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

Previously, you have already gotten familiar with 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.

### ❏ Constant vs. variable force in cables

The top image shows cables that have a constant force (if we neglect friction in the pulley sheave). The bottom image shows a cable that has a different magnitude of tension in each straight-line 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.

### ❏ A beam is used to support a 30 pound load

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.

### ❏ A cable is used to support a 30 pound load

This FBD of a discrete section of the cable is in static equilibrium. It's a concurrent force problem (all three vectors in the system intersect at a single point).

### ❏ FBD of a discrete section of the 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.

## 🟦 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 (uniform or constant line 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.

### ❏ Flipbook: a cable is a shape-changer

## 🟦 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:

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 internal 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: 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 asymmetric 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. You only need to know one vertical "sag" dimension. In this example, it's the known vertical distance of 23" between A and B.

### ❏ Flipbook: solving an asymmetric cable

## 🟦 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.

This image should remind you of solving trusses with the Method of Joints.

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:

Tension is not constant in the entire cable system.

Tension is constant in each straight-line segment.

The horizontal force is constant throughout the entire system, in each and every FBD.

You discovered that at each node, the difference in the vertical forces in the cables is equal to the applied load at that node.

The internal tensions in each cable segment can be computed using either the Pythagorean Theorem or through similar triangles.

The maximum tension in a cable corresponds to the segment that has the steepest slope. This is because horizontal force is constant, and the steeper the slope, the larger the vertical component of the reaction.

## 🟦 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.

The maximum tension is located at the steepest slope (at the highest support).

The minimum tension occurs at the vertex of the parabola.

Tension is a function of location (or position).

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.

### ❏ Parabolic cable: 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:

plug into the cable equation and solve for H

construct a FBD of half of the cable.

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).

We can use vertical force equilibrium to solve for the vertical component of Tmax (it's wL/2).

We can sum moments about the highest point on the FBD to write: wL/2 * L/4 = Tmin *h

### ❏ Parabolic cable: FBDs at cuts

Key takeaways:

The internal tension is always aligned to the tangent (instantaneous slope) of the cable geometry

The internal tension is perpendicular to the cut that exposes the circular cross-section of the cable itself

The amount of resultant force on a FBD is the line load multiplied by the x-dir. length of the FBD.

## 🟦 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.

Fall 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:

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 each segment of the cable.

Vertical forces = 250#

Horizontal forces = 750#

Tension = 791#

Problem 2.

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

Vertical = 15#

Horizontal = 30#

Tension = 33.5#

Problem 3.

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

Vertical = 15#

Horizontal = 15#

T1 = T2 = 21.2#

T3 = 15#

Problem 4.

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.)

Partial answer: tension in DE = 111.7 kN.

Problem 5.

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.

Ay = 10#, Cy = 20#, H = 20#. Tension in AB is 22.4#. Tension in BC = 28.3#.

FALL 2024 students: you can stop after Problem 5. We'll do the others (below, to be created later) in class together.

Problem 6.

point load problem coming

Problem 7.

distributed load problem coming

Problem 8.

distributed load problem coming

Problem 9.

distributed load problem coming

Problem 10.

distributed load problem coming