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Structural Analysis Fundamentals / Arches

The cable–arch analogy

Why an arch is an upside-down cable, and what a funicular shape means.

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This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

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These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

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Why this matters

Masonry has limited tensile capacity. Arches, vaults and domes can carry suitable loading mainly through compression, but the thrust line, support conditions and material resistance still need checking.

By the end of this lesson you should be able to

  • State the cable–arch analogy
  • Define a funicular shape
  • Explain why an arch thrusts outwards

What you should already know

  • Cables and the constant horizontal component (Module 5)
  • Equilibrium and taking moments (Module 2)

A hanging chain finds an equilibrium shape under its applied loading. Invert that geometry and reverse the internal force signs while retaining corresponding downward loads: the same equilibrium can be carried by compression. This is the cable–arch analogy. It establishes an ideal funicular shape, not its resistance to buckling, material failure or support movement.

A shape is funicular for a loading when it carries that loading through axial force without bending. Proportional increases in the same load pattern retain the funicular shape in first-order analysis. A change in the distribution may introduce bending.

For example, a parabola is funicular for a vertical load uniform per horizontal metre. A circular axis generally develops bending under that loading.

Predict first

A parabolic arch spans 40 m with a rise of 8 m and carries 15 kN/m. A cable of the same span and sag carries the same load. How do their horizontal thrusts compare?

A cable hanging in a shallow curve between two level supports, the shape which when inverted becomes the funicular arch.w = 15 kN/mdip 8 mL = 40 mH = 375 kN · V = 300 kN · Tmax = 480 kN
The hanging shape under a uniform load is a parabola. Invert it and you have the funicular arch for that same load.

Practice

A parabolic arch spans 30 m with a rise of 6.0 m and carries a uniform load of 20 kN/m over the full span. What is the horizontal thrust at each abutment, in kN?

Practice

For the same arch, what is the vertical reaction at each springing, in kN?

Practice

The same arch is rebuilt with the rise increased to 12 m. What is the new horizontal thrust, in kN?

Summary

  • An arch is an inverted cable: same equilibrium, compression instead of tension
  • A funicular shape carries one particular loading with no bending
  • The arch thrusts outwards, and the supports must resist that
  • Depart from the funicular shape and bending appears