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.

Cable shape follows load description

Concentrated loads create polygonal segments, uniform load per horizontal length produces a parabola, and uniform self-weight per cable length produces a catenary.

$Ideal Cable$

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.

Simulation

Controls

24 m

5.00 m

0.0 m

18 kN

10.0 m

taut equilibrium

Horizontal component H

21 kN

Maximum tension

23.48 kN

Left vertical reaction

10.5 kN

Right vertical reaction

7.5 kN

Segment S1

T = 23.48 kN

Segment S2

T = 22.3 kN

T_i=\sqrt{H^2+V_i^2},\qquad H=\text{constant},\qquad V_{left}-V_{right}=P

$Concept Check 1$

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.

Simulation

Controls

24 m

5.00 m

0.0 m

18 kN

10.0 m

12 kN

17.0 m

taut equilibrium

Horizontal component H

28 kN

Maximum tension

32.25 kN

Left vertical reaction

14 kN

Right vertical reaction

16 kN

Segment S1

T = 31.3 kN

Segment S2

T = 28.28 kN

Segment S3

T = 32.25 kN

T_i=\sqrt{H^2+V_i^2},\qquad H=\text{constant},\qquad V_{left}-V_{right}=P

$Concept Check 2$

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.

Simulation

Controls

24 m

5.00 m

0.0 m

18 kN/m

taut equilibrium

Horizontal component H

259.2 kN

Maximum tension

337.4 kN

Left vertical reaction

216 kN

Right vertical reaction

216 kN

Lowest point location

12.000 m from left

Support tensions

337.4 / 337.4 kN

H=\frac{wL^2}{8f},\qquad y_{sag}(x)=\frac{4fx(L-x)}{L^2}

$Concept Check 3$

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.

Simulation

Controls

24 m

5.00 m

18 kN/m

taut equilibrium

Horizontal component H

273.01 kN

Maximum tension

363.01 kN

Left vertical reaction

239.25 kN

Right vertical reaction

239.25 kN

Catenary parameter a

15.167 m

Cable length

26.583 m

Lowest point location

12.000 m from left

Support tension

363.01 kN

f=a\left[\cosh\left(\frac{L}{2a}\right)-1\right],\qquad H=w_c a,\qquad T_s=H\cosh\left(\frac{L}{2a}\right)

$Concept Check 4$

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.

Simulation

Controls

24 m

5.00 m

0.0 m

18 kN/m

taut equilibrium

Horizontal component H

259.2 kN

Maximum tension

337.4 kN

Left vertical reaction

216 kN

Right vertical reaction

216 kN

Lowest point location

12.000 m from left

Support tensions

337.4 / 337.4 kN

Deck-load resultant

432 kN

Load-transfer model

Deck → hangers → main cable

H=\frac{wL^2}{8f},\qquad y_{sag}(x)=\frac{4fx(L-x)}{L^2}

$Concept Check 5$

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.

Cable Model Selection Workflow

Select the cable idealization from the actual load measure and support conditions, then solve equilibrium and boundary conditions without confusing polygonal, parabolic, and catenary models.

Identify support geometry, load direction, and how load intensity is measured → Only concentrated loads, with negligible distributed cable weight between them?; Only concentrated loads, with negligible distributed cable weight between them? — Yes → Use straight polygonal cable segments; Only concentrated loads, with negligible distributed cable weight between them? — No → Loading solely vertical and uniform per horizontal projection, with other effects negligible?; Loading solely vertical and uniform per horizontal projection, with other effects negligible? — Yes → Use a parabolic cable model; Loading solely vertical and uniform per horizontal projection, with other effects negligible? — No → Cable self-weight is the only significant load between supports?; Cable self-weight is the only significant load between supports? — Yes → Use a catenary model; Cable self-weight is the only significant load between supports? — No → Use the general cable equilibrium equations or a numerical model; Use straight polygonal cable segments → Apply equilibrium with the required support, sag, length, or lowest-point conditions; Use a parabolic cable model → Apply equilibrium with the required support, sag, length, or lowest-point conditions; Use a catenary model → Apply equilibrium with the required support, sag, length, or lowest-point conditions; Use the general cable equilibrium equations or a numerical model → Apply equilibrium with the required support, sag, length, or lowest-point conditions; Apply equilibrium with the required support, sag, length, or lowest-point conditions → Solve geometry and tension; constant H requires applied loads with no horizontal component; Solve geometry and tension; constant H requires applied loads with no horizontal component → Tension-only state, geometry, and support reactions physically admissible?; Tension-only state, geometry, and support reactions physically admissible? — Yes → Cable solution verified; Tension-only state, geometry, and support reactions physically admissible? — No → Revise the assumed cable state, active span, or boundary conditions; Revise the assumed cable state, active span, or boundary conditions → Identify support geometry, load direction, and how load intensity is measured

Identify support geometry, load direction, and how load intensity is measured: terminator
Only concentrated loads, with negligible distributed cable weight between them?: decision
Use straight polygonal cable segments: process
Loading solely vertical and uniform per horizontal projection, with other effects negligible?: decision
Use a parabolic cable model: process
Cable self-weight is the only significant load between supports?: decision
Use a catenary model: process
Use the general cable equilibrium equations or a numerical model: process
Apply equilibrium with the required support, sag, length, or lowest-point conditions: process
Solve geometry and tension; constant H requires applied loads with no horizontal component: process
Tension-only state, geometry, and support reactions physically admissible?: decision
Revise the assumed cable state, active span, or boundary conditions: process
Cable solution verified: terminator

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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