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Civil
Horizontal reaction of a three-hinged arch
Inward horizontal thrust at the supports of a three-hinged arch with supports at the same level and a crown hinge at height h above the chord.
- Ref
- CIV-38
- Rev
- v1
- Field
- Civil
- Result
- H [N]
In a three-hinged arch the crown hinge cannot carry any moment. The moment at the crown from the vertical loads and the reactions is therefore M0(c) − H·h = 0, where M0(c) is the moment at the same position in an equivalent simply supported beam. Solving gives the thrust H = M0(c)/h, so a lower rise h means a larger thrust.
Inward horizontal thrust at the supports of a three-hinged arch with supports at the same level and a crown hinge at height h above the chord. Mc is the sagging moment at the crown position in the equivalent simply supported beam under the same vertical loads. The hinge makes the moment there zero, so H = Mc/h.
Calculator
- Thrust if the rise were halved, 2Mc/h80,000 N
- Thrust if the rise were doubled, Mc/(2h)20,000 N
- Thrust per 100 kN·m of beam moment, 100000/h33,330 N
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Each support must resist an inward horizontal thrust of 40 kN, from a 3 m rise.
- Halving the rise would double the thrust to 80 kN, which is why shallow arches need strong abutments or ties.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N) | Local step | Change in result |
|---|---|---|---|---|---|---|
| Mc Beam moment at the crown position | 1,000 to 5,000,000 N·m | 333.3 | 1,667,000 | 1,666,000 | +10% | +10 % |
| h Crown height above the chord | 0.5 to 20 m | 6,000 | 240,000 | 234,000 | +10% | −9.09 % |
Sweep of Crown height above the chord
| Crown height above the chord (m) | Horizontal thrust (N) |
|---|---|
| 0.5 | 240,000 |
| 2.938 | 40,850 |
| 5.375 | 22,330 |
| 7.813 | 15,360 |
| 10.25 | 11,710 |
| 12.69 | 9,458 |
| 15.13 | 7,934 |
| 17.56 | 6,833 |
| 20 | 6,000 |
Sweep one variable
Variables and how each one contributes
Horizontal thrust
What the equation gives you. Everything else feeds it.
Beam moment at the crown position
The sagging moment at the crown position of the equivalent simply supported beam, in N·m. Raises the thrust in direct proportion. Raising it by 10 % (to 132,000) moves the result by +10 %, from 40,000 to 44,000 N.
Crown height above the chord
The height of the crown hinge above the support chord, in m. A taller arch has a smaller thrust, in inverse proportion. Raising it by 10 % (to 3.3) moves the result by −9.09 %, from 40,000 to 36,360 N.
Worked example with the values above
- Crown condition: M0(c) − H·h = 0.
- H = Mc/h = 120,000 / 3.
- H = 40,000 N.
Assumptions and limits
- Supports at the same level and vertical loads only, for arches and for cables.
- A flexible cable with negligible bending stiffness and a load uniform along the horizontal span (parabola).
Common mistakes
- Using the cable formulas for a cable under its own weight, which hangs as a catenary.
- Using sag measured from the wrong reference line.
- Forgetting the vertical component of tension at the supports.
Statistics and uncertainty
Cable tension is inversely proportional to sag, so the formulas are very sensitive to the measured sag: a 5 % error in sag gives a 5 % error in tension. Real cables also stretch and have some bending stiffness, which shifts results by a few per cent.
Where it is used
- Sizing cables, anchors, abutments and ties.
- Finding thrust and moments in three-hinged arches.
Maths you need
- Funicular shape: the shape that carries load by axial force only
- Three-hinge condition: the moment at the crown hinge is zero
- Horizontal tension: the constant horizontal component along a loaded cable