Cables and Arches - Theory & Concepts

Cables and Arches

Learning Objectives

Explain why an ideal cable carries tension only and assumes a load-dependent shape.
Analyze polygonal cables subjected to one or more concentrated loads.
Distinguish a parabolic cable under uniform horizontal loading from a catenary under self-weight.
Calculate horizontal and vertical tension components, segment tensions, sag, and support reactions.
Detect invalid free-hanging or slack cable configurations.
Relate cable action in tension to the inverted compression action of an ideal arch.

Ideal Cable

A perfectly flexible structural element with negligible bending stiffness that can carry axial tension but cannot carry compression, shear, or bending moment.

Governing Cable Assumptions

Cable tension is tangent to the cable at every point.
The cable is inextensible for the rigid-body statics calculation unless elongation is explicitly modeled elsewhere.
Loads and support elevations determine the equilibrium shape.
Cable elements carry tension only; a solution requiring compression, a downward hold-down reaction in a free support, or a lowest point outside the intended span is invalid for the assumed model.
Civil Engineering units in the simulations are metres, kilonewtons, and kilonewtons per metre.

Cable Model Distinction

A cable under discrete point loads forms straight polygonal segments. A cable under load uniform per horizontal metre is parabolic. A cable carrying self-weight uniform per metre of cable follows a catenary. These models are not interchangeable.

Cable Tension Components

The magnitude of a segment tension follows from its constant horizontal and local vertical components.

T_i=\sqrt{H^2+V_i^2}

Variables

Symbol	Description	Unit
$T_i$	Tension in cable segment i	kN
$H$	Constant horizontal tension component	kN
$V_i$	Vertical tension component in segment i	kN

Guided Example: Choose the Correct Cable Model

STEP-BY-STEP

Identify whether the loading is discrete at hangers, uniform over a horizontal deck projection, or uniform along the cable itself.
Use a polygonal model for discrete point loads, a parabolic model for uniform horizontal loading, or a catenary model for self-weight.
Apply whole-cable vertical equilibrium to calculate support vertical components.
Use the prescribed sag or lowest-point condition to calculate the constant horizontal component $H$ .
Calculate each segment or support tension from horizontal and vertical components.
Reject configurations requiring cable compression, an unintended hold-down force, or a lowest point outside the span.

Polygonal Cable

A cable consisting of straight segments between concentrated-load points, with constant horizontal tension component and a changing vertical component from segment to segment.

One Concentrated Load

For a single point load, the cable has two straight segments. Whole-system equilibrium gives the support vertical reactions, while a specified sag point determines $H$ through the funicular relationship between equivalent beam moment and cable ordinate.

Funicular Relationship for Point-Loaded Cables

The sag below the support chord is proportional to the equivalent simply supported beam moment.

y_{\mathrm{sag}}(x)=\frac{M_{\mathrm{beam}}(x)}{H}

Variables

Symbol	Description	Unit
$y_{\mathrm{sag}}(x)$	Cable sag below the straight support chord	m
$M_{\mathrm{beam}}(x)$	Moment in the equivalent simply supported beam	$kN\cdot m$
$H$	Horizontal cable-tension component	kN

Simulation 1 Instructions: One Concentrated Load

Move the point load and change the prescribed sag and support elevation. Reveal the horizontal component, two segment tensions, and support vertical components.

Interactive engineering simulation

Cable with One Concentrated Load

Use a prescribed sag at the load point to solve two straight cable segments and their tensions.

Tension only

Model distinction: point loads create straight polygonal segments; load uniform per horizontal metre creates a parabola; self-weight uniform per cable metre creates a catenary. These models are never silently substituted for one another.

The prescribed sag is measured below the support chord at the first load location, x=10.00 m.

Horizontal component H

21.00 kN

Maximum tension

23.48 kN

Left vertical component

10.50 kN

Right vertical component

7.50 kN

Horizontal span24 m

10–60

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Sag below support chord5.00 m

1.00–14.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Right support elevation above left0.0 m

-8.0–8.0

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Point load P118 kN

1–50

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Point load P1 position10.0 m

0.5–23.5

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

H

remains constant through every segment, while

T_i=\sqrt{H^2+V_i^2}

changes with segment slope. A negative free-support reaction is reported as a hold-down requirement rather than accepted silently.

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 1

For the same point load and span, why does increasing the prescribed sag reduce the required horizontal tension component?

Multiple Concentrated Loads

Several point loads create a cable polygon whose slope changes at every loaded joint. The vertical component changes by the applied joint load, while $H$ remains constant through the cable.

Simulation 2 Instructions: Multiple Concentrated Loads

Move and resize two point loads. Inspect every straight segment, its slope, vertical component, and tension, and watch for an invalid free-hanging support reaction.

Interactive engineering simulation

Cable with Multiple Concentrated Loads

Create a polygonal funicular cable and inspect the force in every straight segment.

Tension only

The prescribed sag is measured below the support chord at the first load location, x=10.00 m.

Horizontal component H

28.00 kN

Maximum tension

32.25 kN

Left vertical component

14.00 kN

Right vertical component

16.00 kN

Horizontal span24 m

10–60

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Sag below support chord5.00 m

1.00–14.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Right support elevation above left0.0 m

-8.0–8.0

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Point load P118 kN

1–50

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Point load P1 position10.0 m

0.5–23.5

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Point load P212 kN

1–50

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Point load P2 position17.0 m

0.5–23.5

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

H

remains constant through every segment, while

T_i=\sqrt{H^2+V_i^2}

changes with segment slope. A negative free-support reaction is reported as a hold-down requirement rather than accepted silently.

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 2

What equilibrium relation connects the change in vertical cable component across a loaded joint to the concentrated load at that joint?

Parabolic Cable

The exact funicular shape of an ideal cable subjected to a load that is uniform per unit horizontal projection.

Symmetric Parabolic Cable Horizontal Tension

For level supports, span L, midspan sag f, and uniform horizontal load w, the horizontal component is inversely proportional to sag.

H=\frac{wL^2}{8f}

Variables

Symbol	Description	Unit
$w$	Load per unit horizontal length	kN/m
$L$	Horizontal span	m
$f$	Sag below the support chord	m

Symmetric Parabolic Cable Shape

Cable sag below the level support chord under uniform horizontal loading.

y_{\mathrm{sag}}(x)=\frac{4fx(L-x)}{L^2}

Variables

Symbol	Description	Unit
$x$	Horizontal coordinate from the left support	m

Simulation 3 Instructions: Uniform Horizontal Loading

Change span, sag, load intensity, and support elevation. Verify the parabolic shape, the lowest-point location, and the support tension components.

Interactive engineering simulation

Cable under Uniform Horizontal Loading

Apply load uniform per horizontal metre and verify the parabolic cable model.

Tension only

Horizontal component H

259.20 kN

Maximum tension

337.40 kN

Left vertical component

216.00 kN

Right vertical component

216.00 kN

Horizontal span24 m

10–60

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Sag below support chord5.00 m

1.00–14.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Right support elevation above left0.0 m

-8.0–8.0

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Uniform horizontal load18 kN/m

kN/m

1–50

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

H=\frac{wL^2}{8f}

y_{\mathrm{sag}}(x)=\frac{4fx(L-x)}{L^2}

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 3

Why is the cable parabolic when the load is uniform over the horizontal projection rather than uniform along the curved cable length?

Catenary

The exact equilibrium shape of a perfectly flexible cable subjected to constant self-weight per unit length measured along the cable.

Catenary Equation

With the origin at the lowest point, the catenary rise is expressed with the hyperbolic cosine function.

y=a\cosh\left(\frac{x}{a}\right)-a

Variables

Symbol	Description	Unit
$a$	Catenary parameter equal to H divided by cable self-weight intensity	m
$x$	Horizontal coordinate from the lowest point	m
$H$	Horizontal tension component	kN

Catenary Horizontal Component

The horizontal component is the product of self-weight intensity per cable length and catenary parameter.

H=w_c a

Variables

Symbol	Description	Unit
$w_c$	Cable self-weight per unit cable length	kN/m

Simulation 4 Instructions: Self-Weight Catenary

Change span, sag, and self-weight per cable length. Reveal the solved catenary parameter, cable length, horizontal component, and support tension.

Interactive engineering simulation

Cable under Self-Weight

Apply self-weight uniform per cable metre and solve the exact level-support catenary.

Tension only

Horizontal component H

273.01 kN

Maximum tension

363.01 kN

Left vertical component

239.25 kN

Right vertical component

239.25 kN

Horizontal span24 m

10–60

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Sag below support chord5.00 m

1.00–14.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Self-weight per cable length18 kN/m

kN/m

1–50

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Catenary parameter a

15.167 m

Cable length

26.583 m

Support tension

363.01 kN

y=a\cosh\left(\frac{x}{a}\right)-a,\qquad H=w_c a

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 4

Why does replacing the catenary by a parabola become less accurate as the sag-to-span ratio increases?

Suspension-Bridge Main Cable Approximation

When the deck load transferred by closely spaced hangers is approximately uniform per horizontal metre, the main cable between towers is commonly modeled as parabolic for statics. Tower elevation differences shift the lowest point and make the two support vertical components unequal.

Simulation 5 Instructions: Suspension Bridge

Change deck load, span, sag, and tower elevation difference. Predict whether horizontal tension increases or decreases, then reveal anchor tension and invalid hold-down warnings.

Interactive engineering simulation

Suspension-Bridge Sag and Tension Explorer

Change deck load, tower elevation, and sag to study horizontal and anchor tension.

Tension only

Horizontal component H

259.20 kN

Maximum tension

337.40 kN

Left vertical component

216.00 kN

Right vertical component

216.00 kN

Horizontal span24 m

10–60

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Sag below support chord5.00 m

1.00–14.00

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Right support elevation above left0.0 m

-8.0–8.0

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

Uniform horizontal load18 kN/m

kN/m

1–50

Drag for exploration or enter an exact value. Press Enter to apply; Escape restores the current value.

H=\frac{wL^2}{8f}

y_{\mathrm{sag}}(x)=\frac{4fx(L-x)}{L^2}

Model scope and verification

Use the displayed units and idealizations, then verify the governing balance or compatibility equation before interpreting the result.

Concept Check 5

Why does reducing cable sag make the bridge profile shallower but greatly increase the horizontal force transferred to towers and anchorages?

Arch

A curved structural member that carries a matching funicular load primarily through axial compression and may be understood as the compression counterpart of a cable.

Cable-Arch Analogy

An ideal cable adopts a tension-only funicular shape for its loading. Inverting that shape produces an ideal compression-only arch for the same load pattern. Real arches also require shear and bending capacity because their shape cannot match every moving or unsymmetrical load case.

Three-Hinged Arch

A statically determinate arch with pins at both supports and an internal crown hinge where bending moment is zero.

Concise Three-Hinged Arch Analysis

STEP-BY-STEP

Apply whole-arch equilibrium to obtain three independent reaction relations.
Separate the arch at the crown hinge.
Use the zero crown-hinge moment condition to obtain the additional reaction equation.
Apply section equilibrium to find axial force, shear, and bending moment at a selected arch section.

Common Cable Mistakes

Using one shape equation for point loads, uniform horizontal load, and self-weight.
Treating cable tension as constant in magnitude; only the horizontal component is constant under vertical loading.
Ignoring support elevation differences when determining slopes and vertical reactions.
Accepting a negative free-support vertical reaction without recognizing the need for a hold-down device.
Allowing cable compression or calling a slack segment an equilibrium cable.
Using $H=wL^2/(8f)$ for a self-weight catenary without clearly labeling it as an approximation.
Confusing load per horizontal metre with load per metre measured along the cable.

Key Takeaways

Ideal cables carry tension only and assume a shape dictated by load distribution and support geometry.
Point loads produce polygonal cable segments; uniform horizontal loading produces a parabola; self-weight produces a catenary.
The horizontal tension component is constant for vertically loaded ideal cables, while vertical components and segment tensions vary.
Greater sag generally reduces horizontal tension for fixed span and loading.
Invalid free-hanging configurations must be reported rather than silently forced into a cable model.
An ideal arch is the compression counterpart of a funicular cable, but real variable loading introduces bending and shear.

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