Three-hinged parabolic arch — horizontal thrust for partial uniformly distributed load A three-hinged parabolic arch (hinged at both springings and the crown) has a horizontal span L = 4.8 m and a central rise f = 1 m. It carries a uniformly distributed load of 0.75 tonne per metre over the left half of the span only. Determine the horizontal thrust H at the supports.

Difficulty: Medium

Correct Answer: 1.08 tonnes

Explanation:


Introduction / Context:
Three-hinged arches are statically determinate. For a parabolic arch, the bending moment at the crown hinge must be zero. This condition allows direct calculation of the horizontal thrust H by comparing the bending moment at midspan for the equivalent simply supported beam with the arch rise at the crown. This is especially handy for unsymmetrical loading (such as a uniform load over half the span).


Given Data / Assumptions:

  • Span L = 4.8 m; rise at crown f = 1 m.
  • UDL w = 0.75 t/m applied over left half: 0 ≤ x ≤ L/2 = 2.4 m.
  • Arch is three-hinged (hinges at supports and crown), parabolic in profile.


Concept / Approach:
At the crown hinge of a three-hinged arch, internal bending moment is zero. Therefore:
H * y_crown = M_beam at crown (with the same loading on a simply supported beam).
Thus, H = M_beam(crown) / f. Compute the simply supported beam's reaction forces for the given partial UDL, then find M at midspan, and divide by f to obtain H.


Step-by-Step Solution:

Total load W = w * (L/2) = 0.75 * 2.4 = 1.8 t acting at 1.2 m from the left support.Simply supported reactions: R_A = W * (L − a)/L = 1.8 * (4.8 − 1.2)/4.8 = 1.35 t, R_B = W − R_A = 0.45 t.Beam bending moment at midspan x = 2.4 m (within loaded region): M(mid) = R_A * x − w * x^2 / 2 = 1.35 * 2.4 − 0.75 * (2.4)^2 / 2 = 3.24 − 2.16 = 1.08 t·m.Rise at crown f = 1 m, so horizontal thrust H = M(mid)/f = 1.08 / 1 = 1.08 t.


Verification / Alternative check:
Because only half the span is loaded, there is a nonzero vertical shear at midspan in the beam analogy, but the crown-hinge moment condition remains valid. Units check: t·m divided by m gives t, appropriate for thrust.


Why Other Options Are Wrong:

  • 1.8 t and 0.8 t: incorrect magnitudes relative to the computed beam moment and rise.
  • 10.8 t: off by a factor of 10; likely from unit or decimal misplacement.
  • None of these: unnecessary since 1.08 t is correct.


Common Pitfalls:
Forgetting to use the beam moment at the crown location, misplacing the resultant load of the partial UDL, or confusing rise f with total depth. Also, mixing tonne and kiloNewton units can cause errors if conversions are not tracked.


Final Answer:
1.08 tonnes

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