Parabolic three-hinged arch under uniform load For a three-hinged parabolic arch carrying a uniformly distributed load over the entire span, what internal action does every section primarily resist?

Introduction / Context:
Funicular shapes transmit loads primarily through axial action. A parabolic three-hinged arch is the classic example: when subjected to a uniform load across its span, it closely follows the funicular (cable) polygon for that load.

Given Data / Assumptions:

• Arch geometry is parabolic.
• Loading is uniformly distributed across the entire span.
• Three hinges: at both springings and at the crown.

Concept / Approach:
For a parabolic arch under a full-span uniform load, the bending moment at every section vanishes because the arch axis coincides with the thrust line (funicular line). The line of compression stays within the depth, so internal forces reduce to normal thrust (compression) with negligible bending. Shear may be present at supports in global equilibrium, but the key sectional action is axial compression.

Step-by-Step Solution:
Identify funicular shape for UDL → parabola.Match arch axis to funicular → bending moment M ≈ 0 throughout.Internal force left is axial compression (normal thrust) at each section.

Verification / Alternative check:
Using conventional arch equations shows M(x) = 0 when the arch profile equals the funicular; influence lines also confirm zero bending for uniform loading in the ideal case.

Why Other Options Are Wrong:
Tensile force is incorrect because arches work in compression. Shear force is not the defining action at each section for this ideal case. Bending moment is specifically minimized to zero for the funicular arch.

Common Pitfalls:
Assuming any arch always has bending; geometry–load matching can eliminate bending ideally.

Final Answer:
compressive force