Roller screw systems typically have smaller footprints and more compact designs than their hydraulic counterparts. One motivation to replace hydraulics with roller screws is to reduce bulky and auxiliary equipment and its associated maintenance without compromising performance.
Roller screw shafts come in a variety of diameters and load capacities. Some are 3.5 mm in diameter with a 500-lb load capacity, while others can be 120 mm with a 500,000-lb capacity — enough to lift the Statue of Liberty. Regardless of size, roller screw systems perform well in high-force applications and simplify integration, setup, operation, and maintenance compared to complex hydraulic systems.
So, how might a design engineer determine which size screw will work for a given application?
Generally, there are four major selection criteria for choosing the right roller screw size: cycle or operation life, peak load, speed, and physical space requirements.
The first and probably most important consideration is a machine build’s cycle life or operation life before fatigue occurs. So, we compare how many cycles the proposed roller-screw-based system can make to how many cycles the application requires. Different-sized screws can handle a given peak load but yield different cycle lives, so it’s essential to know both the peak load and the loading over a typical axis cycle.
The nominal lifespan is typically calculated as revolutions or hours. There are different calculations for screws with single nuts and screws with preloaded nuts. For single, non-preloaded nuts, the general calculation is:
L10 = (C/Fm)3 x 106
Where L10 = nominal lifespan
C = dynamic load capacity
Fm = mean load
This calculation is common among roller screws and is based on a 90% probability of survival. The nominal lifetime can be adjusted for greater reliability by multiplying L10 by a reliability factor. For instance, if an application requires 95% reliability, we can multiply L10 by 0.62 to get the modified lifespan.
Note that if the load is constant throughout the cycle, the mean load Fm equals the actual or peak load. However, if the load varies throughout the cycle, Fm can be calculated based on the load applied for each distance traveled.
For example, say we want to replace the hydraulic system in a car elevator or service jack. The hydraulic system is 1,500 psi with a 5-in. cylinder. The peak load is 25,000 lb, and the car elevator lifts the peak load during the entire cycle. When sizing a screw-based actuator, the screw would must be 48 mm in diameter with a 20-mm lead, and the cycle life would be about 1.1 million cycles. We can convert the number of cycles into hours, years, weeks, or other units of measure to ensure the system aligns with the target life.
However, say we have the same peak load of 25,000 lb and the same 5-in. hydraulic cylinder at 1,500 psi in a press. Unlike the car elevator that lifts the peak load throughout the cycle, the peak load of the press is only present for a small portion of the stroke.
For instance, in a system with a 12-in. stroke, say the load starts at 5,000 lb for the first 8 in., and the peak load of 25,000 lb is pressed only during the final 4 in. of the stroke. As the system retracts 12 in., the load returns to 5,000 lb. Since we still must output the same 25,000-lb load as in the car elevator example, we could select the same 48-mm-diameter screw with a 20-mm lead. Though instead of 1.1 million cycles in the car elevator, the screw would make 6.4 million cycles in the press.
In this case, we need to define the target life or specify how many parts we want to make. If 6.4 million cycles convert to 6.4 million parts, and we only need to make 1 million parts, we can choose a smaller screw. Instead of the 48-mm diameter screw, we can drop to a 36-mm diameter screw with a 20-mm lead that yields about 1.2 million cycles.
When considering the target cycle life or number of parts to make, screw speed is often another selection criterion. Though the distance that the screw needs to travel, and the load applied during that distance, are necessary for determining the average or mean load during the cycle, the speed across such distance impacts production.
Though the maximum rotation speed limits the application speed, it rarely has an effect when the correct screw end bearings are selected. Nonetheless, the maximum rotation speed is generally calculated as follows:
Nmax = 108 x 106 x (do/L2) x Fkr
Where Nmax = maximum rotation speed
do = screw diameter
L = screw length
Fkr = correction factor for bearing type
Lastly, the application’s physical space affects roller screw sizing, especially when replacing systems in compact environments. For example, to replace a hydraulic cylinder underneath a scissor lift, the new roller screw must be packaged well to fit into the small space. Because roller screws vary in diameter and load capacity, spatial constraints influence which options physically fit each system and application.
Check out the on-demand webinar relating to this topic at Design World.
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