Statics

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.

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).

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.

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)

Here is an example solution for a symmetric cable. 

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

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.

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.

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.

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