Three-Hinged Arches MCQ with answer and solution

Concept

Horizontal thrust in three-hinged parabolic arch

Formula

H = (V_B × (L/2)) / h

Step-by-step solution

1. Taking moments about right support B: VA × 20 = 80 × 15 -> VA = 60 kN. 2. Vertical equilibrium: VB = 80 - 60 = 20 kN. 3. The crown hinge C is located at midspan (x = 10 m, y = 5 m). 4. The load is to the left of the crown; taking moments of all forces to the right of C about C gives MC = VB × (L/2) - H × h = 0. 5. 20 kN × 10 m - H × 5 m = 0 -> 200 = 5·H -> H = 40 kN. Answer: B

Explanation

In a three-hinged parabolic arch of span L = 20 m and rise h = 5 m with a point load of 80 kN placed 5 m from the left support, the vertical reactions are VA = 60 kN and VB = 20 kN. Equating bending moment at the crown hinge to zero gives H × 5 = 20 × 10, resulting in a horizontal thrust of 40 kN.

Why the other options are wrong

A) 30 kN results from an arithmetic error in vertical reaction calculations. C) 50 kN corresponds to an erroneous rise assumption. D) 60 kN is the vertical reaction VA.

Shortcut

H = (VB × L/2) / h = (20 × 10) / 5 = 40 kN.

Common mistake

Taking moments about the left side without including the 80 kN concentrated load.