For an ideal three-hinged parabolic tied arch under uniform vertical line load over its horizontal span, elementary equilibrium gives horizontal thrust H=wL^2/(8f) and equal vertical reactions V=wL/2. This calculator illustrates that single symmetric load case only.
Timber Cord and Arch Interactive Calculator
Set ideal span, rise, uniform line load and a stress divisor. Results report symmetric equilibrium quantities and a force/stress area quotient, not timber or tie-member design.
Equation Used
- The arch axis is parabolic with a crown hinge and two springing hinges.
- Springings are level and joined by an ideal tie carrying the horizontal equilibrium force.
- Uniform vertical line load acts over the full horizontal span.
- Equilibrium is first-order and two-dimensional.
- Self-weight distribution, partial loading, load combinations, bending, shear, stability, deflection, creep, connections, material resistance and safety factors are omitted.
Ideal Three-hinged Parabolic Tied Arch
The USDA technical bulletin The Glued Laminated Wooden Arch shows the equilibrium polygon for a three-hinged parabolic arch under uniform vertical load over the full span. It identifies equal vertical reactions wL/2 and horizontal thrust wL squared/(8f).
An ideal tie between level springings can carry the equal and opposite horizontal equilibrium force. This is an analytical load case, not a complete timber structure.
What This Calculator Does
The tool evaluates the symmetric funicular load case and displays its parabolic axis, uniform load, horizontal force and vertical reactions.
It does not verify that a real arch follows the thrust line under other loads, nor does it design the arch rib, tie, hinges, shoes, foundations or connections.
Symmetric Equilibrium
V = wL/2
R = sqrt(H2 + V2)
Aq = 1000H/sigma
H is the ideal horizontal reaction and tie force, V is the reaction at each springing, and R is their vector resultant. Aq is only the numerical quotient of force by the entered stress divisor.
Worked Ideal Example
For L=40 m, f=8 m and w=5 kN/m, H=125 kN, each vertical reaction is 100 kN and the springing resultant is about 160.1 kN. Dividing 125 kN by a 250 MPa stress divisor gives a 500 mm squared quotient.
The quotient is not a specified tie area.
Model Boundary
- Calculated: first-order symmetric equilibrium for one three-hinged parabolic full-span load case.
- Displayed: an ideal tie balancing horizontal thrust at level springings.
- Not calculated: dead/live/snow/wind/seismic load combinations, partial loading, non-parabolic geometry, bending, shear, lateral stability, buckling, deflection, vibration, creep, moisture, fire, fatigue, timber grading, tie resistance, net section, connections, anchorage, foundations or code compliance.
Frequently Asked Questions
Does the formula apply to every arch?
No. It is the stated symmetric three-hinged parabolic case.
Is the arch rib in pure compression under all loads?
No. The funicular result applies only to the matching full-span uniform load case.
Is the horizontal force always the tie force?
Only for the ideal level tie and equilibrium model shown.
Is the area quotient a required tie area?
No.
Does the calculator design a timber arch?
No.
Can it be used for construction?
No. A complete structural design and applicable code checks are required.
References
- U.S. Department of Agriculture, The Glued Laminated Wooden Arch, Technical Bulletin 691, figure 61: three-hinged arch equilibrium under full-span uniform vertical load, including H=wL squared/(8f) and V=wL/2.
- University of Maryland, Arch Structures: parabolic-arch reactions for uniform vertical loading and the limits of the matching funicular case.
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