Arches and Cables

Arches and cables are crucial in structural analysis for understanding load distribution in curved structures.

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Why it matters

Arches and cables are fundamental components in many structures such as bridges, roofs, and domes. Understanding their behavior under various loads is crucial for designing safe and efficient structures.

Key ideas

  • Arches: Curved structures that primarily carry loads through compression. They are efficient in spanning large distances and are commonly used in bridges and viaducts.
    • Types of arches include fixed, two-hinged, and three-hinged arches.
    • The shape of the arch affects its load-carrying capacity and stability.
  • Cables: Flexible elements that carry loads through tension. They are used in suspension bridges and cable-stayed structures.
    • Cables can only carry tensile forces and are often used in combination with other structural elements.
    • A cable under uniform load per unit actual cable length forms a catenary; uniform vertical load per unit horizontal span gives a parabola. Ideal point loads give straight cable segments between load points.

Formulas

  • Arch thrust: H = w·L² / (8·f)
    • H: Horizontal thrust (N)
    • w: Uniform load per unit length (N/m)
    • L: Span of the arch (m)
    • f: Rise of the arch (m)
  • Horizontal cable tension component: H = w·L² / (8·d)
    • H: Constant horizontal component (N), not total tension everywhere
    • w: Uniform load per unit length (N/m)
    • L: Span of the cable (m)
    • d: Sag of the cable (m)

These formulas apply to equal-level supports and a full-span vertical UDL measured per horizontal length; the arch result uses a three-hinged arch with its crown hinge at midspan. For the parabolic cable, support tension is √(H² + (wL/2)²), and minimum tension H occurs at the lowest point.

Worked example

Given: A three-hinged arch with a span of 20 m and a rise of 5 m carries a uniform load of 10 kN/m.

  1. Calculate the horizontal thrust using the formula: H = w·L² / (8·f) H = 10 kN/m · (20 m)² / (8 · 5 m) H = 10 · 400 / 40 H = 100 kN

Answer: The horizontal thrust is 100 kN.

Common mistakes

  • Confusing the rise and span of an arch.
  • Incorrectly assuming that cables can carry compressive forces.
  • Misapplying formulas for different types of arches and cables.

For GATE CE

Questions often involve calculating forces in arches and cables, analyzing stability, and understanding the effects of different load types. Practice problems involving different arch types and cable configurations.

Quick check

  1. What type of force do cables primarily carry?
  2. What is the shape of a cable under uniform load?
  3. How does the rise of an arch affect its horizontal thrust?

Answers: 1. Tension 2. Catenary for uniform load per cable length; parabola for uniform load per horizontal span. 3. A higher rise reduces horizontal thrust.

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