Fundamentals of Cable-Supported Structures

Cable

Cables are structural elements used in various constructions including suspension bridges, support wires, transmission lines, electrical networks, stadiums and covered sports halls.

• Made of high-strength steel wires.
• Completely flexible.
• Tensile strength: 4–5 times greater than structural steel.

• High strength-to-weight ratio.
• Suitable for long-span structures (e.g., suspension bridges, arena/convention hall roofs).

To use cable construction effectively, the designer must deal with two problems:

1)    Preventing large displacements and oscillations from developing in cables that carry live loads whose magnitude or direction changes with time.

2)    Providing an efficient means of anchoring the large tensile force carried by cables.

 

Characteristics of Cables

Composition & Strength

• Made of high-strength wires twisted into strands (spiral pattern).
• Ultimate tensile strength ≈ 270 kips/in² (1862 MPa).

Mechanical Properties

• Wire drawing process: raises yield strength but reduces ductility.
• Elongation capacity: 7–8% (vs. 30–40% for structural steel, 36 kips/in² ≈ 248 MPa).
• Modulus of elasticity: ~26,000 kips/in² (179 GPa) (lower than 29,000 kips/in², 200 GPa, for structural steel).
• Lower modulus due to uncoiling of spiral structure under load.

Behavior Under Load

• Cable carries only axial tension.
• Resultant axial force (T) acts tangentially to cable axis.
• No flexural rigidity → highly flexible.

Design Considerations

• Must prevent large deflections and vibrations under live loads.
• Early failures due to oscillations.
• Example: Tacoma Narrows Bridge (1940) — 5939 ft span, destroyed by oscillations up to 28 ft amplitude.

Modern Applications

• Widely used in bridge engineering.
• Examples:
• Cable-stayed bridges (stayed cables).
• Suspension bridges (main cables & suspenders).

 

Variation of Cable Force

Cable with vertical loads only:

Fig: Vertically loaded cable with inclined chord and free body of the cable segment

• The horizontal component (H) of cable tension (T) is constant along the cable.
• Applying equilibrium equation: ∑Fx = 0 for any cable segment. (Figure above).
• If the cable tension is expressed in terms of the horizontal component H and the cable slope θ,

T = H/cosθ

• At a horizontal point on cable (e.g., see point B in above figure a), θ equals zero.
• Since cos θ = 1, above equation, shows that

T = H.

• The maximum value of tension, T typically occurs at the support where the cable slope is largest.

 

Analysis of a Cable Supporting Concentrated Gravity Loads

 

·        When a cable has negligible self-weight and supports only concentrated vertical loads, it takes the shape of a funicular polygon (a piecewise linear curve).

·        A cable under vertical loads is statically determinate: there are 4 unknown support reactions and 4 independent equilibrium equations. [Three equations of static equilibrium and a condition equation, ∑Mz = 0]

·        Since the cable carries only tension (no bending), the bending moment at every section is zero.

·        The condition equation comes from the geometry of the cable: at any section as long as the cable sag.

·        Typically, the designer sets the maximum sag to ensure both a required clearance and an economical design.

Illustration of the Computation

To illustrate the computations of the support reactions and the forces at various points along the cable axis, analyze the cable in Figure a. The cable sag at the location of the 12-kip load is set at 6 ft. In this analysis, assume that the weight of the cable is trivial (compared to the load) and neglect it.

After H is computed, we can establish the cable sag at C by considering a free body of the cable just to the right of C (Figure c).

To compute the force in the three cable segments, we establish θA , θB, and θC

General Cable Theorem

At any point on a cable supporting vertical loads, the product of the cable sag h and the horizontal component H of the cable tension equals the bending moment at the same point in a simply supported beam that carries the same loads in the same position as those on the cable. The span of the beam is equal to that of the cable.

The relationship above can be stated by the following equation:

Hhz = Mz

Where,

·        H = horizontal component of cable tension

·        hz = cable sag at point z where Mz is evaluated

·        Mz = moment at point z in a simply supported beam carrying the loads applied to the cable

ü To verify the general cable theorem, take an arbitrary point z on the cable axis the product of the horizontal component H of cable thrust and the cable sag hz equals the moment at the same point in a simply supported beam carrying the cable loads (shown in Figure below).

ü Also assume that the end supports of the cable are located at different elevations.

ü The vertical distance between the two supports can be expressed in terms of α, the slope of the cable chord, and the cable span L as

                       y = L tan α

Directly below the cable shows a simply supported beam to which apply the cable loads. The distance between loads is the same in both members. In both the cable and the beam, the arbitrary section at which cable theorem will evaluate, is located at distance x to the right of the left support.

For Cable

·        where ∑mB represents the moment about support B of the vertical loads (P1 through P4) applied to the cable.

Considering a free body to the left of point z, we sum moments about point z to produce a second equation in terms of the unknown reactions Ay and H.

·        where ∑m z represents the moment about z of the loads on a free body of the cable to the left of point z.

Solving Equations

Substituting Ay in above equation and simplifying

For Beam

For beam to evaluate the bending moment M z, at point z

To evaluate RA

Substituting RA

So, from the final equations of both the cable and the beam, we can equate the left-hand sides-

Hhz = Mz  (Proved)

 

Example

Determine the reactions at the supports produced by the 120-kip load at midspan (Figure a) using the equations of static equilibrium and (b) using the general cable theorem. Neglect the weight of the cable.

Since supports are not on the same level, we must write two equilibrium equations to solve for the unknown reactions at support C.

⟳+ ∑MA = 0

0 = 120(50) + 5H - 100Cy

Consider right portion from point B

[slope of chord = 5/100 = 0.05

At mid-point = 50x0.05 = 2.5

8.25 – 2.5 = 5.5

5.5+5 = 10.5]

⟳+ ∑MB = 0

0 = 10.5H - 50Cy

H = [50/10.5] * Cy

Substitute H

Substituting Cy

Using the general cable theorem

At midspan where the cable sag hz = 8 ft

Mz = 3000 kip⋅ft

 

Cable subjected to a uniform distributed load

Consider the cable in figure below. The location where the cable is horizontal, zero slope, will always be where the maximum change in elevation occurs. The ends of the cable don’t need to be at the same elevation.

To predict the displaced shape of the cable and find the axial force anywhere along the cable, start with an observation: if the load is applied only vertically, then the horizontal component of the cable force is constant everywhere along the cable.

Find the reaction by constructing a free body diagram of the cable from the point of maximum elevation change, where the cable has zero slope, to the end and then applying equilibrium.

Fig: cable carrying uniform distributed load with ends at different elevations

Fig: Freebody diagram of a uniformly loaded cable from the point of zero slope to the end.

So, we can find the horizontal component of the cable force if we know where the cable is horizontal and what the maximum change in elevation is.

To find the horizontal point consider the cable shown in Figure below. The point where the cable is horizontal is labeled O. Let's assume the peak elevation change. The peak elevation measured from the left support is h1​. The peak elevation measured from the right support is h2​.

Fig: The point where the cable is horizontal will not be at midspan of a cable that has supports at different elevations.

We know that the horizontal component of the cable force is the same on both sides of the cable, so the following must be true:

Since we know the total span of the cable, we can find the location where the cable is horizontal.

To find the vertical position of the cable anywhere along its length, let's draw a free body diagram of a piece of the cable. Specifically, cut at the point where the cable is horizontal, since there will be no vertical component of force at that end. The other end of the piece will be an arbitrary distance x from the horizontal point.

Fig: Free body diagram of a piece of uniformly loaded cable.

Applying vertical and moment equilibrium

This is the equation of parabola.

Substitute the expression H into the above equation

At any point along the cable, the cable force is the vector resultant of the horizontal and vertical components.

This expression shows that the maximum cable tension occurs at a support, where x is largest. Calculating the cable force at position l gives the following expression:

Point Loads with Uniform Loads

When a cable carries both a uniform load and a point load, their effects can be combined through superposition, provided the uniform load is the dominant load. In that case, the cable’s shape remains quadratic and the location of the zero-slope point is known. With this information, the additional effect of the point load can be isolated. Consider the cable, in Figure a, the free-body diagram in Figure b shows the portion of cable from the zero-slope point to the support, including only the point load.

By applying moment equilibrium to this segment, the horizontal component of cable tension caused by the point load can be determined.

Figure: Cable experiencing a uniform load and a point load: (a) If the uniform load dominates, the displaced shape is quadratic; (b) Free body diagram of part of the cable considering only the point load; (c) Free body diagram of the cable with quadratic shape but considering only the point load.

To find the vertical reactions at the ends of the cable, need to consider a free body diagram of the whole cable with just the point load (Figure c). Moment equilibrium gives the vertical reaction at one end of the cable.

Vertical equilibrium gives the vertical reaction at the other end of the cable.

To find the net effect on the cable due to the uniform and point loads, superimpose the horizontal components and vertical components in the cable.

 

Example

The zip line shown in Fig. below. Determine the maximum tension in the cable due to its self-weight.

Find the location of zero slope:

 

Example

The main span of a historic suspension bridge carries a uniform dead load along its length. The bridge dimensions are known (span length, sag, etc.), and the uniformly distributed load per unit horizontal length has already been calculated.

(a)  Determine the horizontal component of the cable tension at the anchorage due to the dead load.

(b)  Determine the maximum tension in the cable at the anchorage.

Analysis:

If the main cable passes smoothly over the pier, the horizontal component (H) remains constant. With uniform load (w), both the main and end spans take the same parabolic shape.

The horizontal component from the main span:

the vertical component at the anchorage:

The end span has the same parabolic shape as the main span

But notice in the figure that the cable does not start with a zero slope at the anchorage — instead, it is already sloped upward because the shorter end span is “pulled” by the longer adjacent span.

At the anchorage, the location x is measured from the lowest point (the vertex) of the main parabolic cable profile.

·         If the cable were symmetric and started at zero slope, the vertex would be midspan, i.e. at x=lmainspan .

·         But here, the end span is shorter (lendspan < lmainspan​), so the lowest point of the parabola lies beyond the end span by an offset.

That offset is exactly:

X = lmainspan – lendspan

Their difference gives how much “extra” horizontal distance the vertex is shifted away from the anchorage.

At the anchorage,

X = 875 – 717 = 158 ft

Example

A cable-supported roof carries a uniform load w = 0.6 kip/ft (Figure a). If the cable sag at midspan is set at 10 ft, what is the maximum tension in the cable (a) between points B and D and (b) between points A and B? use general cable theorem.

(a)

Apply the uniform load to a simply supported beam and compute the moment M z at midspan (Figure c). Since the moment curve is a parabola, the cable is also a parabola between points B and D.

Apply cable theorem

The maximum cable tension in span BD occurs at the supports where the slope is maximum.

To establish the slope at the supports, differentiate the equation of the cable

y = 4hx²/L²

(b)

If we neglect the weight of the cable between points A and B, the cable can be treated as a straight member. Since the cable slope θ is 45°, the cable tension equals

 

Example

The cable in Fig. a supports a girder that weighs 12 kN/m. Determine the tension in the supporting cable at A, B, and C.

At point B, the slope is zero as B is the lowest point on the cable.

The parabolic equation for the cable is:

Let, the point B is located x′ from C

At point A,

Hence, F_H = 1.0 * (12.43)² = 154.4 kN

From Parabolic equation

y = [6/154.4] * x²

y = [6/154.4] * x²

y = 0.03886 * x²

dy/dx = 0.07772 * x

At point A,

x = - (30 -12.43) = -17.57 m

tanθ_A =  dy/dx = 0.07772 * (-17.57) = -1.366

or, θ_A = -53.79⁰

At point B,

x = 0

or, θ_B = 0⁰

At point C,

x = 12.43

tanθ_A =  dy/dx = 0.07772 * (12.43) = 0.9660

or, θ_C = 44⁰

so,

T_A = F_H/cos θ_A = 261 kN

T_B = F_H/cos θ_B = 154 kN

T_C = F_H/cos θ_C = 215 kN

Example

The suspension bridge in Fig. a is constructed using the two stiffening trusses that are pin connected at their ends C and supported by a pin at A and a rocker at B. Determine the maximum tension in the cable IH. The cable has a parabolic shape and the bridge is subjected to the single load of 50 kN.

Free body diagram of the cable-truss system

Taking Moment about B

Consider half the suspended structure

Taking Moment about the pin at C

 

Solving the above equations

F_H = 28.125 kN

As we know,

F_H = wl²/2h

w  = 2 F_H h / l² = 2 x 28.125 x 8 / (12)² = 3.125 kN/m

Hence,

Example

A light suspension bridge is constructed to carry a pathway 3m broad over a channel 21m wide. The pathway is supported by six equidistant suspension rods. The cable has central dip of 2m. The load on the platform is 10kN/m2. Find the maximum tension in the cable.

Total load = 21 × 3 × 10 = 630 kN

Load on each cable = 630 /2 = 315 kN

Load on each suspension rod = 315/6 = 52.5 kN

R_A = R_B = total load /2 = 52.5 x 6 /2 = 157.5 kN

Now, Taking Moment at C (left/right portion)

H_B x 2 - R_B x 10.5 + 52.5 x (1.5 + 4.5 + 7.5) = 0

H_B =  472.5 kN = H_A

T_max = √( H_A² + R_A²) = 498 kN

 

Self-Weight of Cables

For situations like suspension bridges (Figure a), the weight of the bridge is much greater than the weight of the cable. In those cases, uniform load over the horizontal projection is a reasonable idealization.

But in some cases, like power lines (Figure b), the uniform load acts over the length of the cable.

·         Uniform load on horizontal projection (bridges) → cable shape = parabola.

·         Uniform load on cable length (power lines) → cable shape = catenary.

Fig: The dominant load (carrying uniform loads) for the main cable of a suspension bridge acts over the horizontal projection

 

Fig: The dominant load for a power line cable is its self-weight, which acts along its length.

Anchorage of suspension cables

Suspension of cables are provided with two types of supports at the ends.

·        Cables over pulleys

·        Cable provided with saddles

 

Cable over Pulleys

When the cable passes over a pulley on the pier and is stayed at the back, the pier will be subjected to horizontal force as well as vertical force. The tension in the cable and back-stay will be same at the pier.

Total vertical force on the pier

=Tcosα+Tcosβ

=T(cosα+cosβ)

But

T sinα=H or T= H/sinα

H is horizontal tension in the cable.

Total vertical force on pier

= H/sinα ​[cosα+cosβ]

= H (cotα + cosβ cosecα).

Net horizontal force on the pier

= T sinα – T sinβ

= T(sinα−sinβ)

= [H/sinα] ​(sinα−sinβ)

= H (1−sinβ cosecα).

Cable provided with saddles

When the cable passes over saddles and then it is back-stayed, there will be no horizontal force acting on the pier, as saddle allows free movement and thus horizontal components of tension in cable and tension in back-stay must be equal.

Let

T be the tension in the cable at the pier and T′ be the tension in the back-stay.

α be the angle made by the cable with the vertical at the pier and β be the angle made by back-stay with the vertical at the pier.

H =T sinα =T′ sinβ

T′ = T sinα​ / sinβ = H / sinβ​

Vertical force on the pier

= T cosα + T′ cosβ

= [H / sinα] ​cosα + [H / sinβ] ​cosβ

= H (cotα + cotβ).

 

Home Work

1.     Determine the reactions at the supports, the magnitude of the cable sag at joints B and E, the magnitude of the tension force in each segment of the cable, and the total length of the cable in Figure 1.

2.     The cable in Figure 2 supports four simply supported girders uniformly loaded with 6 kips/ft. (a) Determine the minimum required area of the main cable ABCDE if the allowable stress is 60 kips/in.2. (b) Determine the cable sag at point B.

3.     The cable in Figure 3 supports girder DE uniformly loaded with 4 kips/ft. The supporting hangers are closely spaced, generating a smooth curved cable. Determine the support reactions at A and C. If the maximum tensile force in the cable cannot exceed 600 kips, determine the sag hB at midspan.

4.     Determine the reactions at supports A and E and the maximum tension in the cable in Figure 4. Also Establish the cable sag at points C and D.

5.     Compute the support reactions and the maximum tension in the main cable in Figure 5. The hangers can be assumed to provide a simple support for the suspended beams.

6.     What value of θ is associated with the minimum volume of cable material required to support the 100-kip load in Figure 6? The allowable stress in the cable is 150 kips/in.2.

7.     The cables in Figure 7 have been dimensioned so that a 3-kip tension force develops in each vertical strand when the main cables are tensioned. What value of jacking force T must be applied at supports B and C to tension the system?

8.     Compute the support reactions and the maximum tension in the cable in Figure 8.

9.     A uniformly distributed load on beam ABC in Figure 9 causes it to sag. To counteract this sag, a cable and post are added beneath the beam. The cable is tensioned until the force in the post causes a moment equal in magnitude, but opposite in direction, to the moment in the beam. Determine the forces in the cable and the post, and determine the reactions at A and C.

10. Compute the support reactions and the maximum value of w if the allowable tension force in the cable in Figure 10 is 200 kN.

11.  The cable in Figure 11 is capable of carrying a tensile load of 180 kips and the pin supports are capable of providing a horizontal reaction of 150 kips. Determine the shape of the cable subjected to the loading shown.

12.  A cable ABCD is pulled at end E by a force P (Figure 12). The cable is supported at point D by a rigid member DF. Compute the force P that produces a sag of 2 m at points B and C. The horizontal reaction at support F is zero. Compute the vertical reaction at F.

13.  Compute the support reactions and the maximum tension in the cable in Figure 13. The sag at midspan is 12 ft. Each hanger can be assumed to provide a simple support for the suspended beam. Determine the sag at points B and D.

14.  Determine the location of the 40-kN load such that sags at points B and C in Figure 14 are 3 m and 2 m, respectively. Determine the maximum tension in the cable and the reactions at supports A and D.