Cables and Arches - Theory & Concepts

Cables and Arches - Theory & Concepts

Cables and arches are specialized structural elements that carry loads primarily through axial forces. They are highly efficient over long spans and are extensively used in modern civil engineering structures like suspension bridges and stadiums.

Analysis of Cables

A cable is a perfectly flexible structural member that can only resist axial tension. It offers zero resistance to bending or compression.

Assumptions for Cable Analysis

Perfect Flexibility: The cable has no bending stiffness. As a result, the internal tension force is always perfectly tangent to the curve of the cable at any point.
Inextensibility: The change in length of the cable due to the applied tension is usually small enough to be neglected.
Concentrated vs. Distributed: The shape the cable assumes depends entirely on the type of load it supports.

Cables Subjected to Concentrated Loads

When a cable of negligible weight supports a series of discrete, concentrated vertical loads, it assumes the shape of a series of straight-line segments.

Cables Subjected to Distributed Loads

If the load is distributed continuously, the cable takes the form of a smooth curve. There are two primary cases depending on how the load is applied.

Parabolic Cables

If the distributed load

w

is uniformly distributed along the horizontal span (e.g., a suspension bridge deck), the cable assumes the shape of a parabola.

y = \frac{w_0}{2T_0} x^2

Where:

$T_0$ is the constant horizontal tension at the lowest point.
$w_0$ is the uniform horizontal load intensity ( $\text{N/m}$ ).
The maximum tension occurs at the supports: $T_{\text{max}} = \sqrt{T_0^2 + (w_0 x)^2}$ .

Cable Length for Parabolic Cables

For a parabolic cable with a uniform horizontal load, the exact length

S

of the cable spanning between two supports at the same elevation, separated by distance

L

and with a maximum sag

h

, can be determined by integrating

ds = \sqrt{1 + (dy/dx)^2} dx

. For practical engineering purposes where the sag-to-span ratio (

h/L

) is relatively small, the length is often approximated using the series expansion:

S \approx L \left[ 1 + \frac{8}{3}\left(\frac{h}{L}\right)^2 - \frac{32}{5}\left(\frac{h}{L}\right)^4 + \dots \right]

This formula is critical for determining the initial un-tensioned length of cables required during construction.

Catenary Cables

If the distributed load is uniformly distributed along the length of the cable itself (e.g., the self-weight of a power transmission line), the cable assumes the shape of a catenary.

y = \frac{T_0}{w_0} \left( \cosh \left( \frac{w_0}{T_0} x \right) - 1 \right)

Where

w_0

is the uniform weight per unit length of the cable.

Interactive Cable Simulation

Explore how different types of loading (Concentrated, Parabolic, Catenary) change the shape and the internal tension forces of a cable spanning a distance. Note how increasing the sag dramatically decreases the tension forces.

Cable Forces Simulation

Cable Type

Distributed Load w (kN/m) : 10

Sag h (m): 5

Analysis Results

Min Tension ($T_0$ at center):100.0 kN

Max Tension ($T_max$ at supports):141.4 kN

Analysis of Arches

An arch is essentially an inverted cable. While a cable carries load purely in tension, an ideally shaped arch carries load purely in compression. However, because practical arches must support variable loads, they are designed to resist some bending and shear as well.

Three-Hinged Arches

A three-hinged arch is a common statically determinate structure. It consists of two arch segments pinned together at a central crown hinge and pinned at both supports.

Analysis Procedure

Overall Free-Body Diagram: Draw the FBD of the entire arch. There are 4 unknown reaction components (two at each pin support).
Equilibrium Equations: Apply $\Sigma F_x = 0$ , $\Sigma F_y = 0$ , and take moments about one of the supports ( $\Sigma M = 0$ ). This is not enough to find all 4 unknowns.
Crown Hinge Condition: The internal bending moment at the central hinge is zero ( $M = 0$ ). Make an imaginary cut at the crown hinge and draw the FBD of either the left or right segment.
Solve for Unknowns: Take the moment about the crown hinge for the isolated segment to generate the 4th independent equation needed to solve for all support reactions.
Internal Forces: Once reactions are known, use the method of sections to find the internal normal force, shear force, and bending moment at any point along the arch.

Cables with Concentrated Loads

A cable carrying concentrated loads takes the shape of a series of linear segments forming a polygon.

Analysis of Cables with Concentrated Loads

If the loads are vertical, the horizontal component of the cable tension ( $T_x$ ) is constant throughout the entire length of the cable.
The maximum tension occurs in the segment with the steepest slope.
The shape of the cable can be found by applying equilibrium equations to sections of the cable or individual joints (nodes).

Cables with Distributed Loads

A cable subjected to a continuous distributed load forms a curve. Two common cases exist:

Parabolic and Catenary Cables

Parabolic Cable: If the distributed load is uniform along the horizontal projection (like the deck of a suspension bridge), the cable assumes a parabolic shape ( $y = \frac{w_0 x^2}{2T_0}$ ).
Catenary Cable: If the distributed load is uniform along the length of the cable itself (like the weight of the cable hanging under its own weight), the cable assumes a catenary shape involving hyperbolic functions ( $y = a \cosh(\frac{x}{a})$ ).

Key Takeaways

Cables can only resist tension. They are perfectly flexible and assume a shape dictated by their loading.
For vertical loads, the horizontal component of tension ( $T_x$ or $T_0$ ) is constant throughout the cable.
A uniform load distributed horizontally (like a bridge deck) creates a parabolic cable profile.
A uniform load distributed along the cable length (like self-weight) creates a catenary profile.
The maximum tension always occurs at the highest support point where the slope is steepest.
A three-hinged arch is statically determinate. The zero-moment condition at the central hinge provides the extra equation needed to solve for the four support reactions.

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