Three-Hinged Arches MCQ with answer and solution

Concept

A parabolic three-hinged arch carrying a uniformly distributed load over the entire span is subjected to axial compression only, resulting in zero bending moment throughout the arch.

Formula

H = wL² / (8h)

Step-by-step solution

1. Consider a symmetric three-hinged parabolic arch of span L and rise h subjected to a UDL w over the entire span. 2. The horizontal thrust H is obtained by taking moments about the crown hinge: H × h = (w × L/2) × (L/4) - (w × L/2) × (L/4) + ... yielding H = wL² / (8h). 3. The bending moment at any section of a parabolic arch carrying UDL is given by M = M_s - H y, where M_s is the simply supported beam moment and H y is the moment due to horizontal thrust. 4. For a parabolic profile, M_s = wLx/2 - wx²/2 = Hy at all points, hence the resultant bending moment M = 0.

Explanation

For a symmetric three-hinged parabolic arch subjected to a uniformly distributed load (UDL) over the entire span, the horizontal thrust is given by wL²/(8h). Since the shape of a three-hinged arch carrying a UDL follows the exact shape of the bending moment diagram of a simply supported beam under the same loading, the bending moment at any section of the arch is zero.

Why the other options are wrong

A: wL²/8 is the maximum bending moment for a simply supported beam, not the arch. B: These values do not correspond to the standard elastic analysis of three-hinged parabolic arches under UDL. D: Incorrect denominator and dimensions.

Shortcut

Remember that a parabolic arch under a full UDL is an antifunicular curve, meaning it carries the load purely in compression with zero bending moment.

Common mistake

Students often assume that arches experience high bending moments everywhere, forgetting that a parabolic arch under UDL has zero bending moment.