## 🟦 9.1 What cables are covered in this lesson?

In prior lessons, we studied cables that carry a constant tensile force that aligns to the cable geometry, such as the two examples depicted here:

a pulley changes the direction of the tension force in a cable; when friction is neglected, tension in the cable is constant

an inclined cable is a two-force member that exerts a pull on the loaded pin; tension in the cable is constant

In this lesson, we will focus on cables that are loaded perpendicular to the span (the distance between supports).

This subcategory of cables works in an interesting and unique way.

As a cable is only effective in carrying (transferring) a tension force, it changes shape when loaded. It "finds" static equilibrium by changing geometry until the line of action of the tension force in each portion of the cable aligns with the cable geometry itself.

As a result of the change in shape, tension in this type of cable is not constant for the entire length of the cable. The tension force in the cable is only constant in each straight-line cable segment.

## 🟦 9.2 Comparison: cables and beams

Let's say that we need to design a structure that spans about 20 inches, and supports a point load of 30 pounds or so.

We could support the load by using a solid member, like a ruler or a yardstick. This member would be modeled as a beam.

Alternatively, we could use a cable. We pick up a piece of rope and hold it with two hands. Cables sag under their own weight, taking the shape of a catenary curve.

We don't study catenary curves in this class. They look like parabolas, but they are actually a trigonometric function (a hyperbolic cosine).As we suspend the 30 pound weight from the rope, the cable geometry morphs into two straight, taut segments. The cable deforms (changes its geometric shape). It is only through deformation that the cable attains static equilibrium.

In general, statics is the study of rigid bodies (meaning that the deformed geometry typically doesn't affect our analysis). This lesson is the one exception to that rule. While we will not be calculating the deformed geometry of cables, we must recognize that the equations of equilibrium are only applicable when the cable geometry is drawn in the deformed state.

## 🟦 9.3 Cables in the wild

In this lesson, we'll explore cables that are loaded with point loads, such as the zipline shown below.

After that we will investigate cables with a loading that approximates a uniform line load, such as the loading experienced by the main cables in a suspension bridge, such as the Golden Gate Bridge.

## 🟦 9.4 Deformed geometry is responsive to applied loads

In this flipbook, you will see how the deformed geometry of a cable is responsive to the loads that are applied.

When a cable supports only point loads, it morphs into a series of straight segments.

As we apply more and more evenly spaced point loads to the cable, we approach a distributed load. A distributed load creates the deformed shape of a parabola.

Symmetry is important too: if the loading and supports are symmetric, then the deformed geometry is also symmetric.

## 🟦 9.5 Fundamentals of cable analysis: a single point load

In this flipbook, we step through the analysis of a cable that supports a single, symmetric load.

The example is simple, but the take-aways are important:

we can determine reactions by using the E.o.E. on the global structure

we can cut the cable in various locations to expose the tension force, and use the E.o.E. on those FBDs

we can use Newton's Third Law to solve a cable to systematically moving from node to node (exactly as you did in the Method of Joints for trusses)

### ❏ Flipbook: cable supporting 1 point load

## 🟦 9.6 Cables that support multiple point loads

In the image below, four different cables have been solved for you. All support a total load of 30 pounds.

The symbol V is used as shorthand for the vertical reactions. The symbol H is used as shorthand for the horizontal reactions.

Inspect these solutions carefully, and try to answer these questions by inspection:

how does symmetry affect the reactions for each cable?

how does the cable geometry (the green triangles) affect the reactions?

when are the vertical reactions equal? when are they unequal?

when are the horizontal reactions equal?

We see that the vertical reactions are identical when the cable geometry and loads are perfectly symmetric. We see that the horizontal reactions are identical in all cables (both symmetric and asymmetric cables), due to global force equilibrium in the x-direction. We also see that the horizontal and vertical reaction components must resolve into a resultant vector that is collinear with the segment of the cable nearest that support.

This next image shows the same 4 cable systems, but this time, nodal FBDs have been constructed for you. Note that each FBD is a concurrent system of forces. What patterns can you discern related to horizontal force equilibrium? How about the vertical force equilibrium?

You discovered that the horizontal force is constant throughout the entire system, in each and every FBD, due to Newton's Third Law pairs. You also discovered that at each node, the difference in the vertical forces in the cables is equal to the applied load at that node.

Now, inspect one last drawing. Here, the internal tensions in each cable segment have been provided. These can be computed using either the Pythagorean Theorem or through similar triangles. Look closely at the magnitudes of the internal tension forces. Tension is not constant in these types of cables globally, but it is constant in each straight-line-segment of the cable.

Use your calculator and the Pythagorean Theorem to verify a few of the internal tension forces for yourself.

What do these drawings teach you about the location of maximum tension in a cable?

You may have been able to deduce that the maximum tension in a cable occurs in the segment that has the steepest slope. This is due to the fact that each cable segment acts as a two-force member, and that horizontal force is constant throughout the cable system. You may have also gathered that tensile forces in a multi-segment cable system increase as you move from the low-point of the cable towards one of the two supports.

## 🟦 9.7 Symmetric cables that support a line load

Earlier in this lesson, we developed an intuitive sense for why a cable carrying a line load (or a uniformly distributed load) takes the shape of a parabola in order to ensure that each particle experiences pure tension.

This flipbook is a proof, or a derivation, of that fact.

NOTE TO SELF: ADD A SECTION THE EXPLAINS THE WX1, WX2, WX3 ETC. IDEA IN THE GRAPHIC BELOW. THIS IS THE FIRST TIME THIS CONCEPT APPEARS AND IT NEEDS TO BE EXPLAINED VERY EXPLICITLY HERE.

Cables subjected to line loads (uniformly distributed loads) take the shape of a parabola. Therefore, there are no straight-line segments. That means that internal tension in the cable is not constant.

In this image, I have cut a cable at six different planes. Note that the internal tension force at each plane is a vector with a different direction and magnitude.

The tension force is always:

aligned to the tangent (instantaneous slope) of the cable geometry

perpendicular to the cut that exposes the circular cross-section of the cable itself

## 🟦 9.8 Asymmetric cables supporting line loads (uniformly distributed loads)

We will do this in class (but the flipbook is a good preview).

Note that there are just three FBDs that are useful:

the global (entire) system

the left side

the right side

For those latter two, you want to cut through the lowest point of the cable, so that the internal tension force is perfectly horizontal.

After you cut your FBD, just use the E.o.E. to solve for unknowns.

## ➜ Practice Problems

Problem 1. Solve for the reaction forces. Then, determine the internal tension in cables 1 and 2.

You can check your solution by referencing section 9.6.

Problem 2. Solve for the reaction forces. Then, determine the internal tension in cables 1, 2, and 3.

You can check your solution by referencing section 9.6.

Problem 3. Let's say that L = 100 meters, and that P = 20 kN.

Solve for the reaction forces. Then, determine the internal tension in segments AB, BC, CD, DE, and EF. (You are encouraged to use symmetry.)

Problem 4. Solve for the reactions. Then, solve for the internal tension forces in cable segments AB and BC.

You can check your solution by referencing section 9.6.

Problem 5. This is a tough problem. You have just enough information to solve this problem, but there isn't any extra information. That means that you have to be strategic in your problem-solving process. Brainstorm a strategy before you start writing.

Your task (again) is to solve for the reactions at A and D and determine the internal tension in segments 1, 2, and 3.

You can check your solution by referencing section 9.6.

We will do some distributed load problems in class together in Summer II 2023 (Amsterdam). I plan to get some problems posted here online later.