Three Hinged Arch - 1

Instructional Objectives:

After reading this chapter the student will be able to

• Define an arch.
• Identify three-hinged, two-hinged and hingeless arches.
• State advantages of arch construction.
• Analyse three-hinged arch.
• Evaluate horizontal thrust in three-hinged arch.

Introduction

For long spans a beam becomes uneconomical because the maximum bending moment increases with the square of the span. In such cases an arch is often more efficient since it develops horizontal reactions that reduce bending moments in the member and transfer much of the load as compression.

As an illustration, for a simply supported beam under a concentrated load the bending moment at the loaded section may be 3PL/16. For a two-hinged symmetrical arch of the same span and loading the vertical reactions can be obtained from the equilibrium equations and the horizontal reaction by an appropriate compatibility (for example, the method of least work). The bending moment below the load in an arch is less than that in the corresponding beam because of the presence of horizontal thrust.

The funicular shape (loaded cable form) is the shape a cable takes under a particular loading. If an arch is built with the inverted funicular shape corresponding to a given loading, the arch will carry that loading in pure compression (i.e. with zero bending moment). In practice the arch shape and the actual loading usually do not match perfectly, so arches generally carry both axial compression and bending. Because arches are mainly in compression they must be designed to resist buckling as well as the compressive stresses.

Historically, masonry arches and vaults (stone and brick) were common for spanning openings. In modern construction arches are built in reinforced concrete and steel and are used frequently in bridges, large doorways and long-span roofs.

It is important to understand that a structural element is classified as an arch on the basis of how it carries lateral load (primarily by compression and with the development of horizontal thrust), not purely by its geometric shape. A structure that does not develop horizontal reaction behaves more like a beam; bending moments are not reduced in that case.

Types of arches

Common types of arches used in practice are:

• Three-hinged arch - a hinge at each support and a hinge at the crown making the structure statically determinate.
• Two-hinged arch - hinges at the supports only; the structure is statically indeterminate to the first degree.
• Fixed-fixed (hingeless) arch - no hinges; the structure is statically indeterminate to a higher degree.

For indeterminate arches (two-hinged and fixed-fixed) the redundant reactions are found by the method of least work, flexibility methods or matrix methods. This lesson focuses on the three-hinged arch, which is statically determinate and simpler to analyse using equilibrium alone.

Analysis of three-hinged arch

A three-hinged arch has three hinges: one at each support and one at the crown. Because of the crown hinge the structure is statically determinate and all support reactions and internal forces can be found from equilibrium.

Consider a three-hinged arch subjected to a concentrated load P as shown in the figure.

There are four reaction components: horizontal and vertical reactions at each support (but horizontal reactions at the two supports are equal and opposite). Three independent equilibrium equations for the whole structure are not sufficient to find four unknowns. An additional equation is obtained by taking moments about the crown hinge (or about any hinge) for the free part of the structure; the moment about a hinge must be zero because a hinge cannot resist moment. Writing moments about the crown hinge for the left and right segments supplies the extra equation needed to determine the reactions.

Taking moments about hinge A of all the forces acting on the left portion of the arch gives

(32.1)

(32.2)

Taking moments of all forces to the right of hinge C about hinge C yields

(32.3)

Applying ∑Fx = 0 for the whole structure gives

The bending moment at a section below a concentrated load can be written as

(32.4)

(32.5)

For comparison, the bending moment under the same concentrated load for a simply supported beam of the same span is

(32.6)

For the particular case illustrated the arch reduces the bending moment by 66.66% relative to the simply supported beam.

Example 32.1

A three-hinged parabolic arch of uniform cross section has a span of 60 m and a rise of 10 m. It is subjected to uniformly distributed load of intensity 10 kN/m as shown in Fig. 32.6 Show that the bending moment is zero at any cross section of the arch.

Reactions

Compute the total uniformly distributed load on the span.

The total load = intensity × span = 10 kN/m × 60 m = 600 kN.

By symmetry the vertical reactions at the two supports are equal.

Vertical reaction at each support = 600 kN / 2 = 300 kN.

To determine the horizontal thrust, take moments about the left hinge A for the right-hand portion or use the condition that the moment about the crown hinge is zero for each half. Writing the moment equilibrium (the detailed algebra is represented by the image) gives

(1)

Taking moments of forces left of hinge C about C gives

(2)

Evaluating the algebraic expression yields the horizontal thrust at a support

H = 450 kN.

From ∑Fx = 0 one could write H_b = 450 kN.

The shear force at the mid-span is zero by symmetry.

Bending moment

Consider a section at distance x from the left support. The bending moment at that section due to the internal forces can be written as (see image for the algebraic form used)

(3)

The equation of the three-hinged parabolic arch (vertical coordinate of the arch as a function of x, measured from left support) is

(4)

Substituting the expression for the ordinate of the arch into the bending moment expression and simplifying gives

= 300x - 300x + 5x² - 5x² = 0

Thus the bending moment at any section M(x) = 0.

Therefore a three-hinged parabolic arch subjected to a uniformly distributed load over the span is in pure compression at every cross section when the arch follows the appropriate parabolic funicular shape for that loading. The arch supports the load without bending moments.

Can you explain why the moment is zero at all points in a three-hinged parabolic arch?