Three-Hinged Arches MCQ with answer and solution

Concept

Find vertical reactions from equilibrium. Find horizontal thrust from zero moment at the crown hinge. The resultant is the vector sum of the two.

Formula

H = (V_B × L/2) / h ; R_A = √(V_A² + H²)

Step-by-step solution

1. Total load W = 30 × 8 = 240 kN. It acts at 4 m from A. 2. Take moments about B: V_A × 16 = 240 × 12, so V_A = 180 kN. 3. V_B = 240 − 180 = 60 kN. 4. Moment at crown C = 0 (right part): V_B × 8 = H × 3, so H = 480/3 = 160 kN. 5. R_A = √(180² + 160²) = √58000 = 240.8 kN ≈ 241 kN.

Explanation

The vertical reaction at A comes from simple equilibrium of the whole arch. The horizontal thrust comes from zero bending moment at the crown hinge. The resultant thrust at A combines both: about 241 kN.

Why the other options are wrong

271 kN, 141 kN and 171 kN do not match √(V_A² + H²) with V_A = 180 kN and H = 160 kN.

Shortcut

For UDL w on half span: H = wL²/(16h) = 30 × 256/(16 × 3) = 160 kN, and V_A = 3wL/8 = 180 kN.

Common mistake

Giving only H (160 kN) or only V_A (180 kN) as the answer, or using the full-span formula H = wL²/(8h).