Three-Hinged Arch

What is modelled

A parabolic arch of span L and rise f has hinges at both supports and at the crown. The crown hinge carries no moment, which gives the extra equation needed to find the horizontal thrust H without knowing the arch stiffness. Loads are a point load P at distance a from the left support and a uniform load w over the whole span.

Equations

• Reactions: VA = P (L − a)/L + w L/2, VB = P a/L + w L/2.
• Crown moment of the left half is zero: H = M0(L/2) / f, with M₀ the moment of the same loads on a simple beam.
• Arch axis y(x) = 4 f x (L − x)/L²; moment M(x) = M0(x) − H y(x).
• Uniform load alone: H = w L²/(8 f) and M(x) = 0 everywhere, because the parabola is the funicular shape.

Checks run in the page

• Vertical equilibrium: V_A + V_B = P + w L.
• Thrust from a linear solve of the four support-reaction equations against the closed form.
• Uniform load: thrust equals w L²/(8 f) and the moment diagram is zero.
• Moment at the crown hinge is zero for any load position.

Honest simplifications

Rigid supports, a parabolic axis, small deflections and no self-weight unless entered as the uniform load; no axial or shear deformation, buckling or temperature effects. It teaches the statics of the three-hinged form, not a bridge design.

• Choose Uniform load only and see the moment diagram collapse to zero.
• Choose Point load moving and watch the maximum moment change with position.
• Lower the rise and watch the thrust grow in inverse proportion.