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Queensferry

Module 6 · Lesson 6.2

The three-pinned arch

Using the crown hinge to make an arch determinate, and finding the internal actions.

Try it

The thrust line, and why an arch carries almost no moment

An arch works by pushing sideways. Where the thrust line follows the arch axis there is no bending at all — and you can see exactly what pulls them apart.

40m
8.0m
20kN/m
0kN
25% of span
Arch axis against the thrust line, with the resulting bending momentcrown pinH 500arch axisthrust linebending moment in the arch (kN·m)thrust H 500 kN · reactions 400 / 400 kNthrust line lies ON the axis - the arch carries NO bending, only compressiona beam of the same span would carry 4000 kN·m
Span / rise
5.00
Horizontal thrust H
500 kN
Vertical reaction, left
400 kN
Vertical reaction, right
400 kN
Largest moment in the ARCH
0.0 kN·m
Largest moment in a BEAM
4000 kN·m
Moment the arch avoids
100.0 %
Funicular for this loading
yes
Thrust / total vertical load
0.63 ×

Things worth trying

  • Start with the uniform load only and no point load. The thrust line sits exactly on the arch axis and the moment diagram is flat at zero — the arch carries pure compression.
  • That is not a coincidence. A parabolic arch is the FUNICULAR shape for a uniform load: H·y(x) exactly cancels the beam moment at every section, because both are parabolas of the same proportions.
  • Read the two moment rows. The beam of the same span carries thousands of kN·m; the arch carries zero. That is what an arch is FOR, and it is bought entirely with horizontal thrust.
  • Now drag the rise down. The thrust climbs steeply — H = wL²/8r, the same inverse law as a cable's sag — while the moment stays at zero. A flat arch is still funicular; it just pushes very much harder.
  • Take the rise to 1 m on a 40 m span. The thrust exceeds 2000 kN. Every bit of that goes into the abutments, and it is why a shallow arch needs rock or a tie.
  • Now add a point load. The thrust line lifts away from the axis and the moment diagram comes alive. A single concentrated load is enough to destroy the funicular condition.
  • Move the point load along the span and watch where the moment peaks. The worst position is the QUARTER POINT — 1125 kN·m at 25% or 75% of the span, against 750 at the crown and only 600 at 40%. That is why arch bridges are checked for load on half the span rather than for load at mid-span.
  • Note what the departure means physically: the gap between the two lines, multiplied by the thrust, IS the bending moment. Moment = H × eccentricity, and that is the whole of arch analysis.
  • Set the uniform load to zero and leave the point load. Now nothing is funicular and the arch behaves much more like a bent beam — the shape was only ever right for one loading.

Why this matters

An arch with pinned supports has four unknown reactions but only three equations. Adding a third pin at the crown makes it solvable by hand — and gives a structure that tolerates foundation movement and temperature change without distress.

By the end of this lesson you should be able to

  • Explain why a three-pinned arch is determinate
  • Find the reactions using the crown hinge condition
  • Calculate M, N and V at an arch section using the beam analogy

A pin at each springing gives two reaction components each — four unknowns — against the three equations of plane equilibrium. The arch is indeterminate to the first degree.

Put a third pin at the crown and you gain one extra piece of information: the bending moment at the crown is zero. Taking moments about the crown hinge for one half of the arch supplies the fourth equation, and the structure becomes statically determinate.

From first principles

Internal actions in a three-pinned arch

We want to show: the thrust H from the crown condition, and the bending moment as M = Mbeam − H·y.

Here is the neatest way to think about an arch. Imagine a simply supported beam of the same span carrying the same vertical loads. It would have some bending moment at every point — call it Mbeam. Now the arch has the same vertical reactions as that beam, because vertical equilibrium and moments about a support do not care about the shape. What the arch has in addition is the horizontal thrust H. At a point where the arch rises a height y above the springings, that thrust acts at a lever arm y and provides a moment H·y in the opposite sense. So the arch moment is just the beam moment reduced by H·y. If the shape is such that H·y exactly cancels Mbeam everywhere, the arch carries no bending at all — and that is precisely what funicular means.

A three-pinned parabolic arch of 40 m span and 8 m rise under a uniform load of 15 kN per metre, with the horizontal thrust marked and the moment a beam of the same span would carry shown beneath for comparisonw = 15 kN/mmoment the same span would carry as a beamcrown hingeH = 375 kNrise 8 mparabolic arch under uniform load — no bending anywhere
The arch of the worked example. The shaded region beneath the springing line is the bending moment the same span would carry as a beam. The arch removes it entirely — a parabolic arch under a uniform load carries no bending at all, only thrust. That claim is not an assertion here: the figure computes M = Mbeam − H·y at every station and finds nothing to draw.

Worked example

Three-pinned parabolic arch under a uniform load

Given

  • Span 40.0 m between level springings, rise 8.0 m
  • Parabolic arch axis, pinned at both springings and at the crown
  • Uniformly distributed load of 15 kN/m over the whole span

Find

The reactions, and the bending moment and axial force at the springing.

Assumptions

  • Loads vertical
  • Small deflections
  • Parabolic axis
    The same three-pinned parabolic arch carrying a single 120 kN point load at the quarter point, showing the residual bending moment along the arch axis120 kNcrown hingeH = 75 kNrise 8 mthe wrong shape for the load — bending returns
    The same arch, now with a single point load. The shading along the axis is the bending moment that remains — the arch shape no longer matches the load, so the thrust line departs from the axis and real bending appears. It is still zero at the crown hinge, which is what makes the arch solvable.

    Worked example

    The same arch under a point load

    Given

    • The same 40.0 m span, 8.0 m rise, three-pinned parabolic arch
    • A single point load of 120 kN at 10.0 m from the left springing

    Find

    The reactions and the bending moment under the load.

    Assumptions

    • Same as before; the arch shape is unchanged

      Practice

      A three-pinned parabolic arch spans 30.0 m with a rise of 5.0 m and carries a uniform load of 20 kN/m. What is the horizontal thrust at each springing?

      Check yourself

      Why is a three-pinned arch statically determinate?

      Practice

      A three-pinned parabolic arch spans 40 m with a rise of 8.0 m and carries a uniform load of 15 kN/m over the whole span. What is the bending moment at the quarter point, in kNm?

      Worked example

      Plotting the bending moment along an arch

      Given

      • Three-pinned parabolic arch, span 40.0 m, rise 8.00 m
      • Single point load of 120 kN at the quarter point, x = 10.0 m

      Find

      The bending moment at several stations along the arch, and where it peaks.

        Practice

        A three-pinned parabolic arch spans 40.0 m with a rise of 8.00 m and carries 120 kN at x = 10.0 m. What is the horizontal thrust, in kN?

        Practice

        For that arch, what is the bending moment directly under the 120 kN load at x = 10.0 m, in kNm?

        Summary

        • Two pins give four unknowns; the crown hinge supplies the fourth equation
        • H = Mbeam,crown / h
        • March = Mbeam − H·y — the beam analogy
        • A parabola is funicular for a uniform load: zero moment and zero shear
        • Change the loading and bending appears; the thrust must always be carried to the ground
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        This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint