Two-hinged parabolic arch of span l and rise h carries a triangular load that varies from zero at the left support to ω (per unit length) at the right support. What is the horizontal thrust H at the supports?

Introduction / Context:
Horizontal thrust in a two-hinged parabolic arch depends on the load shape and the arch geometry. Unlike a three-hinged arch, indeterminacy requires using energy or funicular principles to obtain H for non-uniform loading. Here, the loading is triangular from zero to ω across span l with rise h.

Given Data / Assumptions:

• Parabolic two-hinged arch, span = l, rise = h.
• Distributed load w(x) varies linearly from 0 at left to ω at right: w(x) = (ω/l) * x.
• Small-slope approximation for the strain energy integral; ds ≈ dx.

Concept / Approach:
For a two-hinged arch, the horizontal thrust H can be obtained by Castigliano’s theorem or the consistent deformation method:
H = (∫ M_0 * y dx) / (∫ y^2 dx)where M_0 is the bending moment of the corresponding simply supported beam under the same load, and y is the arch ordinate measured from the chord line. For a parabolic arch: y = 4h * x * (l - x) / l^2.

Step-by-Step Solution:
1) Reactions for the simply supported beam with triangular load: total load W = (1/2) * ω * l, R_left = W/3 = ω l / 6, R_right = 2W/3 = ω l / 3.2) Beam bending moment function: M_0(x) = R_left * x − (ω/(6l)) * x^3.3) Arch ordinate: y(x) = 4h * x * (l − x) / l^2.4) Compute numerator I1 = ∫_0^l M_0(x) * y(x) dx.5) Compute denominator I2 = ∫_0^l y(x)^2 dx.6) Evaluate H = I1 / I2, which simplifies to H = (ω * l^2) / (16 * h).

Verification / Alternative check:
The expression has correct dimensions of force since ω has units of force/length and l^2 / h is length, yielding force. It also trends correctly: larger h reduces H, larger l or ω increases H.

Why Other Options Are Wrong:
Values with 8h or 12h correspond to other load cases (e.g., uniform load) and are not valid for a zero-to-ω triangular distribution.Expressions with l or l^3 in the numerator have incorrect dimensional consistency for this case.

Common Pitfalls:

• Using the UDL arch thrust formula H = ω l^2 / (8 h) instead of integrating for the triangular loading.
• Forgetting the correct centroid and reactions for the triangular load.
• Using the wrong arch ordinate function.

Final Answer:
H = (ω * l^2) / (16 * h)