### Calculate Tension and Reactive Forces in Suspended Cables

Spanning large distances without vertical columns has always been a challenge for engineers, due in large part to the force of gravity acting on spanning members. Bridge-building is a field of engineering that is particularly concerned with the spanning of great distances. In recent years, the use of thick steel cables to “suspend” a bridge deck over large distances has become popular, known as a suspension bridge. (See my article Bridge-building – Facts and Figures for more on bridges.)

Although distances of up to 7,000 feet have been spanned (Akashi Kaikyo Bridge, Japan) due to the superior and ideal qualities of steel for such purposes, any material can essentially be used to form cable or rope, and suspend weight for any purpose. The amount of weight that can be supported or otherwise manipulated by a hanging cable can be determined by properties such as the following:

• The Young’s modulus (E) of the cable, in pascals (Gpa, Mpa, etc)
• The tensile strength (TS) of the cable, also in pascals
• The cross-sectional area of the cable (A)
• The vertical height of its dip (d)
• The total spanned distance (S)
• The weight-force per unit length along the cable (w) or point load (W)
• The amount of pretension applied
• The angle the cable makes with horizontal after loading

The height of dip is important because dip is essentially determined by the amount of pretension the cable is experiencing. The more pretension, the smaller the dip. Conversely, the less pretension, the bigger the dip. It should be easy to see that the less tension a cable is initially experiencing, the more it can take on before failing. The spanned distance along with the dip can be used to determine the length of the cable.

Calculating Approximate Cable Length from Circular Arc

The approximate length of the cable can be calculated if you know the span and the radius of an imaginary circle of which the hanging cable is a segment. Again, this will only be an approximation as a hanging cable is actually a catenary, not a circular arc. The less dip there is, the closer the approximation will be to the cable’s true length. Here are the equations for determining both the length and radius (be sure to use radians in your calculations):

Calculating Tension for Uniformly Distributed Loads (UDL)

Calculating the length is great and all, but you may want to find out other things such as the tension under a given load, and the reactive forces at the supports. The tension can be calculated by adding the squares of the horizontal and vertical reactions and taking the square root, like this (refer to the list above for symbols and what they stand for):

Deriving Vertical and Horizontal Reactive Forces at Connections

The two equations can then be simplified to this:

Note that the tension can only be calculated using the equations above when the load is uniformly distributed, that is, either hanging freely without any external forces, or the load is equally distributed across the cable. If there is a single point load or several such loads scattered throughout the cable’s length, the tension must be calculated using other equations.

The forces at the supports are equal to the horizontal and vertical reactions expressed in the equation above. These reactions are not only determined by the span and sag, but the weight of the cable along with any additional load (w), expressed in weight per unit length. The numerator “w times S” is simply the weight per unit length multiplied by the total length – implying a uniformly distributed load (UDL) as opposed to a single point load.

By dividing over 2, you get the vertical reaction on one end of the cable. The reactive forces in the horizontal direction are similar to the vertical but depend largely on the dip of the cable. The bigger the dip, the further the cable deviates from parallel, which will in turn lessen the force in the horizontal direction. Remember that these equations assume a UDL.

If the cable supports external loads, you will need to take these into consideration. Depending on where along the cable these loads are applied, the values of sag, tension, and other forces will change. For example, a UDL will result in different values when compared with a single point load, whose values will also vary depending on location along the cable.

Here are a few things you should remember that may help you:

• Vertical reactive forces at the 2 end-connection points of a cable are equal to half the total weight of the cable and any additional load, as expressed in the equation V = wS/2, provided the weight is either a UDL or located at the cable’s midpoint.
• Vertical reactive forces for a single point load that isn’t centered must still add up to the total weight-force of the point load, although they won’t be equal to each other.
• Horizontal reactive forces at the 2 end-connection points of a cable are always equal to each other. This holds true whether or not the load is centered, the two connection points are of the same altitude, or there are multiple loads at random locations.
• Cable tension across segments of the cable will only be uniform if the altitude of the connection points are vertically equal and there is either a uniformly distributed load, or the point load is centered.
• The smaller the angle between horizontal and cable, the more tension for a given load. Conversely, the bigger the angle, the less tension.

Calculating Vertical Reactive Forces for single Non-centered Point Load

To calculate vertical reactive forces resulting from a non-centered single load on connections of equal altitude, use the following equations:

where Ay and By are the left and right vertical force components respectively, a is the horizontal span from A to the load, b is the horizontal span from the load to B, W is the load, and S is the total span.

Calculating Tension and Horizontal Reactive Forces for Single Point Loads

Calculating the horizontal reactive forces when the load isn’t a UDL requires the tension values on both sides of the load – or just one side if the load is centered. This is a good example of a real-life cable-related statics situation that has a fairly high chance of being encountered even on a DIY or hobby level – which is why I’ve chosen to use it as an example.

Let’s take a steel cable connected across 2 points, A and B, of equal altitude. Say we want to hang an object with a weight force of 1kN from its center. Assuming the cable is sufficiently strong enough to support this weight, and inextensible, we go ahead and hang it from the center of the cable. What happens next is that the cable will inevitably sag due to the gravitational force of the object.

Of course, just how much it sags will depend not only on the weight of the object, but also on how much the cable was initially pretensioned. For simplicity’s sake, let’s just say that both ends of the cable form angles of 30 degrees with horizontal after the load is applied. This angle is important because it, along with the applied load, are the 2 known values from which our tension calculations will be based.

As you can see, with a 30 degree angle, the tension in each segment of cable is exactly equal to the weight force of the load itself. Note that the angles and load are the only 2 things we need to perform this calculation. As I covered above, increasing the angle will decrease the tension, and decreasing the angle will increase the tension. You can experiment yourself by replacing the angle value and recalculating the equations.

Now let’s see what happens if the load is NOT centered. As I covered above, when the horizontal position of the load in relation to the fixed connection points is not centered, not only will the tension differ between cable segments, but the vertical reactive forces will differ as well. But if you recall, although the vertical forces differ, they still add up to the weight force of the applied load.

Calculating the tension when the load is uncentered involves the same basic equations as when it is centered. You start out by measuring the angles. In this case we’ll say that A is 60 degrees and B is 30 degrees. Here, with a little trigonometry, we not only assume that the 3rd angle is 90 degrees, but we can also determine the horizontal distance between the load and each connection point if we needed to.

However, we won’t be needing those values for calculating this problem. (Note that the angles are not drawn with precision.)

Up till now we’ve only talked about the tension, but judging from the diagrams, it’s clear that we can also derive the vertical and horizontal reactive forces from the tension values. By multiplying the cosine of the angle a given cable segment makes with the horizontal, with the tension of that same cable segment, you get the horizontal reactive force component. For the vertical force component we do the same thing, except we replace cosine with sine.

Now that you know how to calculate tension values given known angles and load, you can use it to ensure your cable is indeed strong enough. Up to this point, we ran on the assumption that the cable was strong enough, because it wasn’t necessary for the calculation. But in reality, you must ensure the cable can not only support the load but an additional factor of safetyFOS of 3 is considered minimum for most applications.

For our 2 humble examples above, it would be safe to say that we’d need cable with a tensile strength of at least 3 to 5 kNs. Tensile strength between metals and even among steels can vary considerably, so it’s important you either get your facts straight or go for a higher FOS. Considerations that may influence how high your FOS should be include cost, weight, level of danger in event of failure, wear and fatigue, exposure to loads unaccounted for, or at the least not well understood, such as wind, snow, oscillations, etc.