Cable systems are common structures in engineering that are popular because of their relatively lightweight construction and their aesthetic beauty. When someone says, "Picture a bridge in your mind," many people in the United States think of the Golden Gate Bridge in San Francisco, renowned for its glowing red shape and long, slender cable system.
Although you still must observe all the rules of equilibrium I discuss in Part V, cable systems provide a unique set of challenges in that the forces in the cable structures are all dependent on the geometry of the cable system. For the same applied loads, you can get completely different geometrical behaviors!
In this chapter, I explain the three major categories of flexible cable systems (concentrated loaded, uniformly loaded, and catenary)and the properties that make cable systems unique. I show you how to calculate the tension and deflection (known as sag) in cables for each of the different types of system and introduce a shortcut method known as the beam analogy (but try to resist jumping directly to that discussion).
In Chapter 19, I show you trusses, which are systems of multiple members (objects) that are axially loaded (or members whose internal forces are all acting in the direction of their longitudinal axis). Another type of important axially loaded-only member is the flexible cable, which I cover in this chapter. Flexible cables are commonly used in a wide variety of applications, including power and telephone lines, aerial trams, and suspension bridges.
The internal forces in a cable system depend significantly on the following factors:
Cable tension force: Cable tension force is the internal axial tension of the cable. Cables and ropes can only support axial tension — they can't support compression, shear, or moments (which I cover in Chapter 20).
Sag: Sag is a measure of the displacements of a cable system and directly affects the internal forces of the cable. Tension in cables can change dramatically for the same given applied loads with a simple variation in the sag of the cable. The sag for all cable systems varies with position, and you'll never see a suspended cable that is entirely horizontal.
Geometry: Geometry includes factors such as the span of the cable system, the support locations, and the elevations of the cable supports and applied loads.
Note
Cables are assumed to be axial tension-only members and have negligible resistance to bending. End supports of cables are always assumed to be pinned supports(see Chapter 13) which are free to rotate but can't move (or translate) otherwise. You can't have a roller support on a cable system.
In general, you can divide cable systems into three major classifications:
Concentrated load systems: Concentrated load systems are cable systems with point loads(or concentrated forces) acting on them. You can apply multiple concentrated loads, but they must be spread out and not be continuous or distributed loads. (Flip to Chapters 9 and 10 for more on concentrated and distributed loads, respectively.)
Uniform load systems:Uniformly loaded cable systems are those systems that are loaded by a constant, uniformly distributed load acting over a horizontal length. A uniformly loaded cable system is sometimes referred to as a parabolic cable system.
Catenary systems: A catenary system is a cable system that is deflecting under its own self weight (the force created by gravity'seffects on the system's mass) or is subjected to a load that acts along the length of the cable itself (as opposed to a horizontal dimension).
Loads on cables are typically vertical and are applied as either concentrated loads or distributed loads. Self weight of cable systems is often assumed to be negligible because it's usually significantly smaller in comparison to applied concentrated and uniform loads. The load type determines the overall shape of the geometry and is the primary factor in choosing a solution technique. I cover three major types of cable systems in the coming sections.
A cable system subjected to concentrated loads deflects into a shape that resembles a series of straight line segments. The tension in each of these segments may have a different magnitude (or the size of the internal cable tension force) for a given cable segment (section of cable between concentrated loads). Concentrated systems are the easiest to work with because the tension remains constant over a given cable segment and is directly related to the angles of the cable segments, which are based on the sag of the system at the point loads. Figure 22-1 shows a cable system subjected to concentrated point loads.
Tip
You can use simple geometry to calculate the angles of the cable segments, or you can use proportion triangles (which I cover in Chapter 5) to set up relationships. Cable systems lend themselves very well to proportion triangle calculations because you almost always know the vertical dimensions (sag) and the horizontal dimension (cable segment length or distance to point loads). I explain more in "Solving for Tension in Flexible Cables" later in the chapter.
A parabolic cable system develops when a uniform load is applied horizontally along the full length of the cable. An example of this type of loading may be the result of roadway decks on suspension bridges. Figure 22-2 shows a parabolic cable system subjected to uniform loads.
The deflected shape of a cable subjected to a uniform load is a continually curved or parabolic cable-system shape. Suppose you have a parabolic cable system that is loaded with 20 pounds per linear foot (or plf) for a cable that is 15 feet long and tied to supports that are separated by a distance of 10 feet. You can determine the total uniform load (or the resultant) acting on this system with the following equation:
Total Load = (20 plf) · (10 ft) = 200 lb
Note
The parabolic system is only dependent on the intensity of the load and the horizontal projection distance over which it acts.
As I note earlier in the chapter, a catenary cable system deflects under a load along the length of the cable such as ice on power lines or even the weight of the cable itself. Figure 22-3 shows a catenary cable system subjected to uniform loads.
Suppose you have a catenary system loaded with 20 pounds per linear foot (or plf) for a cable that is 15 feet long and tied to supports that are 10 feet apart. You can find the total uniform load acting on this catenary system with the following formula:
Total Load = (20 plf) · (15 ft) = 300 lb
You can see that this catenary system has a significantly higher load because the resultant load (or total load) from this system is directly related to the length of the cable itself.
Warning
This equation looks strikingly similar to the one for parabolic cable systems in the preceding section, but it uses some different terms. Be sure you plug in the correct numbers for horizontal span or cable length depending on the type of system.
For very small-intensity distributed loads (or even very small sag amounts), catenary systems are basically the same as parabolic cable systems.
After you know how to identify the three major types of flexible cable systems (see the preceding section), you can calculate the sag and cable tension for each type.
One of the problems you face when working with cable structures is that you often don't know the sag until after the load has been applied. And if you don't know the sag, you may find it difficult to determine the maximum tension in the cable. And if you don't know the maximum tension in the cable, you can't actually design the cable to hold the loads in equilibrium. And if you can't design the cable, you can't predict the sag. . . . And hence, you see the major problems with cable systems: Where do you start?
Tip
The forces in the cables become directly dependent on the geometry of the system. As the geometry changes, the forces in the systems change. To get around this issue, you generally assume one parameter, such as the sag or the cable tension values, and then solve the problem for the other value.
Because cable loads are always assumed to be vertical, the horizontal components at the reactions are constant and in opposite directions to each other. This setup means that the horizontal component at every location along a cable is also constant.
You apply a method similar to the method of joints for trusses (which I explain in Chapter 19) at every concentrated load location, treating each cable segment on either side of the load as an individual two force member and drawing a free-body diagram (F.B.D.; see Part IV) of each "joint" (or concentrated load). You can compute the tensions in the cable because only the two unknown force vectors are acting at that point.
Consider the two cases of Figure 22-4, where two cable systems with the same span support the same 100-pound load in the middle. The only difference is that Case 1 has a sag of 6 inches, and Case 2 has a sag of 8 inches.
Using a simple F.B.D., you can see that two cable forces TAB and TBC are acting at Point B. Because the system in this example is symmetrical, you can take advantage of symmetry to note that
Equilibrium (see Part V) in the vertical direction then yields the following equation for the cable tension in Case 1:
Note that the 20.88 inches value is the hypotenuse of the proportion triangle for the forces in the cable segments. (Chapter 5 gives you the lowdown on proportion triangles.) For Case 2 (which has a different hypotenuse value of 21.54 inches now), you have a similar equation:
As you can see from this example, a small change in the amount of sag can make a significant difference in the tension in the cable.
Tip
Larger cable tensions usually occur near concentrated loads. Large tension loads can also occur at locations with very small sags. If you don't believe me, try it out in the nearby "Relating tension to sag" sidebar.
The first step to solving parabolic cable problems is to locate an origin for a coordinate system at the location of maximum sag. At this point, you must align the x-axis with the horizontal direction and the y-axis with the vertical.
If you remember your basic calculus, you know that the tangent to a curve at a point has a slope of zero at a location of maximum or minimum value. For cables, this means that the axial force is acting horizontally at the point of maximum sag, which means that the tensile force To is horizontal, and the vertical component of the cable tension is zero at that location. In calculus terms, this information means you can create an expression:
which is the differential equation that governs the behavior of flexible parabolic cables.
The reason for this shift of the coordinate system location is that the shape of the cable becomes symmetric at the point of maximum sag. This symmetry doesn't mean that the dimensions LA and LB are necessarily the same value, but rather that the overall parabolic shape is symmetrical about that point. Based on this new origin, you can define y = 0 at x = 0 (this setup is called a boundary condition). By integrating the governing differential equation and applying a little bit of algebra to the known boundary condition, you can produce the expression for the deflected shape of the cable:
So with this expression, all you need to know is the vertical load applied, w, and the horizontal component of the tension in the cable, To, and you can determine the amount of sag y at any point x measured from the origin.
Figure 22-5a shows a parabolic cable with a uniform load of intensity acting on a horizontal length.
The F.B.D. of segment BC in Figure 22-5b shows that for a given segment, the horizontal component of the tension at all points must be equal to To, or
As you move further from the origin location, the horizontal component remains the same, but the vertical component increases to its maximum value at the support locations. Thus, the maximum magnitude of the tension occurs at the support location.
Tip
This characteristic also means that the horizontal component To must be the same magnitude as the horizontal pin support reactions (if the applied loads are all vertical).
You can use a system's known sag to figure out how much tension it's supporting. Consider the suspension bridge in Figure 22-6a, which has a span between the towers of 500 feet (assumed to be pinned at the towers), and a maximum sag of 35 feet at the midpoint. A deck load of 1 kip per foot is applied over the horizontal length of 500 feet. To begin your analysis, you draw an F.B.D. (as shown in Figure 22-6b) of the cable between the towers and treat their supports as pinned at both ends.
The first step to find the horizontal tension component (To) in the cable is to locate the coordinate system at the point of maximum sag at a distance of 250 feet from both Point A and Point B, measured horizontally. Thus, at a distance of x = 250 feet from the origin, y = 35 feet (which is the maximum sag).
Rearranging the basic parabolic cable equation and solving for horizontal cable component To gives you the following equation:
The horizontal component has the same magnitude as the support reactions, Ax and Ay, so
If you sum moments at Point A, you can then solve for By, the vertical reaction at Point B:
Now that the vertical and horizontal components of the support reactions have been determined, you can compute the maximum magnitude of the tension in the cable:
The solution process for calculating sag from tension is basically the reverse of the tension-from-sag calculation in the preceding section. Suppose that for the suspension bridge shown in Figure 22-6a, the design engineer tells you that you're limited to a maximum of 750 kip of tension in the cable. You can show that the vertical component of the reaction remains unchanged at 250 kip. (See the preceding section for this calculation.)
Rearranging the magnitude equation, you can compute the horizontal component of the cable tension as
Recall that the horizontal component of the cable tension is the same value as the horizontal reactions (assuming that all loads are vertical, as they are in this example). Utilizing the boundary conditions for this problem at x = 250 ft, y = maximum sag, you can create the following equation:
By comparing the results in the examples in this section, you can see that by allowing the cable to sag almost 10 additional feet, you can reduce its tension from 927 kip to 707 kip (or by roughly 25 percent).
Note
Catenary cable problems are a little more mathematically challenging because the uniform load applied is now a function of the cable length. The more sag a cable system has, the more cable length is present to carry the load.
The derivation for a catenary problem is very similar to the derivation you use for parabolic cables (see "Parabolic cable systems" earlier in the chapter). You still must define your coordinate system such that the origin is at the location of maximum sag and the horizontal component of tension To is equal to the cable tension at that point.
Figure 22-7a shows the same cable as Figure 22-5, except now the load is applied per length of cable, not on a horizontal basis. To clarify this difference, I've changed the applied load intensity from a w symbol to a µ symbol. Also, the length over which the load acts is now the arc length of the cable µs. The origin is placed at the location of maximum sag, (at Point C) as shown in Figure 22-7b.
Applying a little calculus gives you the following governing differential equation for the catenary cable:
By comparing this equation to the parabolic version, the load of the parabolic system equations, w(Dx), has been replaced with µ(ds). The major mathematical difference is that the incremental arc length term ds is now also a function of both the sag (y) and position (x).
Solving the differential equation and applying a bit of algebra, you can determine that the sag of a catenary cable can be expressed by the following equation:
where the cosh term (which you probably remember from trigonometry) represents the hyperbolic cosine trig function or
In the catenary equation, the a term in the cosh definition is
Finally, the magnitude of the tension T at any point can be shown to be related to the sag y at that point:
If the horizontal component of the tension To in Figure 22-6 is the same 893 kip, the load applied to the cable is still 1 kip per foot, and the span between supports is still 500 feet, you can now compute the sag of a catenary system at a distance of x = 250 feet from the new origin with the following equation:
where the parameter a is given by
which lets you then solve for the sag in the catenary cable:
Finally, you can compute the maximum tension at a sag of 35.23 feet on the catenary cable by using the equation T = 893 k + 1 klf · 35.23 ft = 928.23 k. That's slightly higher than the tension of the parabolic case. This discrepancy makes sense because you actually have slightly more total load acting on the catenary system due to the fact that the arc length of the cable is always longer than its horizontal dimension.
The sag on the parabolic system was given as 35.0 feet, so the same geometry conditions applied to the catenary cable only increased the sag by 0.23 feet.
Note
For this example, the cable is much longer in comparison to the amount of sag. For problems with very small sag values, the arc length of the cable is very nearly the same as the horizontal distance between the supports. As the sag decreases, the total load on the catenary cable structure becomes more similar to the horizontal load of the parabolic cables. The difference in catenary behavior comes when the sag of the system becomes much larger.
When working with cable systems, the first major piece of information you need to determine is the horizontal component of the cable force which occurs at the location of maximum sag. The beam analogy for flexible cables is a simplified technique for determining the horizontal tension component.
Note
Before you use the analogy, though, keep the following basic assumptions in mind:
All loads must be vertical, but they can be either concentrated or uniformly loaded.
Any distributed load must be acting on a horizontal length, so this method doesn't work for catenary cables.
The beam analogy, as the name implies, requires that you create an analogous beam with the same loads as the cable structure. To implement this process, you start by following three simple steps:
Create a horizontal beam having the same length as the cable system between support reactions.
Apply all external loads from the cable structures on the beam at the same horizontal location and show the vertical reactions of the beam.
You can't compute the horizontal reactions yet, but that's okay.
Draw the moment diagram for the loadings of the beam.
From the moment diagram (see Chapter 20), you can then make use of the following relationship:
The maximum sag occurs at the location of the horizontal tension component To. The moment MBEAM isn't necessarily the maximum moment on the moment diagram; rather, it corresponds to the moment at the point on the moment diagram where To is to be calculated.
Consider the cable structure shown in the real system portion of Figure 22-8, with sag as shown. Two 30-kip forces are applied at 15-foot increments, one each at both Point B and Point C.
Now you're ready to create the moment diagram for the analogous beam as shown in the analogous beam portion of Figure 22-8. The beam has two support reactions Ax and Ay acting at Point A, and two support reactions Dx and Dy acting at Point D, and an applied 30-kip point loads acting at both Point B and Point C.
From equilibrium, you then compute both of the vertical reactions (you need both to draw the moment diagram). I choose to sum moments at Point A in order to find the vertical reaction at Point D, Dy first.
Tip
You can use symmetry to discover that Ay = Dy = 30 k.
Finally, the shear and moment diagrams are drawn, and you see that at the location of maximum sag (Point B), the applied moment MBEAM is 450 kip-feet. After you have this moment computed, you can then calculate the internal tension To in the cable by using the given sag at the location (or 20 inches in this example).
After you compute the tension To, you can then proceed on a segment-by-segment base and determine the magnitude of the tension in each segment (or at all locations).
Tip
If you know the cable tension and want to find the maximum sag, you can also compute that from the previous equation by entering the appropriate value of To and then solving for the unknown maximum sag value!