Three-hinged arch under an eccentric point load: A three-hinged arch of span 10 m has the crown 1 m above the springings. An isolated load of 1000 kg acts 2.5 m horizontally from the crown. Compute the horizontal thrust.

Introduction / Context:
Three-hinged arches are statically determinate. The horizontal thrust H can be obtained from moment equilibrium at the crown hinge by using the bending moment there as if for a simply supported beam and dividing by the rise.

Given Data / Assumptions:

• Span = 10 m; rise at crown h = 1 m.
• Point load P = 1000 kg located 2.5 m from crown (toward the right).
• Supports at the same level.
• Small deflection, linear elasticity, determinate analysis.

Concept / Approach:
For a three-hinged arch, the bending moment at the crown hinge is zero. Taking the structure as a simply supported beam to compute the “beam moment” M_beam at the crown section due to external loads, the horizontal thrust satisfies H * h = M_beam, hence H = M_beam / h.

Step-by-Step Solution:
Location of load from left support: 5 m (to crown) + 2.5 m = 7.5 m.Simply-supported reactions: R_A = P * (L − a) / L = 1000 * (10 − 7.5) / 10 = 250 kg.Bending moment at crown section (x = 5 m from left, load to the right): M_beam = R_A * 5 = 250 * 5 = 1250 kg·m.Horizontal thrust: H = M_beam / h = 1250 / 1 = 1250 kg.

Verification / Alternative check:
Symmetry about midspan is broken by the eccentric load; using the right reaction would give the same crown moment via the right segment.

Why Other Options Are Wrong:
125 kg and 750 kg underpredict; 2325 kg and 2500 kg overpredict; they do not satisfy M = H h with the correct beam crown moment.

Common Pitfalls:
Including the load contribution in the left free-body for x = 5 m (it lies to the right and should be excluded in that cut), or using the full span moment instead of the crown section moment.

Final Answer:
1250 kg.