Three-hinged parabolic arch (span l, rise h) under a uniformly distributed load w per unit span over the entire horizontal span: identify the correct statements about internal actions.

Introduction / Context:
For a three-hinged parabolic arch carrying a uniform load over the span, the arch shape can be chosen as the funicular (thrust) line for that loading. When the geometry matches the funicular for the given loading, bending moments vanish everywhere along the arch rib, leaving only axial thrust and shear at sections.

Given Data / Assumptions:

• Three-hinged parabolic arch with hinges at the two springings and at the crown.
• Uniformly distributed load w per unit horizontal span over the full span l.
• Rise at crown = h; small-deflection, elastic line analysis with standard arch statics.

Concept / Approach:

At any section x from the left springing, beam moment due to UDL on the simply supported equivalent is M_beam(x) = w x (l − x) / 2. Arch bending moment is M_arch(x) = M_beam(x) − H y(x). For a parabolic arch, ordinate y(x) = 4 h x (l − x) / l². Choosing H so that M_arch(x) = 0 for all x gives the required horizontal thrust and proves zero bending moment everywhere.

Step-by-Step Solution:

Verification / Alternative check:

Arch handbooks list H = w l² / (8 h) for a three-hinged parabola under UDL, with zero bending moment throughout because the rib follows the funicular polygon for that load.

Why Other Options Are Wrong:

Shear force is generally non-zero (option b false). Option e claims all statements are correct and is therefore wrong.

Common Pitfalls:

Assuming “three-hinged” alone implies zero moment; it is the match of parabolic shape to UDL that causes M = 0 everywhere.

Final Answer:

Both (a) and (c) are correct