically determinate.
  • Differentiate between parabolic, circular, and tied arches.
  • Calculate the support reactions for a three-hinged arch subject to various loads.
  • Determine the internal shear, normal force, and bending moment at any given section of the arch.

This lesson covers the fundamentals of three-hinged arches. It details how the introduction of a third hinge at the crown renders the structure statically determinate, discusses different arch geometries, and explains the procedure for calculating both global reactions and internal forces.

Three-Hinged Arch

A three-hinged arch is a major structural system used to span large distances, such as bridges and auditoriums, when an unobstructed space is required. It is characterized by having three pin (hinge) connections: one at each support and one at the crown.

Crown

The crown is the highest point of an arch. In a three-hinged arch, this is typically where the third internal hinge is located, allowing the structure to articulate and avoiding internal moments at that point.

Tied Arch

A tied arch is a structural variation where a horizontal tension tie (like a steel cable or beam) connects the two bottom pin supports. The tie entirely resists the horizontal thrust, allowing the arch to be supported vertically by simple columns or rollers without needing massive foundation abutments.

Characteristics and Determinacy

The defining feature of a three-hinged arch is its static determinacy.

A typical arch with two pinned supports has four unknown reaction components (two at each pin). The three standard equations of equilibrium (ΣFx=0\Sigma F_x = 0, ΣFy=0\Sigma F_y = 0, ΣM=0\Sigma M = 0) are insufficient to solve for four unknowns, making a two-hinged arch statically indeterminate to the first degree.

By introducing a third hinge (usually at the crown), an additional condition equation is created: the internal bending moment at the hinge is zero (ΣMhinge=0\Sigma M_{\text{hinge}} = 0). This provides the necessary fourth equation, rendering the entire structure statically determinate and solvable using only the principles of statics.

Advantages: Because they are statically determinate, three-hinged arches are not subject to internal stresses caused by thermal expansion/contraction or slight settlement of the supports, which would normally induce large forces in a statically indeterminate structure.

Types of Arches: Parabolic vs. Circular

Arches can be categorized based on their geometry:

  • Parabolic Arches: Designed so their shape naturally follows the funicular polygon for a uniformly distributed horizontal load. Under a uniform gravity load (like a road deck), a parabolic arch experiences purely axial compression with zero bending moment and zero shear force anywhere along its length. It is the most efficient shape for this type of loading.
  • Circular Arches: Often chosen for architectural or aesthetic reasons, or ease of construction. Because its shape does not perfectly match the funicular curve of gravity loads, a circular arch will develop internal shear forces and bending moments, requiring a larger cross-section to maintain stability.

Parabolic Arch Equation

The geometric equation describing the curve of a parabolic arch with its origin at one of the supports.

y=4hL2x(L−x)y = \frac{4h}{L^2} x (L - x)

Variables

SymbolDescriptionUnit
yyThe vertical elevation of the arch curve at a given distance xm
xxThe horizontal distance from the origin (usually the left support)m
hhThe maximum height (rise) of the arch, located at the crownm
LLThe total horizontal span length of the archm

Interactive Simulation

Use the simulation below to explore how the reactions of a three-hinged arch change as you move a concentrated point load across its span.

Three-Hinged Arch Analysis

Real-time Reactions:

AyA_y: 75.0 kN

ByB_y: 25.0 kN

Thrust (HH): 50.0 kN

Internal Forces in Arches

Like beams, arches must be designed to withstand internal shear (VV) and bending moment (MM). However, because of their curvature and the horizontal reactions at the supports, they also develop significant internal normal (axial) forces (NN).

Arches convert vertical loads primarily into compressive forces along the arch rib and significant outward horizontal thrusts at the supports.

Analysis Procedure for Reactions

The first step in analyzing a three-hinged arch is to determine the global support reactions.

  1. Draw a free-body diagram of the entire arch.
  2. Apply global equilibrium equations (ΣFx=0\Sigma F_x = 0, ΣFy=0\Sigma F_y = 0, ΣMsupport=0\Sigma M_{\text{support}} = 0) to establish relationships between the reactions. Often, summing moments about one support will yield the vertical reaction at the other support.
  3. Separate the arch into two segments at the internal hinge (the crown).
  4. Apply the condition equation (ΣMhinge=0\Sigma M_{\text{hinge}} = 0) to either the left or the right segment to solve for the horizontal thrust forces.

Calculation of Internal Forces

To find the internal normal force (NN), shear force (VV), and bending moment (MM) at a specific point in the arch:

  1. Determine the global support reactions as outlined above.
  2. Cut the arch at the point of interest and consider the equilibrium of one of the segments (either the left or right portion).
  3. Sum moments about the cut point to find the internal bending moment (MM).
  4. Determine the angle θ\theta of the tangent to the arch curve at the cut point. For a parabolic arch, this requires differentiating the arch equation to find the slope (tan⁡θ=dydx\tan \theta = \frac{dy}{dx}).
  5. Resolve the horizontal and vertical forces acting on the cut section into components parallel and perpendicular to the tangent of the arch curve.
  6. Sum the parallel components to find the internal normal force (NN). Sum the perpendicular components to find the internal shear force (VV).

Sign Conventions for Internal Forces

When calculating internal forces at a cut section, ensure you are using consistent sign conventions. Typically, internal normal forces are considered positive in tension (though arches are mostly in compression), shear forces follow standard beam conventions, and bending moments that cause compression on the top fibers of the arch rib are considered positive.

Key Takeaways
  • A three-hinged arch is statically determinate due to the presence of an internal hinge (usually at the crown) which provides the condition ΣMhinge=0\Sigma M_{\text{hinge}} = 0.
  • Parabolic arches subjected to uniform horizontal loading experience purely axial compression, making them highly efficient.
  • The analysis involves alternating between global equilibrium equations for the entire arch and local equilibrium equations for the disassembled segments at the hinge.
  • Determining internal forces at a specific point requires finding the tangent angle θ\theta at that location to correctly resolve forces into normal (NN) and shear (VV) components.
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