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Three-hinged arches

Intermediate Structural Analysis Reviewed 24 Oct 2026 2 min read

An arch carries load mainly by compression along its curve instead of by bending. Masonry bridges have used this for two thousand years because stone is strong in compression and weak in tension. A three-hinged arch has pins at both springings and one at the crown. The crown hinge provides an extra condition, zero moment there, so the arch is statically determinate even though its supports give four reactions.

To analyse it, find the vertical reactions from moments of the whole arch about each springing, then take moments about the crown for one half to get the horizontal thrust H. The bending moment at any section is then the beam moment minus H times the arch rise at that section, M = M_beam − H y. If the arch's shape follows the funicular (thrust) line for the load, the moment is zero everywhere and the arch is in pure compression. For a uniform load over the span, that shape is a parabola, which is why parabolic arches suit bridge decks carrying evenly spread loads.

Diagram: Indiacon original, not to scale.

Because it is determinate, a three-hinged arch is not stressed by temperature change or support settlement, an advantage on poor ground. Two-hinged and fixed arches are stiffer and more common in modern concrete and steel bridges but need indeterminate analysis and must cope with thermal and rib-shortening effects. Horizontal thrust must always be resisted: by abutments, by rock, or by a tie as in a bowstring arch.

Key formulas
  • Horizontal thrust (UDL, parabolic arch): H = w L² / (8 h)
  • Moment at any section: M = M_beam − H y
  • Parabola: y = 4 h x (L − x) / L²
  • Normal thrust: N = H cos θ + V sin θ
Assumptions and limits
  • Zero bending holds only when the arch axis matches the funicular line of the load; partial or moving loads still cause bending.
  • Temperature and settlement cause no stresses in a determinate arch, but they do change its geometry (the crown rises or falls).
Solved example

Problem. A three-hinged parabolic arch spans 30 m with 6 m rise and carries 20 kN/m over the whole span. Find the reactions and the moment at quarter span.

  1. Vertical reactions = 20 × 30 / 2 = 300 kN each.
  2. H = w L² / (8 h) = 20 × 900 / 48 = 375 kN.
  3. At x = 7.5 m: y = 4 × 6 × 7.5 × 22.5 / 900 = 4.5 m; beam moment = 300 × 7.5 − 20 × 7.5² / 2 = 2 250 − 562.5 = 1 687.5 kNm.
  4. M = 1 687.5 − 375 × 4.5 = 0. The parabola is funicular for this load.

Answer. V = 300 kN, H = 375 kN, moment at quarter span = 0

Quick quiz

1Why is a three-hinged arch determinate?

The crown hinge adds a zero-moment condition to the three equilibrium equations.

2Funicular shape for a uniform load?

A parabola.

3Is a three-hinged arch stressed by temperature change?

No, it is determinate.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • R. C. Hibbeler, Structural Analysis, Pearson, latest edition.
  • C. S. Reddy, Basic Structural Analysis, McGraw Hill Education (India), latest edition.
  • Devdas Menon, Structural Analysis, Narosa, latest edition.
  • A. K. Chopra, Dynamics of Structures, Pearson, latest edition.

Free and official sources

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