# Statics

## 🟦  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:

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.

## 🟦  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 won'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. That means that we generally do not need to understand the deformed shape of a body in order to do a statics analysis. This cables lesson is the one exception to that rule. We will apply the equations of equilibrium to the deformed geometry of a given cable.

## 🟦  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.

Image source:By Ashley98lee - Own work, CC BY-SA 4.0

## 🟦  9.4 A cable's 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 continues to be important. If the cable loading and supports are symmetric, then the deformed geometry is also symmetric.

## 🟦  9.5 Analysis of point-loaded cables

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:

### ❏ Flipbook: solving a simple cable

Here is an example solution for a symmetric cable.

The goal is to determine the internal tension forces in all of the cable segments.

### ❏ Flipbook: solving a symmetric cable

Here is an example solution for an symmetric cable.

The goal is to determine the internal tension forces in all of the cable segments.

This type of cable problem is quite a bit more involved, but the principles are the same as before. Note that you don't necessarily have to have all of the geometric information given to you in order to solve a cable problem.

## 🟦  9.6 Compare and contrast: four point-loaded cables

Consider these four different cables.

The cables have different spans. The top cables are loaded symmetrically and the bottom cables are asymmetrically loaded.

They all support the same load of 30 pounds.

Here are the reactions for the same cables.

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

The symmetric cables have symmetric reactions.

The horizontal reactions are a function of the ratios depicted in the green triangles.

Here are nodal equilibrium diagrams for the four cables.

Look at the horizontal force in each cable. It remains constant across the entire structure.

And here are the tension forces in each cable segment.

Note that within a cable system, the greatest amount of tension force lies in the cable segment with the steepest slope.

Takeaways from this exercise:

## 🟦  9.7 Analysis of cables that support a uniform 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.

We can also write the equation of the deformed shape of the parabola:

y = (wx^2)/2H

where w = the uniform line load

and H = the horizontal force in the cable

Remember that throughout this cables unit we are only focused on the deformed geometry.

Since cables subjected to line loads (or uniformly distributed loads) take the shape of a parabola, there are no straight-line segments.

That means that internal tension in the cable is not constant.

We often set the origin of the coordinate system at the vertex of the parabola.

Note also that it is customary to draw the line load either above or below the loading diagram.

Here is a FBD of the entire cable.

We can convert the line load into an equivalent point load (or resultant force).

The maximum tension occurs at the two equal-height supports.

The vertical component of the maximum tension is simply half of the total load: wL/2.

To get the horizontal component of the maximum tension, we can either:

We'll do the FBD method below.

### ❏ Parabolic cable: global FBD

The cable has been cut into two separate FBDs.

The left FBD is between arbitrary points A and B. The key concept in this image is that when you cut a FBD, you're also cutting the load function. That's why the resultant force for segment AB is equal to wx₁.

The right FBD is extremely useful! We cut the cable right at the vertex. This exposes the horizontal force in the cable (which is equal to Tmin for a cable with equal-height pin supports).

h is defined as the y-dir. height of the FBD

Key takeaways:

## 🟦  9.8 Practice problems for cables supporting line loads

It's easy to analyze a cable that has two equal-height supports and supports a line load.

The analysis of a cable that has supports at different heights is a little harder.

Summer 2024: we will mostly do this in class (but the flipbook is a good preview).

Note that there are just three FBDs that are useful:

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

Summer 2024 students:

I'm going to be creating and posting more problems here between June 6th and the end of the term. To distinguish the ones I have added from the originals I assigned you, I have decided to use letters for the new ones (A, B, C, etc.).

The ones originally assigned (that you are responsible for) are numbered 1, 2, 3, and 4. These are the ones you're responsible for working in your Learning Notebook. (The others are optional).

Problem A. Solve for the reaction forces. Then, determine the internal tension in each segment of the cable.

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

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

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.