Cables, Arches and Funicular Structures

Introduction and context

Cables and arches are structural systems whose geometry is closely related to the way loads are transmitted. A flexible cable carries load primarily through tension, while an arch carries load primarily through compression. Both can be shaped so that their centreline follows a funicular curve—the geometry that permits a structure to transmit a particular loading pattern mainly through axial force, with little or no bending.

These systems are important in roofs, bridges, stadiums, industrial buildings, pedestrian structures and temporary works. Their efficiency results from aligning structural form with force flow. Their disadvantages include sensitivity to geometry, support movement, temperature change, construction tolerances and lateral stability.

This chapter develops methods for analysing cables, suspension systems and three-hinged arches under common loading conditions. The treatment assumes statics, idealised supports and small-deformation structural behaviour unless stated otherwise.

Learning objectives

By the end of this chapter, the reader should be able to:

1. Explain the distinction between cable, arch and funicular action.
2. Identify the assumptions used in ideal cable and arch analysis.
3. Determine support reactions and cable tension for concentrated and uniformly distributed loads.
4. Derive and use the parabolic cable equation for horizontal projected loading.
5. Analyse three-hinged arches using equilibrium and the zero-moment condition at the crown hinge.
6. Interpret thrust, sag, rise, stability and construction implications.
7. Evaluate practical limitations and select appropriate analysis methods.

1. Fundamental concepts

1.1 Cable action

A cable is flexible and has negligible bending stiffness. It cannot resist compression or bending; therefore, its internal force is tangent to its shape and is tensile. The cable changes shape until its geometry is compatible with the applied loads and support conditions.

A cable suspended between two supports is commonly described by:

• span, L: horizontal distance between supports;
• sag, f: vertical distance from the support chord to the cable at a specified point, often the lowest point;
• tension, T: axial tensile force in the cable;
• horizontal component, H: constant horizontal component of cable tension for vertical loading;
• vertical component, V: vertical component of tension at a section;
• support reaction: force exerted by a support on the cable.

For a cable carrying only vertical loads, the horizontal component H is constant along the cable. The total tension at any section is:

Here T, H and V are forces in newtons or kilonewtons. This equation follows directly from resolving the tangent force into horizontal and vertical components.

1.2 Arch action

An arch is a curved compression member. It transfers vertical loads to supports while also producing a horizontal reaction called horizontal thrust. In a two-hinged or fixed arch, the horizontal thrust depends on stiffness and deformation compatibility. In a three-hinged arch, it can be found from statics alone because the crown hinge supplies an additional moment condition.

An ideal arch is most efficient when its centreline coincides with the funicular curve for the applied loading. If the actual arch departs from that curve, bending moments develop. Real arches must also resist bending caused by self-weight, nonuniform loading, temperature effects, support settlement and imperfections.

1.3 Funicular structures

A funicular structure is shaped according to the equilibrium curve generated by a specified load system. A hanging cable under a load becomes a tension funicular. If that geometry is inverted, it becomes a compression funicular, such as an ideal arch.

The inversion principle is useful but must be applied carefully. A cable and an inverted arch have corresponding axial-force patterns only when the loading is transformed consistently and the supports provide the required reactions. The arch must also be stable against buckling and lateral movement.

2. Assumptions used in elementary analysis

The following assumptions commonly support hand calculations:

1. Loads are static and known.
2. Supports are ideal pins, rollers or hinges.
3. The cable is perfectly flexible and inextensible, unless elastic extension is explicitly considered.
4. The cable has negligible self-weight, or its self-weight is represented by a stated distributed load.
5. For a parabolic cable, the load is uniformly distributed per unit horizontal projection, not necessarily per unit cable length.
6. Connections transmit the forces assumed by the model.
7. Deformations are small enough that equilibrium may be considered in the original or prescribed geometry.

These assumptions should always be checked against the actual structure. For long-span cables, geometric nonlinearity and sag changes can be significant. For slender arches, second-order effects and buckling may govern.

3. Equilibrium of a cable

Consider a cable under vertical loading. Isolate a segment between two sections. Since no horizontal distributed load acts, horizontal equilibrium gives a constant horizontal component H. Vertical equilibrium gives the change in V between the two sections.

If the tangent angle to the horizontal is θ, then:

and therefore:

These expressions apply to a cable under vertical loading when H is constant. At a support, the vertical component of cable tension balances the corresponding support reaction, while the horizontal component contributes to the support thrust.

3.1 Cable carrying a single concentrated load

Suppose a cable has supports A and B at the same level, span L, and a point load P at a horizontal distance a from A and b from B, where:

Let the vertical reactions be R_A and R_B. Global equilibrium gives:

At the load point, let the cable sag below the support chord be f. Taking moments about the load point for the left portion gives:

Thus:

The tension immediately to the left and right of the load is:

The cable has a change in slope at the point load. The tension is not generally the same on both sides because the vertical component changes abruptly.

#### Worked example 1: point-loaded cable

A cable spans 30 m between level supports. A 60 kN load acts 12 m from the left support. The sag at the load is 3 m. Determine the support reactions, horizontal component and tensions adjacent to the load.

Here a = 12 m and b = 18 m.

Therefore:

The large horizontal component illustrates an important design fact: reducing sag increases cable tension. The reactions are 36 kN and 24 kN vertically, while each support also receives a horizontal force of 144 kN directed toward the cable span.

3.2 Cable carrying a uniformly distributed load

When a cable carries a load w uniformly distributed per unit horizontal distance, where w has units kN/m, the cable takes a parabolic shape if the supports are at the same level and the load is vertical.

Let x be measured horizontally from the lowest point of the cable and y be the vertical ordinate measured upward from that lowest point. For a symmetric cable with span L and sag f, the equation is:

for the half-span, with 0 ≤ x ≤ L/2. If x is measured from the left support and y is measured downward from the support chord, a convenient equation is:

At x = 0 and x = L, y = 0; at x = L/2, y = f. The second form is especially useful for calculating ordinates at any point along a level-span cable.

For a uniformly distributed load over the horizontal span, global equilibrium gives each vertical support reaction as wL/2. At the lowest point, the vertical component is zero. Taking moments of half the cable about the lowest point gives:

This is the horizontal cable force at the supports and throughout the cable. The support tension is:

The formula H = wL²/(8f) applies only to a level-span cable under a load uniformly distributed per unit horizontal projection. It should not be used without modification for a load uniform per unit cable length, unequal support levels or substantial cable self-weight.

#### Worked example 2: uniformly loaded cable

A cable spans 40 m, has a sag of 5 m, and carries 2.5 kN/m uniformly over its horizontal projection. Determine H and the support tension.

The vertical reaction at each support is:

Therefore:

The cable’s maximum tension occurs at the supports for this symmetric loading condition because the vertical component is greatest there.

3.3 Unequal support levels

For unequal support elevations, the cable is not symmetric and the lowest point is displaced from midspan. The general equilibrium method remains valid, but the simple symmetric equations do not. A practical procedure is:

1. Establish the support coordinates and loading.
2. Determine vertical reactions from global equilibrium.
3. Use the constant horizontal component H and the cable geometry to relate ordinates and slopes.
4. Solve for H using the known support elevations or specified sag.
5. Calculate tension from T = √(H² + V²).

For large-span engineering design, numerical analysis is normally preferred because cable geometry, elasticity and self-weight interact.

4. Suspension systems

A suspension bridge typically consists of a main cable, suspenders, stiffening girder or truss, anchorages and towers. The main cable carries tension; vertical suspenders transfer deck loads to the cable; towers transfer cable forces to foundations; anchorages resist the horizontal cable components.

If the deck load is transferred through closely spaced suspenders and the stiffening system maintains an approximately uniform load per horizontal distance, the main cable may be approximated by a parabola. If the cable’s own weight dominates and is uniform per unit cable length, the theoretical unloaded cable is closer to a catenary.

The catenary equation for an ideal cable carrying only its own uniformly distributed weight per unit cable length is commonly expressed as:

where a = H/w_c, H is the horizontal tension component and w_c is cable weight per unit cable length. The hyperbolic cosine function is cosh u = (e^u + e^-u)/2. For shallow cables, the catenary and parabola may be numerically close; they are not conceptually identical.

A suspension system requires more than cable-strength calculations. Engineers must check:

• lateral wind stability and aerodynamic response;
• hanger fatigue and vibration;
• cable corrosion and water ingress;
• anchorage and foundation capacity;
• construction-stage geometry;
• differential temperature and support movement;
• redundancy and inspection access.

5. Three-hinged arches

A three-hinged arch has hinges at the two supports and at the crown or another internal point. The three hinges make the structure statically determinate under ideal static loading. The crown hinge has zero bending moment:

This condition allows the horizontal thrust to be determined from equilibrium.

Consider a three-hinged arch with span L and crown rise h. Let M₀C be the bending moment at the crown section of a simply supported beam carrying the same vertical loads. Let H be the horizontal thrust. The moment at the corresponding arch section is:

Since M_C = 0:

This equation assumes the crown is at height h above the springing line and that the equivalent simply supported beam has the same span and vertical loading. It is a general and powerful relation for a three-hinged arch.

At any section x measured horizontally from the left support, if y is the arch ordinate above the springing line, the arch moment is:

where M₀x is the bending moment in the equivalent simply supported beam at x. The sign convention must be used consistently. The arch axial force and shear can then be obtained by resolving the section force components relative to the arch tangent.

5.1 Reactions of a three-hinged arch

For vertical loading only, the vertical reactions equal those of the equivalent simply supported beam. The horizontal reactions are equal and opposite, each having magnitude H. Thus, at the left support, the reaction generally has components V_A and H; at the right support, V_B and H in the opposite horizontal direction.

The support resultant is:

and similarly for the right support.

#### Worked example 3: three-hinged arch under a central point load

A three-hinged arch has a 24 m span and a crown rise of 6 m. A 120 kN point load acts at midspan. Find the support reactions and horizontal thrust.

By symmetry:

For the equivalent simply supported beam, the crown moment is:

Therefore:

Each support reaction resultant is:

The crown hinge has zero moment. Because the load is at the crown hinge, the vertical shear changes there, while the horizontal thrust remains constant in the ideal arch model.

5.2 Uniformly distributed loading on a three-hinged arch

For a level-span arch carrying a uniform load w per unit horizontal distance over the full span, the equivalent simply supported beam has:

If the arch is parabolic with rise h:

Then:

Consequently:

at every section. This demonstrates the defining property of a funicular arch: under the specified loading, it carries the load in pure compression in the idealised model. In practice, bending remains because of self-weight, nonuniform loads, stiffness, construction deviations and support movements.

6. Cable–arch analogy and structural interpretation

The cable and arch equations reveal a direct relationship. For a given loading, the cable geometry is a tension funicular. Its inverted form is a compression funicular. The same load pattern can therefore guide preliminary form-finding for an arch.

However, “pure compression” does not mean “no design risk.” A compression arch may fail by:

• crushing or excessive compressive stress;
• in-plane buckling;
• out-of-plane buckling;
• hinge formation under changing loads;
• sliding or overturning at the supports;
• foundation movement;
• local instability at spandrels, ribs or connections.

The line of thrust is the locus of the resultant compressive force through the arch. An arch remains stable when the line of thrust lies within an acceptable region of the section, often described through the middle-third concept for simplified masonry analysis. The precise allowable region depends on material, joint behaviour, tensile capacity and code requirements.

7. Applied case: preliminary concept for a pedestrian footbridge

Background

This illustrative case concerns a proposed 40 m pedestrian footbridge over a roadway. The architectural concept includes a shallow suspension cable supporting a lightweight deck. The preliminary load transmitted to the main cable is estimated as 3.0 kN/m uniformly over the horizontal span. The design team is considering a sag of either 4 m or 6 m.

Problem

Compare the horizontal cable force for the two sag alternatives and interpret the structural consequences. The preliminary calculation uses the parabolic cable model:

Analysis

For f = 4 m:

For f = 6 m:

The vertical reaction at each end is 60 kN in both alternatives. The approximate support tensions are:

Approach and solution

The deeper cable reduces horizontal thrust and total support tension, but increases the cable’s vertical clearance envelope and may require taller towers or longer suspenders. The shallow cable provides a flatter visual profile but imposes larger anchorage forces. The final design must include cable self-weight, wind, pedestrian-induced vibration, asymmetric loading, fatigue, construction sequence and code load combinations.

Outcome

The 6 m sag option is structurally more favourable with respect to cable force and anchorage demand, while the 4 m option may be preferable architecturally or where clearance constraints require a shallow profile. Neither option can be selected from the cable equation alone. Geometry, foundations, accessibility, durability, erection method and lifecycle cost must be considered together.

Learning points

1. Sag is a primary structural design variable.
2. A small sag can create large horizontal forces.
3. Preliminary statics supports concept selection but does not replace code-based design.
4. Construction-stage and serviceability conditions may govern a flexible structure.

8. Construction, design and professional practice

8.1 Geometry control

Cable and arch forces depend strongly on geometry. Surveying, setting-out and construction tolerances should be treated as structural issues. For suspension systems, the cable is often erected to a target profile before the deck and permanent loads are applied. The final geometry must account for elastic stretch, temperature and staged loading.

8.2 Foundations and anchorages

Horizontal thrust is often the controlling action. A cable anchorage or arch foundation must resist horizontal and vertical reactions without unacceptable sliding, overturning or settlement. Ground investigation is therefore inseparable from structural analysis.

8.3 Load combinations and changing force flow

A funicular shape is exact only for a specified load pattern. Partial live load, asymmetric snow, wind, maintenance loads or crowd loading may move the line of thrust away from the intended centreline. Stiffening girders, ties, bracing or multiple load paths may be needed.

8.4 Analysis software and digital methods

Finite-element software can model nonlinear cable behaviour, large displacement, staged construction, contact, prestress and dynamic response. A sound workflow is to:

1. establish a transparent hand-calculation model;
2. define geometry, material properties, supports and load paths;
3. perform linear checks for initial equilibrium;
4. introduce geometric nonlinearity and construction stages;
5. compare reactions and key forces with independent calculations;
6. conduct sensitivity studies for sag, stiffness, temperature and support movement.

AI-assisted tools may help generate parametric geometries, identify unusual results or automate data checks, but professional judgement remains essential. Inputs, assumptions, model verification, version control and human approval must be documented. An attractive digital form is not evidence of structural safety.

9. Limitations and common errors

Several mistakes recur in cable and arch analysis:

• using the parabolic cable equation for a catenary without checking the loading basis;
• confusing sag with rise;
• omitting the horizontal thrust from support design;
• treating a three-hinged arch as indeterminate;
• applying H = M₀C/h when the crown height or load arrangement has been misidentified;
• using metres for geometry and newtons for force without consistent conversion;
• assuming a funicular arch remains moment-free under all load cases;
• ignoring support settlement, temperature and construction sequence;
• reporting only axial force while overlooking buckling and lateral restraint.

A useful verification routine is to check global force equilibrium, global moment equilibrium, dimensional consistency, symmetry where applicable and limiting behaviour. For example, in H = wL²/(8f), doubling the sag should halve H, while doubling the span should quadruple H if w is unchanged.

Chapter Summary

Cables are flexible tension members whose internal force follows their tangent. Under vertical loading, the horizontal component of cable tension is constant, and total tension is obtained from the horizontal and vertical components. A cable under uniform load per unit horizontal projection has a parabolic profile, with horizontal force H = wL²/(8f). A cable carrying its own uniformly distributed weight follows a catenary more closely.

An arch is a compression structure that develops horizontal thrust. A three-hinged arch is statically determinate because the crown hinge provides the condition M_C = 0. Its horizontal thrust is H = M₀C/h, and its section moment is M_x = M₀x − Hy. A parabolic arch under a matching uniform horizontal load can be a theoretical funicular form with zero bending moment.

Funicular geometry improves structural efficiency but does not eliminate design obligations. Real structures require checks for changing loads, buckling, vibration, fatigue, durability, foundations, construction sequence, support movement and regulatory compliance. Hand calculations provide essential insight and independent verification; nonlinear computational models may be required for final design.

Key Terms

Cable: Flexible structural member that carries axial tension.

Arch: Curved structural member that primarily carries compression.

Funicular curve: Equilibrium shape corresponding to a specified loading pattern.

Sag: Vertical distance from the support chord to a cable point, commonly its lowest point.

Rise: Vertical height of an arch crown above its springing line.

Horizontal thrust: Horizontal support force produced by a cable or arch.

Tension: Axial tensile force in a cable or member.

Catenary: Curve of an ideal cable carrying uniform self-weight per unit cable length.

Parabola: Curve approximating a cable under uniform load per unit horizontal projection.

Three-hinged arch: Arch hinged at both supports and at an internal crown hinge.

Crown hinge: Internal hinge, usually at the arch crown, where bending moment is zero.

Line of thrust: Locus of the resultant compressive force through an arch.

Equivalent beam: Simply supported beam used to calculate M₀ for arch analysis.

Support reaction: Force exerted by a support on a structure.

Stiffening girder: Girder or truss that limits deformation in a suspension bridge deck.

Anchorage: System that transfers cable force into the ground or supporting structure.

Geometric nonlinearity: Dependence of equilibrium on the deformed geometry.

Funicular arch: Arch shaped to carry a specified loading primarily through compression.

Review Questions

Short Answer Questions

1. Why is the horizontal component of cable tension constant under purely vertical loading?
2. Define sag and explain its effect on cable force.
3. What is the significance of the crown hinge in a three-hinged arch?
4. Distinguish between a parabolic cable and a catenary.
5. What is meant by the line of thrust in an arch?

Long Answer Questions

1. Derive the horizontal force equation for a cable carrying a uniform load per unit horizontal projection.
2. Explain the assumptions and limitations of the three-hinged arch equation H = M₀C/h.
3. Describe the load path in a suspension bridge from deck to foundations.
4. Explain the cable–arch inversion principle and its practical limitations.
5. Discuss the effects of support settlement, temperature and construction sequence on cables and arches.

Analytical / Application Questions

1. A 36 m cable carries a uniform load of 2 kN/m and has a sag of 4.5 m. Calculate the horizontal component of tension and the support tension.
2. A three-hinged arch has a 30 m span and a 5 m crown rise. A 100 kN point load acts 10 m from the left support. Determine the vertical reactions, the equivalent beam moment at the crown, and the horizontal thrust.
3. A design team proposes reducing the sag of a suspension footbridge by 25% while retaining the same span and uniform load. Predict the percentage change in horizontal cable force and discuss two consequences for design.

Multiple Choice Questions with Answers

Question 1

For a cable under vertical loading only, which quantity is constant along the cable?

A. Vertical component of tension

B. Horizontal component of tension

C. Total tension

D. Cable slope

Correct Answer: B. Horizontal component of tension

Question 2

A cable spans 40 m, carries 2.5 kN/m uniformly per horizontal projection, and has a 5 m sag. What is the horizontal tension component?

A. 50 kN

B. 80 kN

C. 100 kN

D. 200 kN

Correct Answer: C. 100 kN

Question 3

For a single 60 kN load located 12 m from the left support of a 30 m span, what is the left vertical reaction?

A. 24 kN

B. 30 kN

C. 36 kN

D. 40 kN

Correct Answer: C. 36 kN

Question 4

What condition is imposed at the internal hinge of a three-hinged arch?

A. Axial force is zero

B. Shear force is zero

C. Horizontal thrust is zero

D. Bending moment is zero

Correct Answer: D. Bending moment is zero

Question 5

For a three-hinged arch, the horizontal thrust is calculated from which relationship?

A. H = h/M₀C

B. H = M₀C/h

C. H = M₀C × h

D. H = wL/(2h)

Correct Answer: B. H = M₀C/h

Question 6

A three-hinged arch has a crown beam moment of 720 kN·m and a crown rise of 6 m. What is the horizontal thrust?

A. 60 kN

B. 120 kN

C. 4320 kN

D. 720 kN

Correct Answer: B. 120 kN

Question 7

Which loading most directly produces a parabolic cable model?

A. Uniform load per unit horizontal projection

B. Uniform load per unit cable length only

C. A lateral point load only

D. No load with unequal support levels

Correct Answer: A. Uniform load per unit horizontal projection

Question 8

For a parabolic arch under a matching uniform load over its horizontal span, the ideal bending moment is:

A. Maximum at the crown

B. Constant and nonzero

C. Zero throughout

D. Equal to the support thrust

Correct Answer: C. Zero throughout

Question 9

If cable sag is doubled while span and uniform horizontal load remain unchanged, the horizontal cable force will:

A. Double

B. Remain unchanged

C. Become four times larger

D. Be reduced by half

Correct Answer: D. Be reduced by half

Question 10

Which action is especially important for a suspension-system anchorage?

A. Resisting only vertical reaction

B. Resisting horizontal cable force

C. Eliminating all cable tension

D. Preventing the cable from carrying load

Correct Answer: B. Resisting horizontal cable force