Manufacturers of flexible couplings have their own rating systems, each one different than the others. Sort out the confusion between these different systems and learn how to apply them correctly for selecting couplings
Because there is no industry standard, manufacturers of flexible couplings have developed their own methods of rating coupling capabilities, based on different criteria. Some catalogs give torque ratings that are only good for ideal conditions (no misalignment, zero speed, or zero hours of use per day). In such cases, you must apply a number of service factors for normal operating conditions, which reduces the ideal catalog ratings to smaller, but more realistic values.
Others give ratings that are good for normal operating conditions. But even with these ratings, service factors must be applied to account for extreme conditions, such as large fluctuating torques, or large misalignments.
Here’s what to look for in using catalog ratings, and how to apply them correctly so the coupling you select will give satisfactory performance and long life.
Coupling ratings found in catalogs include maximum values for torque, speed, misalignment, and axial displacement. Some of these values can be selected independently of each other, whereas others are interdependent, such as torque and misalignment, or speed and axial displacement.
In most cases the published catalog ratings must be reduced to account for specific operating conditions. The methods for doing so are discussed later in the “Service factors” section.
Torque. The torque rating in a coupling catalog is the maximum value the coupling can transmit under ideal conditions and continuous duty. Unfortunately ideal and continuous are subjective terms, as it will be shown later. Torque may be listed in units of lbin. or hp per 100s of rpm, where Torque, lb-in. 5[hp/(100s of rpm)] 3630.
Peak torque. In every machine there are times when the torque is much larger than normal. For example, when a water pump is started, the motor must provide enough torque to not only pump the water at a certain flow rate and pressure, but also to accelerate the rotating masses of the pump and the previously still mass of water in the suction and output pipes. This peak torque can be many times larger than the normal torque.
Couplings can accommodate, occasionally and for a short time, torques larger than the rated values. Ratings for peak torques are found in most coupling catalogs. They are an important factor in selecting a coupling. A coupling that can transmit a machine’s normal torque, but not the peak torque is a poor choice, so make sure you interpret the catalog data correctly.
Speed. At typical operating speeds (up to about 3,600 rpm), centrifugal acceleration causes elastomeric coupling elements to stretch so the outside diameter becomes larger. This can cause interference with coupling guards. Very few catalogs supply data on coupling diameter at operating speed. To be safe, make the guard ID 1.25 times larger than the elastomeric coupling OD.
At high speeds (above 10,000 rpm), centrifugal acceleration can cause excessive stresses in metallic couplings. Consult manufacturer’s catalogs for any speed limitations on couplings.
Misalignment. Coupling misalignment, Figure 1, is the angle that the coupling center element, or spacer, forms with either one of its connected shafts. This angular misalignment is caused by the lateral displacement, or offset, between shafts. Coupling misalignment should not be confused with shaft offset; the first is measured in degrees, the second in inches. When talking about the alignment process, it is wrong to say coupling alignment. We can only align machines; not couplings.
Four facts should be understood by coupling users:
1. All couplings resist being misaligned. Some resist with larger forces than others, but in all cases the resistance is proportional to misalignment. These resisting forces act on shafts, bearings, and seals. Coupling catalogs seldom give values for these forces.
2. Maximum misalignment given in a catalog is the misalignment under which the coupling will give satisfactory life. However, this maximum misalignment can cause forces that are excessive for the bearings or other components in your machines.
3. Misalignment capability of a coupling comes at a price. If coupling A accommodates twice the misalignment of coupling B, it usually means that coupling A can transmit less torque, will last a shorter time, or it is larger.
4. Using a coupling that accommodates a large misalignment (more than 1 deg), does not give you a license to be sloppier in aligning the machines. Most machine alignment guidelines, as given in books or magazine articles, do not even consider the coupling misalignment rating.
Axial displacement. In addition to shaft offset, flexible couplings also accommodate axial movement between the ends of shafts. Most catalogs include the amount of axial displacement couplings can accommodate.
Just as couplings resist misalignment, they also resist axial movements. The thrust forces generated by couplings in resisting axial displacement are either constant or dependent on the displacement. Most catalogs for couplings with metallic elements give these thrust forces, but this data is seldom available for other types of couplings. Just as machine bearings must resist the misalignment forces, they must also resist the axial forces caused by couplings, which may overstress the bearings in the axial direction.
To realistically compare two couplings, you need to look beyond their basic catalog ratings, which are often based on different operating conditions. First apply to these basic ratings all the service factors that are required for your application. Then compare the final corrected ratings — what is left of the rating values after all the derating service factors are applied. These derating factors may vary significantly from one manufacturer to another, or from one type of coupling to another, even for the same manufacturer.
Service factors are applied in one of two ways:
1. Multiply all of the service factors together. Then divide the published coupling rating by the resultant to get a corrected rating that reflects the service conditions.
2. Multiply all the service factors together. Then multiply the resultant by the torque (or other parameter) required for the application to get a corrected torque that reflects the service conditions. The coupling selected must then meet this higher value.
One exception to this procedure involves torque fluctuations as described in the next section. More information is given in the “Selection example” section.
Torque fluctuations. It is generally accepted that the most important service factor for couplings is the one that accounts for torque fluctuations that occur continuously. These fluctuations are often caused by both the driving and the driven machine. In this case, the service factors of the power generator and the power user are added (do not multiply them).
Example. If we compare two pumps that consume the same amount of power, where one is centrifugal and the other is reciprocating, we find that the reciprocating pump torque varies widely during each revolution. As a result, the centrifugal pump has a service factor of 1, whereas the reciprocating pump has a service factor that can be larger than 2. Therefore, the reciprocating pump requires a coupling that has twice the rating of a coupling for the centrifugal pump.
The torque fluctuations of the pump account for only part of this service factor. The other part is caused by torque fluctuations of the driving equipment. For example, an electric motor has a service factor of zero, whereas a multiplecylinder engine might have a service factor of 0.5. Therefore, if the reciprocating pump is driven by a multiple-cylinder engine, the coupling rating must be adjusted downward by a factor of 2 10.5 5 2.5 for torque fluctuations. This is the only case where factors are added; all other factors are multiplied.
Unfortunately some catalogs make the service factor difficult to understand. For example, even zero torque fluctuations (such as for a centrifugal compressor) may require a service factor of 1.25. Therefore, even for the best conditions, these published torque ratings must be reduced by a factor of 1.25.
As mentioned earlier, service factors vary by coupling type. A good example, Table 1, compares torque service factors for similar elastomeric couplings, one with a rubber element, and the other with a urethane element. The urethane coupling is derated more than the rubber one because urethane transforms more of the torque fluctuations into heat. If guards prevent rapid heat dissipation, this heat can cause premature failures.
Usage time. A few manufacturers require that their couplings be derated as a function of the number of hours per day a coupling is used, Figure 2. In such cases, the published torque values must be reduced even if the coupling is used only 1 hr per day.
Misalignment. The influence of misalignment on coupling ratings is approached by manufacturers in different ways. Most of them publish graphs such as Figure 3 and Figure 4. One graph reduces the maximum torque that can be transmitted if misalignment exceeds the rated value (1 deg in this case); the other reduces the maximum speed at which the coupling can operate if the misalignment exceeds 10 min.(0.17 deg). Therefore, if the misalignment exceeds a certain value, you may need to reduce the value of another parameter to obtain an acceptable coupling life, or choose another coupling.
A few manufacturers use another approach. In Figure 5, the published torque must be derated whenever misalignment is larger than zero.
Type of application. Coupling applications can be as mundane as a 1-hp water pump at a washateria, or as sophisticated as a 50,000-hp gas turbine driving a number of centrifugal compressors at 15,000 rpm in an oil refinery. Obviously, different criteria must be used depending on the application. The American Petroleum Institute (API) established standard No. 671 “Flexible Couplings for Refinery Applications” in 1979. One recommendation given therein is that the coupling rating should be 175% of the normal torque of the machines. This API standard started a new type of service factor known as an application factor.
A few manufacturers of high-speed couplings responded to this challenge by issuing their own application factors.
Be careful never to use the published values for torque without first verifying which service or application factors apply. In some cases, Table 2, the minimum application factor is 1.5.
Axial displacement. Couplings with flexible elements are subjected to two types of stresses: constant and alternating. Constant stresses are generated by torque, speed, and axial displacement (stretch of the flexible element). Alternating stresses are generated by misalignment during rotation. All these stresses are additive: in flexible elements (such as diaphragms and disks) this combination of stresses can cause fatigue failure.
Rotating speeds. The interdependence of parameters is illustrated by Figure 6, which is taken from a diaphragm coupling catalog. This graph reduces the maximum operating speed as axial stretch of the disks increases. Be careful never to use the maximum axial displacement from this graph (0.11 in.) because it is applicable only at zero speed.
Miscellaneous. Some manufacturers have introduced other derating factors, such as operating temperature and number of starts per day. Although both of these factors make sense, only a few manufacturers use them. When sizing a coupling, check the catalog carefully to ensure that all such factors are taken into account.
Some catalogs make coupling selection more complicated (and confusing) than others. For example, they may have two ratings for torque, called “nominal” and “maximum.” They say that the nominal rating is to be used without service factors, and the maximum rating is to be used when service factors are considered. The confusing part of this rating method is that most other manufacturers apply the service factors to their nominal torques. (See following “Selection example.”)
Experience.The last factor to apply to coupling ratings is your own experience factor. This factor is very subjective and cannot be found in catalogs.
Should you use a coupling at its published ratings after all the service factors are taken into account? You may have had good or bad experiences with a particular type of coupling or with a particular manufacturer. Maybe you’re willing to use a coupling at its rated value in a small machine for which a spare is available; but you’re more conservative in the case of a critical, non-spared machine.
Also consider the future use of the machine. For example, one of the most popular industrial gas turbines had its horsepower output gradually increased over the years to more than double its original rating. Are you prepared to change couplings each time you increase the machine ratings?
The answer to these questions becomes the experience factor.
This example illustrates the importance of carefully reading the catalog so that you don’t get confused by different rating methods such as described previously under “Miscellaneous.”
A conveyor operates intermittently: for ½ hour each hour. The conveyor is driven by a 250-hp (at 1,800 rpm), 4-cylinder diesel engine. The torque delivered by the engine is:
T = 250 hp x 63,000/1,800 rpm = 8,750 lb-in.
(63,000 is a constant. For lb-ft, the constant is 5,353).
We decide to use one of two elastomeric element couplings, called Alpha and Beta. According to the manufacturers’ catalogs, a size 39 Alpha coupling is rated at a nominal torque of 23,500 lb-in. whereas a size 5 Beta coupling is rated at a nominal torque of 8,850 lb-in., and a maximum torque (not peak torque) of 22,125 lb-in.
At first glance, one might think that the two couplings have similar ratings (23,500 lb-in. vs. 22,125 lb-in.). But the difference becomes apparent when we apply all the service factors specified by each of the catalogs.
Coupling Alpha requires applying only a service factor for torque fluctuations: 1.5 for a conveyor, plus 1 for the engine gives a total of 2.5 (remember that only torque fluctuation factors are additive). Multiplying the engine torque by this factor, we get 8,750 x 2.5 = 21,875 lb-in. required torque. Because this required torque is lower than the nominal rated value (23,500 lbin.), the coupling is satisfactory for the application.
An alternate way to use service factors is to divide the coupling rating (either nominal or maximum) by the service factor, which, in this case, gives a derated torque of 23,500/2. = 59,400 lb-in. Because this derated torque is higher than the conveyor torque (8,750 lb-in.), the coupling is satisfactory.
Coupling Beta requires three service factors:
1.15 for 12 hr/day work load.
1.75 for 24 starts/day.
1.9 for conveyors driven by 4-cylinder engines (already calculated in the catalog).
The total service factor is then 1.15 x 1.75 x 1.9 = 3.82. Multiplying the engine torque by this factor, we get 8,750 x 3.82 533,460 lb-in. Because this required torque is higher than the rated value for maximum torque (22,125 lb-in.), the coupling is not satisfactory. Therefore, we must select a larger size.
This article is based on the book Flexible Couplings - Their design, selection, and use, by Michael M. Calistrat. This comprehensive 575-page publication can be ordered from Caroline Publishing, P.O. Box 451611, Houston, Texas 77245-1611, Fax 713-437-4656.
Michael M. Calistrat is a consultant on power transmission design and failure analysis and owner of Michael Calistrat and Associates, Missouri City, Texas.